Matter, Mixtures & Water Properties — Print · Science 6: Matter, Energy & Earth Systems, Grade 6
How can particle models and measurements account for matter across changes? Which parts of a stated sand–salt–water mixture can a filter separate, and where does the matter go?
Printable chapter lessons and review.
States of Matter & the Particle Model
Learning goals
- I can describe the arrangement, spacing and motion of particles in each state.
- I can read a melting point and a boiling point as evidence about attraction between particles.
- I can explain why temperature stops rising while a substance is melting.
Readings for this unit
Before this unit
Read the chapter opening and the named observation or diagram. Clarify one word or unit that blocks understanding. Recall the relevant earlier measurement, matter, or energy idea.
Study the unit across three 35–45-minute class sessions. Use the worked examples after an attempt and the supplied records when actual equipment is not available.
Oxygen is a gas in this room, but cool it to minus 183 degrees Celsius and it becomes a liquid, and at minus 218 degrees Celsius it becomes a solid. Iron is a solid here, but heat it past 1538 degrees Celsius and it melts, and past 2862 degrees Celsius it boils. Neither substance changes into anything else along the way.
That tells you the state is not a property of the material but of the material at a particular temperature. Water is familiar in all three states because its melting point of 0 degrees Celsius and boiling point of 100 degrees Celsius happen to sit either side of the temperatures we live in, which is a fact about our planet rather than about water.
The particle model explains all of it with one idea. Particles attract each other, heating makes them move faster, and a state change happens when the movement finally overcomes the attraction. The temperature at which that happens measures how strong the attraction was.
Build your understanding
Where this fits
This chapter opens the study of matter and everything in it rests on the particle model introduced here. The previous chapter ended by insisting that every model comes with a limit, and this one has a famous one you will meet in the fourth section. Properties, chemical change and mixtures all follow in the next three lessons.
Use three flexible 40-45-minute sessions: the three states and what distinguishes them; melting and boiling points as evidence; what happens during a change of state, then the practice set, revision and the exit check. Workshop, review, quiz and exams have separate time.
Part 1 · Three arrangements, one set of particles
In a solid the particles are packed closely in a fixed arrangement and vibrate about their positions without swapping places, which is why a solid keeps both its shape and its volume. In a liquid the particles are still close together, almost as close as in the solid, but they are no longer locked in position and can slide past one another, so a liquid keeps its volume but takes the shape of its container. In a gas the particles are far apart, typically something like ten times further apart than in the liquid, and move freely in all directions, so a gas has neither a fixed shape nor a fixed volume and expands to fill whatever it is in.
The spacing is what you can test most easily. Push the plunger of a sealed syringe full of air and it moves a long way, because there is a great deal of empty space between gas particles for them to be pushed into. Do the same with a syringe full of water and it barely moves at all, because the particles in a liquid are already nearly touching and there is almost nothing to close up. That single comparison is strong evidence for the model: something in the gas is mostly empty, and something in the liquid is not, and you can feel the difference through your thumb.
Pause and explain
A sealed syringe of air compresses easily and one of water hardly at all. What does this show about the two states?
Check your reasoning after answering
That a gas contains a great deal of empty space between its particles while a liquid does not. Pushing the plunger closes gaps, so a large movement means large gaps were available. In the liquid the particles are already nearly touching, leaving almost nothing to compress, which is why the volume barely changes however hard you push.
Part 2 · Reading a melting point
A melting point is a number you can look up, and it is better understood as a measurement of how strongly the particles of that substance hold on to each other. Oxygen melts at minus 218 degrees Celsius, which tells you its particles barely attract at all, so even the feeble motion left at that temperature is enough to break the solid apart. Water melts at 0 degrees Celsius and iron at 1538 degrees Celsius, so iron's particles must be gripping each other far more tightly, since the particles have to be moving enormously faster before the arrangement gives way.
The same reasoning applies to boiling points, which measure the attraction that remains in the liquid. Here is a comparison worth sitting with: ethanol melts at minus 114 degrees Celsius and boils at 78 degrees Celsius, while water melts at 0 and boils at 100. Ethanol boils at a lower temperature than water even though both are ordinary liquids at room temperature, and that difference is why ethanol evaporates off your hand so much faster. A list of melting and boiling points is not a set of facts to memorise; it is a table of how firmly different substances hold themselves together. One caution about reading such a table: the values assume ordinary air pressure. Water boils at 100 degrees Celsius at sea level, but on the summit of Mount Everest, where the air is far thinner, it boils at about 71 degrees Celsius, which is why food cannot be properly cooked in boiling water up there. A fixed point is fixed only when the conditions are stated.
Pause and explain
Iron melts at 1538 degrees Celsius and oxygen at minus 218. What does the difference tell you?
Check your reasoning after answering
That the attraction between iron particles is vastly stronger. Melting happens when particle motion overcomes the forces holding the fixed arrangement together, so a substance needing an enormous temperature has strong attractions, while one that melts far below room temperature has very weak ones. The melting point is effectively a measurement of that strength.
Part 3 · The flat part of the graph
Heat a beaker of crushed ice steadily and record the temperature every half minute, and the graph does something that surprises most people. The temperature climbs from minus 10 degrees Celsius up to 0, and then it stops climbing and stays at 0 for several minutes while the ice turns to water, even though the heater is still running. Only once all the ice has melted does the temperature start rising again, towards 100 degrees Celsius, where it flattens out a second time while the water boils.
The energy has not gone missing. Temperature measures how fast the particles are moving on average, and during melting the energy arriving from the heater is going entirely into pulling particles out of their fixed arrangement rather than into speeding them up. The attractions have to be broken before the particles can move faster, and breaking them takes a considerable amount of energy: melting 1 kg of ice takes about 334,000 J, which is roughly the same as heating that water from 0 to 80 degrees Celsius once it has melted. The same thing happens, on a larger scale, at the boiling point. Turning 1 kg of water at 100 degrees Celsius into steam at 100 degrees Celsius takes about 2,260,000 J, nearly seven times the energy needed to melt the same mass of ice. That is why steam causes far worse burns than boiling water: it releases all of that energy again on your skin as it condenses.
Pause and explain
Why does the temperature stay at 0 degrees Celsius while ice is melting, even with the heater running?
Check your reasoning after answering
Because the incoming energy is being used to break attractions rather than to speed particles up. Temperature reports average particle speed, so while energy goes into pulling particles out of the fixed arrangement the reading does not change. Only when all the ice has melted does further energy start increasing the speed again.
Part 4 · What the simple model leaves out
The model as described predicts that any solid should be denser than its own liquid, because the particles are packed more closely. For almost every substance that is exactly right: solid wax sinks in melted wax, and solid iron sinks in molten iron. Water is the famous exception. Liquid water has a density of about 1.00 g/cm3 and ice about 0.92 g/cm3, so ice floats, and the simple model has made a clear prediction that the world contradicts.
The correction is to add the detail the simple picture left out. As water freezes, its molecules lock into an open six-sided arrangement that holds them further apart than they sit when sliding freely as a liquid, so the solid takes up more room than the same mass of liquid. This is worth more than a footnote: because ice floats, a lake freezes downward from the surface and keeps liquid water beneath it, which is why fish survive winters that would otherwise freeze a pond solid from the bottom up. Freezing water also expands with enough force to split things open. Water sitting in a crack in a rock expands by about 9 per cent as it freezes and levers the crack wider, a process that breaks mountains apart over many winters and is the reason burst pipes are a feature of cold weather. You will meet this again in chapter 5 under the name of weathering.
Pause and explain
Why does water freezing to ice contradict the simple particle model, and why does it matter beyond the classroom?
Check your reasoning after answering
The simple picture says solids pack more tightly and so should be denser, yet ice at 0.92 g/cm3 is less dense than liquid water at 1.00 and floats. The missing detail is the open six-sided structure water forms on freezing. It matters because floating ice makes lakes freeze from the top down, leaving liquid water below where fish can survive.
Apply the whole unit · Predict the state
A substance has a melting point of minus 39 degrees Celsius and a boiling point of 357 degrees Celsius. A second substance melts at 801 degrees Celsius and boils at 1465 degrees Celsius. Room temperature is about 20 degrees Celsius.
Your task
State what each substance is at room temperature and compare the attractions between their particles.
What your response should show
- Places room temperature correctly against both fixed points for each substance
- Identifies the state of each substance from that comparison
- Ranks the attractions using the melting points as evidence
- Names the boiling point each substance must pass and notes the temperature plateau
Response guidance — after your attempt
For each substance, compare room temperature with its two fixed points and name the state.
Say which substance has the stronger attractions between particles, with your reason.
Describe what you would have to do to each substance to turn it into a gas.
For the first substance, 20 degrees Celsius sits above its melting point of minus 39 and well below its boiling point of 357, so at room temperature it is a liquid; this is mercury. For the second, 20 degrees Celsius is far below its melting point of 801, so it is a solid; this is common salt. The second substance has much stronger attractions between its particles, because it takes a temperature of 801 degrees Celsius to shake them out of their fixed arrangement while the first gives way at minus 39, a difference of 840 degrees Celsius in the energy needed to break the solid apart. To turn the first into a gas you would heat it past 357 degrees Celsius; to turn the second into a gas you would have to reach 1465 degrees Celsius, and in both cases the temperature would hold steady at the boiling point while the change of state was taking place.
Scientific references for this unit
- Royal Society of Chemistry: states of matter and the kinetic particle model
- National Institute of Standards and Technology: thermophysical properties of fluids
- US Geological Survey: why ice floats and what it means for lakes
- National Physical Laboratory: fixed points of temperature and how they are defined
Parent / teacher teaching notes
Begin with the literal observation, quantity, or diagram relationship. Ask the learner to explain why the evidence supports the claim, check a relevant unit or boundary, then discuss an alternative. The supplied numbers are illustrative, not real observations. A parent can read aloud and allow a spoken rehearsal before writing. Judge science and evidence rather than political, faith, or family agreement. Use the complete data option whenever an actual procedure is unavailable or unsuitable.
The key idea
Solid, liquid and gas are not three different substances but three arrangements of the same particles, differing in how closely they are packed, how strongly they hold each other and how fast they move. Every melting point and boiling point in this lesson is a measurement of how strong that holding is.
Worked examples
Example 1 · A changing state and an open boundary
- Identify which sample exchanges water with the surroundings.
- Distinguish melting from evaporation.
- Explain what remains unmeasured.
Example 2 · Transfer to a different case
- Identify the distinction and conditions in this new case.
- Use the relevant mechanism or quantity; preserve units and the chosen boundary.
- Explain why the answer fits and what further evidence would be needed.
Example 3 · Connect properties with particle arrangement · a focused example
- State the literal condition or observation in the example.
- Apply the distinction explained in this teaching part.
- Check whether the claim stays within the given evidence.
Avoid these mistakes
- Misleading claim: Solid particles are completely motionless.
- Misleading claim: A gas has no mass because it is invisible.
- Omitting units, measurement timing, or a model boundary makes the evidence hard to interpret.
- Treating illustrative classroom data as real field measurements or copying a model is not an independent explanation.
Check your understanding
Does a lower water level prove less water?
Discuss the answer
No. Container shape affects height; compare volume with an appropriate method before claiming loss.
What mainly changes during gas compression?
Discuss the answer
Particle spacing decreases; the model does not require individual particles to shrink.
Must water boil before it can evaporate?
Discuss the answer
No. Surface evaporation occurs below boiling; its rate depends on conditions such as temperature, airflow, and exposed area.
Why is the mass unchanged in this model?
Discuss the answer
The same water remains within the boundary; state and arrangement change, not the amount of matter.
Sustained scientific explanation and revision
These activities are scheduled in the parent plan. Save a first response, a correction, and the reason for the change in the science notebook.
Activity 1 · States of Matter & the Particle Model · Model and explain
Read The Clear Liquid and the Missing Salt and the linked unit teaching; use its complete supplied case and diagrams.
Your task
For States of Matter & the Particle Model, study Parts 3–4 and attempt the worked case: Explain both records using particles and system boundaries. Explain the steps before opening the model.
Success criteria
- An accurate scientific claim answering the actual question.
- Relevant evidence or a labeled model with quantities and units when applicable.
- Reasoning that connects evidence with the mechanism rather than only repeats it.
- A specific method, boundary, scale, or alternative-explanation limit.
Hint
Begin with the actual observation, quantity, or diagram relationship. Explain how it bears on the question before making a broad claim.
Response guidance — discuss after your attempt
Melting in the closed sample changes arrangement while retaining its 50 g of water. The open dish loses 6 g from the measured liquid; evaporation is a possible pathway into the air. The water has not been destroyed. Without further records, the mass loss alone does not exclude a spill or specify the evaporation rate over a particular time.
Activity 2 · States of Matter & the Particle Model · Practice, revise and exit
Read The Clear Liquid and the Missing Salt and the linked unit teaching; use its complete supplied case and diagrams.
Your task
For States of Matter & the Particle Model, complete practice questions 1–6 and both choice checks, then answer independently: A learner says a gas has no mass because it cannot be seen. Explain the error. Save a correction explaining one changed claim or method.
Success criteria
- An accurate scientific claim answering the actual question.
- Relevant evidence or a labeled model with quantities and units when applicable.
- Reasoning that connects evidence with the mechanism rather than only repeats it.
- A specific method, boundary, scale, or alternative-explanation limit.
Hint
Begin with the actual observation, quantity, or diagram relationship. Explain how it bears on the question before making a broad claim.
Response guidance — discuss after your attempt
Visibility is not the definition of matter. Gas particles have mass and occupy space; a suitable comparison or measurement can show these properties.
Words to know
- particle model
- A simplified explanation of matter using tiny moving constituents and their arrangement.
- evaporation
- A change from liquid to gas at a surface, possible below boiling.
- evidence
- A relevant observation, measurement, or record used to evaluate a claim.
- system boundary
- The chosen limit specifying what is included and which exchanges cross it.
- uncertainty
- A stated limit in measurement or knowledge, distinct from a careless mistake.
In the real world
The particle model explains the whole water cycle: the Sun's heat evaporates water from oceans and lakes into vapor, which then cools and condenses high in the sky to form the clouds that bring us rain and snow.
Exit check
A learner says a gas has no mass because it cannot be seen. Explain the error.
Check your explanation after answering
Visibility is not the definition of matter. Gas particles have mass and occupy space; a suitable comparison or measurement can show these properties.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Physical Properties & Physical Changes
Learning goals
- I can measure density and use it to compare two materials.
- I can decide whether a change is physical by asking what substance is present afterwards.
- I can explain why a single property cannot identify an unknown material.
Readings for this unit
Before this unit
Read the chapter opening and the named observation or diagram. Clarify one word or unit that blocks understanding. Recall the relevant earlier measurement, matter, or energy idea.
Study the unit across three 35–45-minute class sessions. Use the worked examples after an attempt and the supplied records when actual equipment is not available.
Gold has a density of about 19.3 g/cm3. Pyrite, the mineral miners called fool's gold, has a density of about 5.0 g/cm3. Both are metallic and yellow, and a great many people in the 1849 California rush carried home sacks of the wrong one.
Colour alone could not separate them, and neither could lustre. What separated them was any second property: pyrite scratches glass and gold does not, pyrite leaves a greenish-black streak on a white tile while gold leaves a yellow one, and a 100 g lump of gold occupies only about 5.2 cm3 while 100 g of pyrite occupies about 20 cm3.
That is the whole lesson in one example. Properties you can measure without destroying the sample are powerful, no single one of them identifies anything, and a change that alters shape or state has not altered what the substance is.
Build your understanding
Where this fits
The last lesson established that particles arrange themselves differently in solids, liquids and gases. This lesson is about the properties that follow from what a substance is rather than from how much of it you have, and about telling a change of form from a change of substance. The next lesson covers changes that do create a new substance.
Use three flexible 40-45-minute sessions: measuring properties without destroying the sample; physical change and what reversibility does not prove; density as a ratio and identifying an unknown, then the practice set, revision and the exit check. Workshop, review, quiz and exams have separate time.
Part 1 · Measuring without destroying
A physical property is anything you can determine while leaving the substance chemically unchanged: colour, hardness, melting point, density, whether it conducts electricity, whether it is attracted to a magnet. Some of these depend on how much you have and some do not, and the distinction matters. Mass and volume change when you take a bigger piece, so neither identifies a material. Density, melting point and hardness stay the same whatever the size of the sample, which is exactly why those are the ones used for identification. A 1 g crumb of aluminium and a 50 kg block both have a density of 2.70 g/cm3 and both melt at 660 degrees Celsius.
Hardness is measured on the Mohs scale, which ranks minerals by what scratches what: talc is 1, gypsum 2, calcite 3, quartz 7 and diamond 10. The test is genuinely practical, because a fingernail is about 2.5, a copper coin about 3.5 and a steel knife about 5.5, so three everyday objects divide the whole range into useful bands. Gold at 2.5 to 3 can be marked with a coin; pyrite at 6 to 6.5 cannot be scratched by a knife and will scratch glass instead. None of these tests consumes the sample, which is the point of calling them physical. Magnetism is worth adding to the list because it is so sharply decisive. Iron, nickel and cobalt are attracted to a magnet and almost nothing else in an ordinary classroom is, so a magnet separates iron filings from sand in seconds and tells you instantly that a shiny grey metal is not aluminium, which a magnet ignores completely.
Pause and explain
Why is density useful for identifying a material when mass is not?
Check your reasoning after answering
Because density does not depend on how much you have. A larger piece of aluminium has more mass and more volume, but the ratio between them stays at 2.70 g/cm3, so the number describes the material itself. Mass changes with every sample size, so it tells you about the particular lump in front of you and nothing about what it is made of.
Part 2 · Changing the form, keeping the substance
A physical change alters state, shape or arrangement while the substance stays what it was. Melting ice gives you water, which is the same substance in a different state. Tearing paper gives you smaller pieces of paper. Dissolving 10 g of salt in 100 mL of water gives you salt water, and the salt is still salt, which you can prove by boiling the water away and weighing what is left: close to 10 g of salt, unchanged. In each case the particles have been rearranged or separated, and nothing new has been made.
The common shortcut, that physical changes are reversible and chemical ones are not, is unreliable in both directions and you should drop it. Tearing paper is physical and essentially impossible to undo. Smashing a plate is physical and permanent. Meanwhile some chemical changes reverse readily under the right conditions. The question that actually works is simpler: after the change, is the same substance present? Ask that rather than asking whether you could get back to where you started, and the awkward cases stop being awkward. Dissolving is the case that causes most argument, so it is worth being explicit. When salt dissolves it seems to vanish, which feels like it must have become something else, but the mass is conserved exactly: 100 g of water plus 10 g of salt gives 110 g of solution, and boiling the water away returns the salt. Nothing was created or destroyed, the particles were simply separated and mixed among the water particles.
Pause and explain
Why is 'can it be reversed?' a poor test for whether a change is physical?
Check your reasoning after answering
Because it gives wrong answers both ways. Tearing paper and smashing a plate are physical changes that cannot practically be undone, while some chemical changes reverse easily in the right conditions. The reliable question is whether the same substance is present afterwards, which is about identity rather than about how difficult the change is to reverse.
Part 3 · Density as a ratio
Density is mass divided by volume, and the units say so: grams per cubic centimetre, written g/cm3, or grams per millilitre for liquids, where 1 mL and 1 cm3 are the same volume. A piece of iron with a mass of 78.7 g and a volume of 10 cm3 has a density of 7.87 g/cm3. Take a piece twice as large and both numbers double, to 157.4 g and 20 cm3, and the ratio is unchanged. That is what it means to say density is a property of the material rather than of the object.
Measuring the volume of an awkward shape is done by displacement, and the method is worth knowing precisely. Fill a measuring cylinder to a marked level, say 50 mL, lower the object in gently so nothing splashes out, and read the new level, say 68 mL. The object displaced 18 mL, so its volume is 18 cm3. Read the cylinder at eye level and at the bottom of the curved surface the water makes, because reading from above can easily shift the value by 1 mL or more, and on a volume of 18 cm3 that is an error of around 6 per cent in the density. For a regular shape you can skip displacement and measure directly instead. A rectangular block 4.0 cm by 3.0 cm by 2.0 cm has a volume of 24 cm3, found by multiplying the three lengths, and that will usually be more accurate than displacement because a ruler can be read more finely than a measuring cylinder. Use displacement when the shape makes the arithmetic impossible, not as a default.
Pause and explain
A stone raises the water in a cylinder from 50 mL to 68 mL and has a mass of 47 g. What is its density, and what could go wrong in the measurement?
Check your reasoning after answering
The stone displaced 18 mL, so its volume is 18 cm3 and the density is 47 divided by 18, about 2.6 g/cm3. The main risks are splashing water out when lowering the stone, which makes the volume look too small, and reading the cylinder from above rather than at eye level, which on 18 cm3 can shift the result by several per cent.
Part 4 · One property is never enough
Return to gold and pyrite. Both are metallic yellow solids, so colour and lustre put them in the same box. Density separates them decisively at 19.3 against 5.0 g/cm3, and so does hardness, 2.5 to 3 against 6 to 6.5, and so does the streak test, where rubbing the mineral on unglazed white tile leaves yellow for gold and greenish-black for pyrite. Any one of those three would have saved a prospector a journey, and the colour that fooled him is a perfectly real property that simply happens to be shared.
This is why identification is always a matter of combining properties and stating how confident you are. A classroom test can usually narrow an unknown to a small group of candidates rather than proving a unique answer, and saying so is part of the result. There is also a safety rule that follows directly: never identify an unknown substance by tasting it, smelling it closely, heating it or breaking it open, because those tests work by exposing you to whatever it turns out to be, which is the one thing you do not yet know. Professional laboratories face the same problem and solve it with instruments rather than senses. A spectrometer identifies a substance from the particular wavelengths of light it absorbs or emits, which is effectively a fingerprint and needs no contact at all. The logic is identical to the classroom version: gather several independent properties, and trust the identification in proportion to how many of them agree.
Pause and explain
Gold and pyrite are both metallic yellow. Name two measurements that separate them and explain why colour is still a real property.
Check your reasoning after answering
Density separates them at 19.3 against 5.0 g/cm3, and hardness at 2.5 to 3 against 6 to 6.5; the streak test also gives yellow against greenish-black. Colour remains a genuine physical property that can be measured and reported, it simply happens to be shared by these two, which shows why identification needs several properties rather than one.
Apply the whole unit · Identify the unknown
A grey metallic block has a mass of 108 g. Lowered into a measuring cylinder, it raises the water level from 50.0 mL to 90.0 mL. A steel knife does not scratch it. Candidate materials: aluminium at 2.70 g/cm3, iron at 7.87, lead at 11.3.
Your task
Calculate the density, choose between the candidates, and state how confident you can be.
What your response should show
- Calculates the volume by displacement and the density with correct units
- Matches the value to a candidate and rules the others out with figures
- Brings the second property in rather than relying on density alone
- States the limit of the identification instead of claiming certainty
Response guidance — after your attempt
Find the volume from the displacement and calculate the density with its unit.
Compare your value with the three candidates and say which it matches.
Use the hardness observation and state what your identification does and does not establish.
The block displaced 90.0 minus 50.0, which is 40.0 mL, so its volume is 40.0 cm3. The density is 108 divided by 40.0, which is 2.70 g/cm3. That matches aluminium exactly and is nowhere near iron at 7.87 or lead at 11.3, so the density alone rules both of those out: a 40 cm3 piece of lead would have a mass of about 452 g, more than four times what was measured. The hardness observation is consistent, since aluminium is harder than a fingernail but is not especially soft. What this establishes is that the block behaves like aluminium on the two properties tested, which is strong evidence given how far apart the three candidates are. What it does not establish is that the block is pure aluminium rather than an aluminium alloy, since many alloys have densities within a few per cent of the pure metal, and separating those would need a test this investigation did not do.
Scientific references for this unit
- US Geological Survey: mineral identification, streak and the Mohs hardness scale
- US Geological Survey Water Science School: density and measuring volume
- National Institute of Standards and Technology: densities of common materials
- Royal Society of Chemistry: physical and chemical change, and safe laboratory practice
Parent / teacher teaching notes
Begin with the literal observation, quantity, or diagram relationship. Ask the learner to explain why the evidence supports the claim, check a relevant unit or boundary, then discuss an alternative. The supplied numbers are illustrative, not real observations. A parent can read aloud and allow a spoken rehearsal before writing. Judge science and evidence rather than political, faith, or family agreement. Use the complete data option whenever an actual procedure is unavailable or unsuitable.
The key idea
A physical property can be measured without turning the substance into something else, and a physical change alters form while leaving identity intact. Because different substances share individual properties, identification always needs more than one.
Worked examples
Example 1 · Mass, volume, and shape
- Divide mass by volume and include units.
- Distinguish shape from chemical identity.
- State the model’s condition about no material exchange.
Example 2 · Transfer to a different case
- Identify the distinction and conditions in this new case.
- Use the relevant mechanism or quantity; preserve units and the chosen boundary.
- Explain why the answer fits and what further evidence would be needed.
Example 3 · Recognize a change in form · a focused example
- State the literal condition or observation in the example.
- Apply the distinction explained in this teaching part.
- Check whether the claim stays within the given evidence.
Avoid these mistakes
- Misleading claim: Every change that is hard to reverse must be chemical.
- Misleading claim: The largest sample must have the greatest density.
- Omitting units, measurement timing, or a model boundary makes the evidence hard to interpret.
- Treating illustrative classroom data as real field measurements or copying a model is not an independent explanation.
Check your understanding
Why is transparency insufficient for identification?
Discuss the answer
Different substances can share that appearance. Additional appropriate measurements and known source information are needed.
Does irreversibility alone prove chemical change?
Discuss the answer
No. A physical change can be difficult to undo; evidence about substance identity is more relevant.
What is the larger piece’s density if it is 60 g and 20 cm³?
Discuss the answer
60 ÷ 20 = 3 g/cm³; size changed but the ratio stayed the same.
How can a second property strengthen identification?
Discuss the answer
It distinguishes possibilities that the first property cannot separate; the conclusion should still match the available evidence.
Sustained scientific explanation and revision
These activities are scheduled in the parent plan. Save a first response, a correction, and the reason for the change in the science notebook.
Activity 1 · Physical Properties & Physical Changes · Model and explain
Read The Clear Liquid and the Missing Salt and the linked unit teaching; use its complete supplied case and diagrams.
Your task
For Physical Properties & Physical Changes, study Parts 3–4 and attempt the worked case: Calculate density and explain why the shape change is physical under the stated model. Explain the steps before opening the model.
Success criteria
- An accurate scientific claim answering the actual question.
- Relevant evidence or a labeled model with quantities and units when applicable.
- Reasoning that connects evidence with the mechanism rather than only repeats it.
- A specific method, boundary, scale, or alternative-explanation limit.
Hint
Begin with the actual observation, quantity, or diagram relationship. Explain how it bears on the question before making a broad claim.
Response guidance — discuss after your attempt
48 ÷ 24 = 2 g/cm³. Reshaping changes outline while the modeled material and measured mass/volume remain. It is described as physical because no new substance is formed in this model, not simply because the shape might be restored.
Activity 2 · Physical Properties & Physical Changes · Practice, revise and exit
Read The Clear Liquid and the Missing Salt and the linked unit teaching; use its complete supplied case and diagrams.
Your task
For Physical Properties & Physical Changes, complete practice questions 1–6 and both choice checks, then answer independently: A material is blue and heavy. Can those words uniquely identify it? Save a correction explaining one changed claim or method.
Success criteria
- An accurate scientific claim answering the actual question.
- Relevant evidence or a labeled model with quantities and units when applicable.
- Reasoning that connects evidence with the mechanism rather than only repeats it.
- A specific method, boundary, scale, or alternative-explanation limit.
Hint
Begin with the actual observation, quantity, or diagram relationship. Explain how it bears on the question before making a broad claim.
Response guidance — discuss after your attempt
No. Blue depends on the observation conditions, and heavy depends on sample size and comparison. Measured properties and additional evidence are needed.
Words to know
- physical property
- A characteristic measured or observed without changing chemical identity.
- density
- Mass per unit volume under stated conditions.
- evidence
- A relevant observation, measurement, or record used to evaluate a claim.
- system boundary
- The chosen limit specifying what is included and which exchanges cross it.
- uncertainty
- A stated limit in measurement or knowledge, distinct from a careless mistake.
In the real world
Jewelers and scientists use density like a fingerprint: because pure gold always has a density of 19.3 g/cm³, measuring a coin's mass and volume can reveal whether it is real gold or a cheaper fake.
Exit check
A material is blue and heavy. Can those words uniquely identify it?
Check your explanation after answering
No. Blue depends on the observation conditions, and heavy depends on sample size and comparison. Measured properties and additional evidence are needed.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Chemical Changes: Evidence & Conservation
Learning goals
- I can decide whether a change is chemical by comparing properties before and after.
- I can explain why bubbles or a colour change are not proof on their own.
- I can account for mass that appears to be gained or lost during a reaction.
Readings for this unit
Before this unit
Read the chapter opening and the named observation or diagram. Clarify one word or unit that blocks understanding. Recall the relevant earlier measurement, matter, or energy idea.
Study the unit across three 35–45-minute class sessions. Use the worked examples after an attempt and the supplied records when actual equipment is not available.
Burn 10.0 g of fine steel wool and weigh the ash. It comes out heavier, often around 13 g. Burn a 10.0 g candle and weigh what is left and it comes out lighter, often by almost the whole 10 g. The two results look like opposite laws of nature, and they are the same law seen from two sides.
In both cases the total mass of everything involved is unchanged. The steel wool combined with oxygen from the air, and oxygen has mass, so the solid that remains weighs more than the metal you started with. The candle also combined with oxygen, but its products, carbon dioxide and water vapour, floated away, so the balance lost track of them.
Antoine Lavoisier settled this in the 1770s by doing his reactions in sealed vessels. Nothing could enter or leave, and the mass before was always exactly the mass after. Mass is conserved; what varies is whether you remembered to weigh everything.
Build your understanding
Where this fits
The last lesson covered changes that leave a substance as it was. This one covers changes that produce something genuinely new, and the evidence that tells the two apart. Conservation of mass, introduced here, is the idea the whole of chemistry is built on and it comes back in chapter 7 when matter cycles through an ecosystem.
Use three flexible 40-45-minute sessions: the signs of chemical change and why they mislead; comparing properties before and after; conservation of mass in open and closed containers, then the practice set, revision and the exit check. Workshop, review, quiz and exams have separate time.
Part 1 · Signs, and why signs are not proof
The usual list of clues is worth knowing: a colour change, bubbles of gas, a temperature change you did not cause with a heater, light being given off, or a solid appearing when two clear liquids are mixed. Each of these does often accompany a chemical change. The trouble is that each also happens without one. Boiling water produces vigorous bubbles and is purely physical. Dissolving copper sulfate crystals turns clear water a deep blue and is purely physical, since evaporating the water returns the blue crystals unchanged. Ice melting absorbs a great deal of energy and cools its surroundings noticeably, and nothing new has been made at all.
So the clues are a reason to look more closely rather than a conclusion. The honest way to use them is as a prompt: something happened here that is worth investigating, and the investigation is a comparison of properties. A student who writes that a change was chemical because bubbles appeared has produced an observation and called it an inference, which is the error the first chapter of this course was about. The bubbles are real; what they mean still has to be established by other evidence. Some signs are more trustworthy than others, and it is worth ranking them. A solid appearing when two clear solutions are mixed is hard to explain physically, and so is a flame. A smell appearing where there was none before is also strong evidence. Colour change and bubbles sit at the weak end of the list, which is unfortunate because they are the two that students reach for first.
Pause and explain
Why is the appearance of bubbles not enough to show that a chemical change has happened?
Check your reasoning after answering
Because bubbles also appear in purely physical changes. Water boiling produces vigorous bubbles of water vapour while remaining water throughout, and a dissolved gas can come out of solution when a liquid is warmed. Gas being released shows that something is happening, not that a new substance has been made, which has to be established by comparing properties.
Part 2 · The test that works
Compare what you have afterwards with what you had before, using the kind of properties from the last lesson. Iron is grey, strongly attracted to a magnet, hard, and has a density of 7.87 g/cm3. Rust is reddish-brown, crumbly, much less strongly magnetic and quite different in density. Those are not the same substance described in two moods; they are two different materials with different properties, so iron turning to rust is a chemical change. Compare that with iron being hammered into a thin sheet: still grey, still magnetic, still 7.87 g/cm3, so nothing new has been made.
The comparison works in every case you will meet. Sugar dissolved in water can be recovered unchanged by evaporating the water, so the sugar was always sugar. Sugar heated until it blackens cannot be recovered, and what is left is carbon, which does not taste sweet, does not dissolve and does not melt at the temperature sugar melts at, so a chemical change has occurred. Notice that this test asks about identity rather than about appearance or effort, which is why it gives the same answer no matter who performs it. One more pair worth having in mind, because it comes up constantly in cooking. Whisking egg white is physical: it traps air and changes the texture completely while the egg stays egg. Frying it is chemical: the clear liquid becomes an opaque white solid that will not return to liquid however it is cooled, because the proteins have been permanently rearranged into something new.
Pause and explain
How do the properties of iron and rust show that rusting is a chemical change?
Check your reasoning after answering
Because the two materials do not share properties. Iron is grey, hard, strongly magnetic and 7.87 g/cm3; rust is reddish-brown, crumbly, far less magnetic and of different density. A substance that has genuinely remained itself keeps its properties, as iron does when hammered flat, so a change in all of them means a different substance is now present.
Part 3 · Nothing is lost
Lavoisier's method was to do the reaction where nothing could escape. He sealed tin and air inside a glass vessel, weighed the whole thing, heated it until the tin turned to a grey powder, let it cool, and weighed it again: identical. Then he opened the vessel and heard air rush in, and the mass rose. The explanation is that the tin had combined with oxygen from the air inside, so the sealed total never changed, and opening the vessel let replacement air in which added mass. The experiment is clean because it removes the only way for the accounting to go wrong.
The steel wool case is the same reaction without the seal. Iron combines with oxygen from the surrounding air, and since the oxygen joins the solid rather than floating away, 10.0 g of steel wool becomes about 13 g of iron oxide. Students almost always predict a loss, because burning feels like destruction and ash looks like a remnant. Weighing it settles the matter, and it is one of the most useful surprises in the course: a belief that felt obvious is overturned in two minutes by a balance. The practical consequences are large. Rusting consumes an estimated several per cent of the world's annual steel production, and the whole business of painting, galvanising and oiling metal exists to keep oxygen and water away from iron. Knowing that rust is iron combined with oxygen rather than iron wearing away is what tells you that a barrier, rather than a repair, is the thing to reach for.
Pause and explain
Why does burning steel wool make it heavier when burning a candle appears to make it lighter?
Check your reasoning after answering
Because of where the products go. In both reactions the substance combines with oxygen, which has mass. The iron oxide stays on the balance, so 10.0 g of steel wool becomes about 13 g. The candle's products, carbon dioxide and water vapour, leave as gases, so the balance no longer holds them and records a loss that has not actually occurred.
Part 4 · Weighing everything
Conservation of mass is a statement about a closed system, which is the idea introduced at the end of chapter 1. If nothing enters or leaves the boundary you have drawn, the mass inside is constant through any chemical change, and this holds no matter how violent the reaction looks. The reason it is hard to believe at first is that almost every reaction you see in ordinary life is open to the air, so gases arrive or depart unweighed and the books appear not to balance.
You can demonstrate it directly. Put vinegar in a flask and a measured amount of baking soda in a small tube inside it, stopper the flask, and weigh the whole assembly, say 150.0 g. Tip the tube so the two mix; the mixture fizzes vigorously and carbon dioxide fills the flask. Weigh again with the stopper still in and the reading is still 150.0 g. Now take the stopper out, let the gas escape, and weigh a third time: the mass has fallen, perhaps to 148.5 g, and that 1.5 g is the carbon dioxide that walked out of your system. It is worth noticing how small that number is compared with the drama of the reaction. The flask fizzes energetically, foam rises, and the whole visible performance accounts for 1.5 g out of 150.0 g, about one per cent of the mass present. Reactions look far more destructive than they are, and the balance is a better guide to what happened than your eyes.
Pause and explain
A stoppered flask holds the same mass during a fizzing reaction but loses 1.5 g once opened. Explain both readings.
Check your reasoning after answering
While the stopper is in, the flask is a closed system: the carbon dioxide produced is still inside the boundary, so the total is unchanged even though a reaction has occurred. Removing the stopper lets that gas leave the system, and the 1.5 g drop measures its mass. No matter was destroyed at any point; it simply crossed a boundary.
Apply the whole unit · Balance the books
A student heats 24.0 g of magnesium ribbon in air. The white powder left behind has a mass of 40.0 g. The student concludes that the balance must be faulty because matter cannot be created.
Your task
Explain the result, state where the extra mass came from, and say what the student should have done.
What your response should show
- Calculates the 16.0 g increase and identifies oxygen as its source
- Explains that the rule covers all reactants, not only what sat on the balance
- Rejects the faulty-balance conclusion using the chemistry rather than by assertion
- Proposes a sealed system and says why that makes the conservation measurable
Response guidance — after your attempt
Calculate the increase in mass and name the substance that supplied it.
Explain why the result does not break conservation of mass.
Describe an experimental change that would make the conservation visible on the balance.
The mass rose by 40.0 minus 24.0, which is 16.0 g. That mass came from oxygen in the air, which combined with the magnesium to make magnesium oxide; the product contains every atom of the original magnesium plus the oxygen that joined it. Conservation of mass is not broken, because the rule applies to everything involved and the student weighed only what was sitting on the balance. The oxygen was drawn from the surrounding air and was never weighed before the reaction, so the accounting was incomplete rather than wrong. To make conservation visible, the student should carry out the reaction in a sealed container holding a fixed amount of air and weigh the whole container before and after: those two readings would match, and the apparent gain would disappear, because now the boundary of the system includes the oxygen that does the reacting.
Scientific references for this unit
- Royal Society of Chemistry: conservation of mass and Lavoisier's sealed vessel experiments
- Science History Institute: Antoine Lavoisier and the chemical revolution
- National Institute of Standards and Technology: chemistry reference data for oxides
- US Department of Energy: corrosion of iron and the cost of rusting
Parent / teacher teaching notes
Begin with the literal observation, quantity, or diagram relationship. Ask the learner to explain why the evidence supports the claim, check a relevant unit or boundary, then discuss an alternative. The supplied numbers are illustrative, not real observations. A parent can read aloud and allow a spoken rehearsal before writing. Judge science and evidence rather than political, faith, or family agreement. Use the complete data option whenever an actual procedure is unavailable or unsuitable.
The key idea
A chemical change makes a substance that was not there before, and the way to tell is to compare properties afterwards rather than to watch for bubbles or colour. However dramatic the change looks, the total mass of everything involved is exactly the same before and after.
Worked examples
Example 1 · A closed-system atom account
- Count A and B before and after.
- Identify the changed particle combinations.
- Use the stated closed boundary and mass record.
Example 2 · Transfer to a different case
- Identify the distinction and conditions in this new case.
- Use the relevant mechanism or quantity; preserve units and the chosen boundary.
- Explain why the answer fits and what further evidence would be needed.
Example 3 · Evaluate signs with an alternative explanation · a focused example
- State the literal condition or observation in the example.
- Apply the distinction explained in this teaching part.
- Check whether the claim stays within the given evidence.
Avoid these mistakes
- Misleading claim: Any bubbles or color change conclusively prove a chemical reaction.
- Misleading claim: A mass change inside an open container means matter was destroyed.
- Omitting units, measurement timing, or a model boundary makes the evidence hard to interpret.
- Treating illustrative classroom data as real field measurements or copying a model is not an independent explanation.
Check your understanding
What does the particle model conserve?
Discuss the answer
The number of each kind of atom. New combinations form, but the represented atoms remain accounted for.
Do bubbles alone prove a new substance?
Discuss the answer
No. A physical state change can also create bubbles; the gas identity and process need further evidence.
Was 4 g of matter destroyed?
Discuss the answer
No. It crossed the chosen boundary as gas. The mass inside differs from the mass of the complete accounted system.
Why does a supplied record still need qualification?
Discuss the answer
The conclusion depends on the record’s stated method and boundaries; illustrative data are not proof that a real procedure was performed.
Sustained scientific explanation and revision
These activities are scheduled in the parent plan. Save a first response, a correction, and the reason for the change in the science notebook.
Activity 1 · Chemical Changes: Evidence & Conservation · Model and explain
Read The Clear Liquid and the Missing Salt and the linked unit teaching; use its complete supplied case and diagrams.
Your task
For Chemical Changes: Evidence & Conservation, study Parts 3–4 and attempt the worked case: Explain new combinations and conservation, then distinguish this evidence from color alone. Explain the steps before opening the model.
Success criteria
- An accurate scientific claim answering the actual question.
- Relevant evidence or a labeled model with quantities and units when applicable.
- Reasoning that connects evidence with the mechanism rather than only repeats it.
- A specific method, boundary, scale, or alternative-explanation limit.
Hint
Begin with the actual observation, quantity, or diagram relationship. Explain how it bears on the question before making a broad claim.
Response guidance — discuss after your attempt
Two A atoms and two B atoms are represented on each side. Their combinations change from A₂ and B₂ to two AB particles. The closed-system mass remains 60 g. This model directly states a rearrangement; a color change alone would not supply the same evidence of substance identity.
Activity 2 · Chemical Changes: Evidence & Conservation · Practice, revise and exit
Read The Clear Liquid and the Missing Salt and the linked unit teaching; use its complete supplied case and diagrams.
Your task
For Chemical Changes: Evidence & Conservation, complete practice questions 1–6 and both choice checks, then answer independently: An exposed metal object gains mass during a stated reaction with oxygen. Does conservation fail? Save a correction explaining one changed claim or method.
Success criteria
- An accurate scientific claim answering the actual question.
- Relevant evidence or a labeled model with quantities and units when applicable.
- Reasoning that connects evidence with the mechanism rather than only repeats it.
- A specific method, boundary, scale, or alternative-explanation limit.
Hint
Begin with the actual observation, quantity, or diagram relationship. Explain how it bears on the question before making a broad claim.
Response guidance — discuss after your attempt
No. Oxygen from the surroundings enters the material. Account for both the metal and the added oxygen rather than treating the exposed object as a closed system.
Words to know
- chemical change
- A change that forms substances with different chemical identities.
- conservation of mass
- Matter remains accounted for when the complete relevant system and exchanges are included.
- evidence
- A relevant observation, measurement, or record used to evaluate a claim.
- system boundary
- The chosen limit specifying what is included and which exchanges cross it.
- uncertainty
- A stated limit in measurement or knowledge, distinct from a careless mistake.
In the real world
Every time you cook you are running chemical reactions: the browning of toast and the rising of bread come from new substances forming out of the reactants in the dough.
Exit check
An exposed metal object gains mass during a stated reaction with oxygen. Does conservation fail?
Check your explanation after answering
No. Oxygen from the surroundings enters the material. Account for both the metal and the added oxygen rather than treating the exposed object as a closed system.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Mixtures, Solutions & Separation
Learning goals
- I can explain why the components of a mixture can be separated but those of a compound cannot.
- I can use solubility figures to predict how much of a solid will dissolve.
- I can choose a separation method by naming the property difference it exploits.
Readings for this unit
Before this unit
Read the chapter opening and the named observation or diagram. Clarify one word or unit that blocks understanding. Recall the relevant earlier measurement, matter, or energy idea.
Study the unit across three 35–45-minute class sessions. Use the worked examples after an attempt and the supplied records when actual equipment is not available.
The air in this room is a mixture: about 78 per cent nitrogen, 21 per cent oxygen, almost 1 per cent argon and about 0.04 per cent carbon dioxide, with water vapour varying from almost nothing to a few per cent. Nothing holds those gases together and nothing fixes the proportions, which is why the carbon dioxide figure has risen from about 0.028 per cent in 1800 to about 0.042 per cent now.
Sea water is a mixture too, carrying roughly 35 g of dissolved salts in every litre. So is the sand on the beach, so is milk, and so is almost everything you meet outside a laboratory. A pure substance with nothing else in it is the rare case rather than the normal one.
Because the components keep their own properties, every mixture can in principle be taken apart, and the method you choose depends on which property the components differ in. That single sentence organises the whole of this lesson.
Build your understanding
Where this fits
The previous lesson distinguished changes that make a new substance from changes that do not. This one deals with substances that are mixed together without reacting at all, which is the state most materials are found in. The techniques here, especially filtering and evaporating, are used again in chapter 7 when water quality is sampled and tested.
Use three flexible 40-45-minute sessions: what makes something a mixture; dissolving and how much will dissolve; matching separation methods to property differences, then the practice set, revision and the exit check. Workshop, review, quiz and exams have separate time.
Part 1 · Parts that stay themselves
Stir iron filings into sand and you have a mixture. The iron is still iron, still grey, still magnetic and still of density 7.87 g/cm3, and the sand is still sand. You can mix them in any proportion you like, 1 g of iron to 99 g of sand or the other way round, because nothing fixes the ratio. That is the signature of a mixture: the components keep every property they had, and the proportions are free. It is also the reason a magnet pulls the iron straight back out, since the property that made the iron magnetic was never surrendered when you stirred it in.
Compare that with iron sulfide, which is what you get if you heat iron and sulfur together until they react. The product is a dull black solid that a magnet ignores, and no amount of stirring, filtering or magnet-waving will get the iron back; separating it needs further chemical reactions. The proportions are fixed too, because the iron and sulfur combine in a set ratio by mass rather than in whatever amounts you happened to put in. A mixture can be unmixed by physical means and a compound cannot, and that is the practical difference between the two. Air makes the same point on a larger scale. Because it is a mixture rather than a compound, its components can be pulled apart physically: industrial plants chill air until it liquefies below minus 190 degrees Celsius and then warm it slowly, so nitrogen boils off first at minus 196 degrees Celsius and oxygen follows at minus 183. That is distillation, and it works only because nothing in air is chemically joined to anything else.
Pause and explain
Why can a magnet separate iron from a sand mixture but not from iron sulfide?
Check your reasoning after answering
Because mixing leaves properties intact while reacting does not. In the mixture the iron is still iron and still magnetic, so a magnet acts on it directly. In iron sulfide the iron has combined with sulfur to form a new substance that is not magnetic, so there is no longer any iron present for the magnet to attract.
Part 2 · How much will actually dissolve
When a solid dissolves, the solid is the solute, the liquid is the solvent and the result is a solution. The solute has not disappeared and has not reacted: its particles have separated and spread evenly among the solvent particles, which is why a solution looks uniform and why the mass is conserved exactly. Add 10 g of salt to 100 g of water and you have 110 g of salt solution, and boiling the water away returns close to 10 g of salt. There is a limit, though. At 20 degrees Celsius, 100 mL of water will dissolve about 36 g of salt and no more; beyond that the extra simply sits at the bottom and the solution is said to be saturated.
Different solids have wildly different limits, and the limits respond differently to temperature. Sugar is far more soluble than salt: about 204 g will dissolve in 100 mL of water at 20 degrees Celsius, rising to something like 487 g at 100 degrees Celsius. Salt barely shifts at all, going from about 36 g to only about 39 g over the same range. That contrast has a practical consequence you can predict without being told: cooling a hot saturated sugar solution forces a great deal of sugar back out as crystals, which is how rock candy is made, while cooling a salt solution produces almost nothing.
Pause and explain
Sugar's solubility rises from about 204 g to 487 g per 100 mL between 20 and 100 degrees Celsius, while salt's rises only from 36 g to 39 g. What does this predict about cooling each solution?
Check your reasoning after answering
That cooling a hot saturated sugar solution will deposit a large mass of crystals, roughly 283 g per 100 mL, because the liquid can no longer hold what it held when hot. A saturated salt solution cooled the same way can only shed about 3 g per 100 mL, so almost nothing appears. Crystallisation by cooling works well for sugar and poorly for salt.
Part 3 · Every method uses a difference
Filtration separates by particle size: sand grains are far too large to pass through the pores of filter paper while dissolved particles pass straight through, so filtering a sand and salt-water mixture leaves sand in the paper and clear salt water in the beaker. Evaporation separates by boiling point, in the crude sense that the solvent leaves as a vapour and the solute does not; boiling off the water leaves the salt behind as crystals. A magnet separates by magnetism, and decanting simply separates by density, pouring a lighter liquid off a settled solid.
Distillation is the one worth understanding properly, because it separates two liquids that are thoroughly mixed. Ethanol boils at 78 degrees Celsius and water at 100, so heating a mixture carefully to just above 78 drives off mostly ethanol vapour, which is then cooled in a condenser and collected as liquid. Chromatography uses a different property again: how strongly each component sticks to the paper compared with how readily it travels in the solvent, which is why a single black ink spot separates into several coloured bands as the solvent rises up the paper. It is worth noticing that these methods are not laboratory curiosities. Filtration and settling are the first two stages of almost every municipal water treatment works, distillation separates crude oil into petrol, diesel and the rest, and chromatography is what a laboratory uses to find out which substances are present in a blood or water sample. The same handful of property differences does nearly all the work.
Pause and explain
Filtration will not separate salt from water. Which property difference does it rely on, and why does salt water defeat it?
Check your reasoning after answering
Filtration relies on a difference in particle size, trapping anything too large to pass through the pores of the paper. Dissolved salt has separated into individual particles spread among the water, and these are far smaller than the pores, so the salt travels through with the water. Separating them needs a method based on boiling point instead.
Part 4 · Choosing in the right order
Take a mixture of iron filings, sand and salt, and the method follows from asking which component differs from the others in a way you can act on. Iron is the only magnetic component, so a magnet removes it first and leaves the other two untouched. Then add water and stir: salt dissolves and sand does not, so that step has turned one hard problem into two easy ones. Filter the result and the sand stays in the paper while the salt water passes through. Finally evaporate the salt water and the salt is left as crystals.
Order matters more than students expect. If you add the water first, the iron filings get wet and begin to rust, and you have introduced a chemical change into what was meant to be a physical separation. If you evaporate before filtering, the salt crystallises among the sand and you are back where you started. A good plan names the property used at each stage, which is also the quickest way to check that your plan can work at all: if you cannot say what difference a step exploits, the step will not separate anything. That test also tells you when a mixture cannot be separated with the equipment you have. Two liquids whose boiling points differ by only a degree or two will not separate cleanly by simple distillation, and two powders of the same size and the same solubility will not separate by filtering or dissolving at all. Recognising that early is more useful than attempting a method that was never going to work.
Pause and explain
Why should the magnet be used before water is added to an iron, sand and salt mixture?
Check your reasoning after answering
Because water starts a chemical change the separation does not want. Wet iron filings begin to rust, turning iron into iron oxide, which is no longer the substance you set out to recover and is not attracted to a magnet in the same way. Removing the iron while everything is dry keeps the whole process physical and reversible.
Apply the whole unit · Plan the separation
A beaker holds 200 mL of water in which 20 g of salt is dissolved, together with 15 g of sand and some small pieces of cork floating on top. Cork has a density of about 0.24 g/cm3.
Your task
Write a separation plan, naming the property difference used at each stage.
What your response should show
- Orders the steps so that no step undoes an earlier one
- Names a specific property difference for each of the three stages
- Predicts the recovered masses from the quantities given
- Accounts for a realistic shortfall rather than claiming perfect recovery
Response guidance — after your attempt
Decide the order of steps and say what leaves the mixture at each one.
Name the property difference each step depends on.
State the mass you expect to recover for the salt and explain any shortfall.
First skim off the cork, which floats because its density of 0.24 g/cm3 is far below the roughly 1.0 g/cm3 of the salt water; the property difference is density. Second, filter the remaining mixture: the sand grains are too large to pass through the filter paper while the dissolved salt passes through with the water, so the difference exploited is particle size, and about 15 g of sand should be recovered from the paper once it is dried. Third, evaporate the filtrate by gentle heating; the water leaves as vapour and the salt, which does not vaporise at these temperatures, stays behind, so the difference is boiling point. I would expect close to 20 g of salt, but slightly less in practice, because a little salt solution soaks into the filter paper and some crystals are lost when scraping the dish. Reporting 18 or 19 g would be a normal result rather than a sign of error.
Scientific references for this unit
- Royal Society of Chemistry: separation techniques and solubility data
- National Oceanic and Atmospheric Administration: why the ocean is salty
- NOAA Global Monitoring Laboratory: the changing composition of the atmosphere
- National Institute of Standards and Technology: solubility and phase change data
Parent / teacher teaching notes
Begin with the literal observation, quantity, or diagram relationship. Ask the learner to explain why the evidence supports the claim, check a relevant unit or boundary, then discuss an alternative. The supplied numbers are illustrative, not real observations. A parent can read aloud and allow a spoken rehearsal before writing. Judge science and evidence rather than political, faith, or family agreement. Use the complete data option whenever an actual procedure is unavailable or unsuitable.
The key idea
In a mixture every component keeps its own properties, which is exactly why mixtures can be separated. Each separation method works by exploiting one property the components do not share, so choosing a method means first finding the difference you can use.
Worked examples
Example 1 · A two-stage mixture model
- State the filter’s modeled role.
- Add water and salt mass after filtration.
- Identify the residue and the water pathway.
Example 2 · Transfer to a different case
- Identify the distinction and conditions in this new case.
- Use the relevant mechanism or quantity; preserve units and the chosen boundary.
- Explain why the answer fits and what further evidence would be needed.
Example 3 · Identify solute, solvent, and concentration · a focused example
- State the literal condition or observation in the example.
- Apply the distinction explained in this teaching part.
- Check whether the claim stays within the given evidence.
Avoid these mistakes
- Misleading claim: A dissolved substance has ceased to exist.
- Misleading claim: Removing visible particles automatically produces safe pure water.
- Omitting units, measurement timing, or a model boundary makes the evidence hard to interpret.
- Treating illustrative classroom data as real field measurements or copying a model is not an independent explanation.
Check your understanding
Does dissolved salt disappear as matter?
Discuss the answer
No. Its particles remain in the solution, contributing mass and properties even though they are not individually visible.
Is the second solution necessarily more concentrated?
Discuss the answer
No. The solute and volume both double, leaving the same ratio under this definition.
Why is the filtered liquid still a mixture?
Discuss the answer
Dissolved salt passes through with the water. Removing visible sand does not establish pure water.
Does recovering less sand prove sand was destroyed?
Discuss the answer
No. Identify retained material, loss pathways, and measurement uncertainty before assigning a cause.
Sustained scientific explanation and revision
These activities are scheduled in the parent plan. Save a first response, a correction, and the reason for the change in the science notebook.
Activity 1 · Mixtures, Solutions & Separation · Model and explain
Read The Clear Liquid and the Missing Salt and the linked unit teaching; use its complete supplied case and diagrams.
Your task
For Mixtures, Solutions & Separation, study Parts 3–4 and attempt the worked case: Track the substances and mass at each stage, with one limitation. Explain the steps before opening the model.
Success criteria
- An accurate scientific claim answering the actual question.
- Relevant evidence or a labeled model with quantities and units when applicable.
- Reasoning that connects evidence with the mechanism rather than only repeats it.
- A specific method, boundary, scale, or alternative-explanation limit.
Hint
Begin with the actual observation, quantity, or diagram relationship. Explain how it bears on the question before making a broad claim.
Response guidance — discuss after your attempt
After removing 3 g sand, 105 g of salt solution remains in the idealized model. After water enters the air, 5 g salt remains. The water is transferred, not destroyed. Real recovery may be incomplete; these assumptions do not show that the liquid was safe to drink.
Activity 2 · Mixtures, Solutions & Separation · Practice, revise and exit
Read The Clear Liquid and the Missing Salt and the linked unit teaching; use its complete supplied case and diagrams.
Your task
For Mixtures, Solutions & Separation, complete practice questions 1–6 and both choice checks, then answer independently: A filter removes visible dirt. Does that prove it removes dissolved chemicals or microbes? Save a correction explaining one changed claim or method.
Success criteria
- An accurate scientific claim answering the actual question.
- Relevant evidence or a labeled model with quantities and units when applicable.
- Reasoning that connects evidence with the mechanism rather than only repeats it.
- A specific method, boundary, scale, or alternative-explanation limit.
Hint
Begin with the actual observation, quantity, or diagram relationship. Explain how it bears on the question before making a broad claim.
Response guidance — discuss after your attempt
No. Filter pore size and other treatment properties matter. Visible clarity alone cannot establish removal of dissolved substances or drinking safety.
Words to know
- solution
- A mixture with dissolved particles distributed through a solvent at the particle scale.
- solute
- The substance dissolved in a solution.
- evidence
- A relevant observation, measurement, or record used to evaluate a claim.
- system boundary
- The chosen limit specifying what is included and which exchanges cross it.
- uncertainty
- A stated limit in measurement or knowledge, distinct from a careless mistake.
In the real world
Cities make tap water safe by separating out impurities — screening and filtering to trap solids, then settling and treating the water — and dry coastal regions use evaporation to pull fresh water out of salty seawater.
Exit check
A filter removes visible dirt. Does that prove it removes dissolved chemicals or microbes?
Check your explanation after answering
No. Filter pore size and other treatment properties matter. Visible clarity alone cannot establish removal of dissolved substances or drinking safety.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Water’s Properties: Density, Floating & Surface Effects
Learning goals
- I can predict whether an object floats by comparing its average density with water's.
- I can explain how a steel ship floats when steel is far denser than water.
- I can explain why coastal places have smaller temperature swings than inland ones.
Readings for this unit
Before this unit
Read the chapter opening and the named observation or diagram. Clarify one word or unit that blocks understanding. Recall the relevant earlier measurement, matter, or energy idea.
Study the unit across three 35–45-minute class sessions. Use the worked examples after an attempt and the supplied records when actual equipment is not available.
Water has a density of 1.00 g/cm3, which is not a coincidence: the gram was originally defined so that 1 cm3 of water would have a mass of 1 g. That makes water a convenient yardstick, because any material with a density below 1.00 floats in it and anything above sinks. Cork at 0.24 g/cm3 floats high, ice at 0.92 floats low, aluminium at 2.70 sinks and lead at 11.3 sinks fast.
Water is also unusually good at absorbing heat without getting much hotter. Raising the temperature of 1 kg of water by 1 degree Celsius takes about 4180 J. The same job for 1 kg of iron takes about 450 J, and for dry sand around 800 J. Water needs roughly nine times the energy iron does.
And water's surface behaves like a stretched skin, strong enough to hold up a steel paperclip laid on it flat and to let insects walk across a pond. All three oddities come from the same cause: water molecules attract each other far more strongly than their small size would suggest.
Build your understanding
Where this fits
This lesson closes the chapter on matter by looking hard at one substance. Water appears in almost every remaining chapter of this course, in the water cycle, in weather, in watersheds and in living things, and the properties set out here are the reason it matters in all of them. The specific heat figure in the last section is the one that explains coastal climates in chapter 6.
Use three flexible 40-45-minute sessions: density and floating; ships, displacement and average density; surface effects and water's heat capacity, then the practice set, revision and the exit check. Workshop, review, quiz and exams have separate time.
Part 1 · What floats and what does not
An object floats in water if its density is less than 1.00 g/cm3 and sinks if its density is greater, and that one comparison settles the whole question. Cork at 0.24 g/cm3 floats with most of itself above the surface. Ice at 0.92 g/cm3 floats, but only just: the fraction submerged is the ratio of the two densities, 0.92 divided by 1.00, so about 92 per cent of an iceberg sits below the waterline and only 8 per cent shows. That is where the expression about the tip of the iceberg comes from, and it is a prediction you can make from two numbers rather than a saying you have to remember.
Oil floats on water for the same reason, with most cooking oils sitting around 0.92 g/cm3. The ordering is strict and you can build a column of layers to show it: honey at about 1.42 g/cm3 at the bottom, then water at 1.00, then oil at 0.92, each floating on the one below because each is less dense than it. Dropping objects into that column sorts them by density too, so a grape at about 1.1 g/cm3 settles between the honey and the water while a cork rides on the very top. Density also sorts things on a planetary scale, which is worth noticing now because chapter 5 depends on it. Earth's interior is layered with the densest material at the centre and the lightest at the surface, for exactly the reason the liquid column layers itself, and the ocean itself is layered too, with colder and saltier water sitting beneath warmer, fresher water.
Pause and explain
Ice has a density of 0.92 g/cm3 and water 1.00. What fraction of a floating iceberg is below the surface, and how do you know?
Check your reasoning after answering
About 92 per cent, because the submerged fraction equals the ratio of the two densities, 0.92 divided by 1.00. The ice must push aside enough water to match its own mass, and since ice is only slightly less dense than water it has to sink almost entirely to do so, leaving roughly 8 per cent visible above the waterline.
Part 2 · How steel floats
Steel has a density of about 7.85 g/cm3, nearly eight times that of water, so a solid steel block sinks immediately. A steel ship floats because what matters is the average density of the whole vessel, hull and cargo and the enormous volume of air enclosed inside it, and that average comes out below 1.00 g/cm3. Flatten the same steel into a sheet and it sinks; fold it into a box and it floats, with no change in the amount of steel. Shape is doing the work by increasing the volume while leaving the mass alone.
The same idea stated the other way round is Archimedes' principle: a floating object pushes aside a mass of water equal to its own mass. A ship with a mass of 50,000 tonnes floats at the depth where its hull has displaced exactly 50,000 tonnes of water, which is why loading cargo makes a ship sit lower rather than making it sink. Ships carry a painted line on the hull marking the safe limit, and a vessel loaded past it has run out of hull to displace with and is genuinely in danger. Submarines use the same principle deliberately in both directions. They carry tanks that can be flooded with sea water to raise the vessel's average density above that of the surrounding water, which makes it sink, and then blown clear with compressed air to lower the average density again so it rises. Nothing about the steel changes; only the proportion of the enclosed volume filled with water rather than air.
Pause and explain
Steel is about 7.85 g/cm3, so why does a steel ship float?
Check your reasoning after answering
Because floating depends on the average density of the whole vessel, not of the steel alone. A hull encloses a very large volume of air, so the total mass divided by the total volume comes out below 1.00 g/cm3. The same steel rolled flat sinks, because the volume collapses while the mass stays the same.
Part 3 · The skin on the surface
Water molecules pull on each other in all directions inside the liquid, but a molecule at the surface has neighbours only below and beside it, so the net pull drags it inward. The result is a surface that behaves like a stretched elastic sheet, which is called surface tension. It is strong enough to support a steel paperclip laid flat, even though steel at 7.85 g/cm3 should sink instantly, because the clip is resting on the surface rather than floating in the water. Push it through the surface and it drops at once, which shows that the skin, not buoyancy, was holding it.
The same effect shapes every water droplet into a sphere, since a sphere has the smallest surface for a given volume and the surface is pulling itself as small as it can. It is why water forms a curved meniscus in a measuring cylinder, which is why the last lesson insisted on reading the bottom of the curve. It is also why adding detergent matters: detergent molecules get between the water molecules and weaken their grip on each other, the surface tension falls, and water that used to bead on a greasy plate now spreads and wets it. Surface tension is not a small force at small scales. A water strider's legs rest in dimples on the surface without breaking through, and the insect can stand on water many times its own mass because its weight is spread across several contact points. Gravity dominates for large objects and surface tension dominates for very small ones, which is why a paperclip can be floated and a spoon cannot.
Pause and explain
Why does a paperclip rest on water when steel is nearly eight times denser than water?
Check your reasoning after answering
Because it is held by the surface rather than floating in the liquid. Molecules at the surface are pulled inward by their neighbours below, producing a skin under tension that can support a small flat object. The clip is not displacing its own mass of water, which is why pushing it through the surface makes it sink immediately.
Part 4 · Water holds on to heat
It takes about 4180 J to raise the temperature of 1 kg of water by 1 degree Celsius, against about 450 J for iron and around 800 J for dry sand. Water therefore warms slowly and cools slowly, while land warms quickly and cools quickly. Anyone who has walked on a beach on a hot afternoon has measured this already: the dry sand is painful underfoot while the sea a few metres away is still cold, even though both have been under the same sun all day.
Scaled up, this governs climate. A town on the coast has a large body of water beside it that changes temperature only slowly, so its summers are cooler and its winters milder than an inland town at the same latitude; the yearly range might be 15 degrees Celsius on the coast against 30 degrees Celsius well inland. It is also why the sea breeze exists, since land heats faster during the day, the air above it rises, and cooler air flows in off the water to replace it. Chapter 6 returns to both of these. The same property does quiet work inside living things. A human body is roughly 60 per cent water by mass, and that water resists sudden temperature changes in exactly the way the sea does, which helps keep internal temperature near 37 degrees Celsius through a hot afternoon or a cold morning. Water's reluctance to change temperature is doing a job in a pond, in a climate and in you.
Pause and explain
Why does a coastal town have a smaller annual temperature range than an inland town at the same latitude?
Check your reasoning after answering
Because water takes about 4180 J to warm 1 kg by 1 degree Celsius against roughly 800 J for sand, so the sea changes temperature far more slowly than land. A large body of water beside the town absorbs heat in summer without warming much and releases it in winter, damping both extremes and leaving a range perhaps half that of an inland site.
Apply the whole unit · Float or sink
A sealed plastic box measures 10.0 cm by 8.0 cm by 5.0 cm and has a mass of 320 g. A student then puts 150 g of sand inside it and seals it again.
Your task
Decide whether the box floats before and after the sand is added, and explain the change.
What your response should show
- Calculates volume and both densities with correct units
- Compares each density against 1.00 g/cm3 to reach the float or sink decision
- Explains that mass changed while volume did not
- States an assumption the prediction depends on
Response guidance — after your attempt
Calculate the volume of the box and its density before the sand is added.
Recalculate the density with the sand inside and compare both with water.
Explain what has changed and what has not, and state one assumption you have made.
The volume is 10.0 times 8.0 times 5.0, which is 400 cm3. Before the sand, the density is 320 divided by 400, or 0.80 g/cm3, which is below 1.00 g/cm3, so the box floats, with about 80 per cent of it submerged. With 150 g of sand inside, the mass rises to 470 g while the volume stays at 400 cm3, because the sand goes into space that was already inside the box. The density is now 470 divided by 400, which is 1.175 g/cm3, above 1.00, so the box sinks. What changed is the mass and therefore the average density; what did not change is the volume, which is why adding contents can sink a container without making it any bigger. I have assumed the box stays sealed and watertight, since a leak would let water in, raise the mass further and sink it regardless of the arithmetic.
Scientific references for this unit
- US Geological Survey Water Science School: water density and its properties
- National Oceanic and Atmospheric Administration: why coasts have milder climates
- National Institute of Standards and Technology: specific heat capacity reference data
- Royal Society of Chemistry: surface tension and the behaviour of water
Parent / teacher teaching notes
Begin with the literal observation, quantity, or diagram relationship. Ask the learner to explain why the evidence supports the claim, check a relevant unit or boundary, then discuss an alternative. The supplied numbers are illustrative, not real observations. A parent can read aloud and allow a spoken rehearsal before writing. Judge science and evidence rather than political, faith, or family agreement. Use the complete data option whenever an actual procedure is unavailable or unsuitable.
The key idea
Water behaves oddly in several ways that matter enormously: its solid floats on its liquid, it holds more heat than almost any common substance, and its surface behaves as though it has a skin. Each of these follows from the same thing, which is how strongly water molecules hold on to one another.
Worked examples
Example 1 · Compare an ice model and a floating hull
- Divide ice mass by its volume.
- Find the mass of displaced water for the floating hull.
- Distinguish material density from whole-object geometry.
Example 2 · Transfer to a different case
- Identify the distinction and conditions in this new case.
- Use the relevant mechanism or quantity; preserve units and the chosen boundary.
- Explain why the answer fits and what further evidence would be needed.
Example 3 · Explain displacement and shape · a focused example
- State the literal condition or observation in the example.
- Apply the distinction explained in this teaching part.
- Check whether the claim stays within the given evidence.
Avoid these mistakes
- Misleading claim: The heaviest or largest object must always sink.
- Misleading claim: Ice floats because freezing removes its mass.
- Omitting units, measurement timing, or a model boundary makes the evidence hard to interpret.
- Treating illustrative classroom data as real field measurements or copying a model is not an independent explanation.
Check your understanding
Which is less dense than the modeled water?
Discuss the answer
The 20 mL object: 12 ÷ 20 = 0.6 g/mL. The second is 2 g/mL; the equal masses do not imply equal floating behavior.
Why can reshaping affect floating without changing the metal?
Discuss the answer
It changes the object’s geometry and enclosed air, allowing different displacement and average density.
Did the freezing model lose mass?
Discuss the answer
No. Mass stays 90 g while volume and density change with arrangement.
Does surface support prove the needle is less dense than water?
Discuss the answer
No. A surface force can matter for a small object; density and buoyancy must be considered separately.
Sustained scientific explanation and revision
These activities are scheduled in the parent plan. Save a first response, a correction, and the reason for the change in the science notebook.
Activity 1 · Water’s Properties: Density, Floating & Surface Effects · Model and explain
Read The Clear Liquid and the Missing Salt and the linked unit teaching; use its complete supplied case and diagrams.
Your task
For Water’s Properties: Density, Floating & Surface Effects, study Parts 3–4 and attempt the worked case: Calculate ice density and explain the hull’s displacement without claiming that every metal floats. Explain the steps before opening the model.
Success criteria
- An accurate scientific claim answering the actual question.
- Relevant evidence or a labeled model with quantities and units when applicable.
- Reasoning that connects evidence with the mechanism rather than only repeats it.
- A specific method, boundary, scale, or alternative-explanation limit.
Hint
Begin with the actual observation, quantity, or diagram relationship. Explain how it bears on the question before making a broad claim.
Response guidance — discuss after your attempt
Ice density is 0.9 g/mL. The hull displaces 60 g of modeled water, consistent with buoyancy balancing its weight. A hollow hull can float because of its overall geometry and average density; a compact piece of the same material may behave differently. Neither result is a rule for every temperature or liquid.
Activity 2 · Water’s Properties: Density, Floating & Surface Effects · Practice, revise and exit
Read The Clear Liquid and the Missing Salt and the linked unit teaching; use its complete supplied case and diagrams.
Your task
For Water’s Properties: Density, Floating & Surface Effects, complete practice questions 1–6 and both choice checks, then answer independently: A learner says the largest object must sink first. What measurements would you ask for? Save a correction explaining one changed claim or method.
Success criteria
- An accurate scientific claim answering the actual question.
- Relevant evidence or a labeled model with quantities and units when applicable.
- Reasoning that connects evidence with the mechanism rather than only repeats it.
- A specific method, boundary, scale, or alternative-explanation limit.
Hint
Begin with the actual observation, quantity, or diagram relationship. Explain how it bears on the question before making a broad claim.
Response guidance — discuss after your attempt
Ask for mass, volume or displacement, shape, enclosed air, and the liquid’s properties. Size alone does not determine average density or buoyancy.
Words to know
- buoyancy
- The upward force a fluid exerts on an immersed or partly immersed object.
- surface tension
- An effect of particle interactions at a liquid surface that can influence small surface objects.
- evidence
- A relevant observation, measurement, or record used to evaluate a claim.
- system boundary
- The chosen limit specifying what is included and which exchanges cross it.
- uncertainty
- A stated limit in measurement or knowledge, distinct from a careless mistake.
In the real world
Exit check
A learner says the largest object must sink first. What measurements would you ask for?
Check your explanation after answering
Ask for mass, volume or displacement, shape, enclosed air, and the liquid’s properties. Size alone does not determine average density or buoyancy.
Explain your reasoning without notes. If a key idea is missing, revisit the lesson and try again.
Problem 1
Answer and explanation
Melting in the closed sample changes arrangement while retaining its 50 g of water. The open dish loses 6 g from the measured liquid; evaporation is a possible pathway into the air. The water has not been destroyed. Without further records, the mass loss alone does not exclude a spill or specify the evaporation rate over a particular time.
Compare the scientific meaning and reasoning after an independent attempt. A different accurate supported response can be correct; human review is required.
Problem 2
Answer and explanation
Visibility is not the definition of matter. Gas particles have mass and occupy space; a suitable comparison or measurement can show these properties.
Compare the scientific meaning and reasoning after an independent attempt. A different accurate supported response can be correct; human review is required.
Problem 3
Answer and explanation
48 ÷ 24 = 2 g/cm³. Reshaping changes outline while the modeled material and measured mass/volume remain. It is described as physical because no new substance is formed in this model, not simply because the shape might be restored.
Compare the scientific meaning and reasoning after an independent attempt. A different accurate supported response can be correct; human review is required.
Problem 4
Answer and explanation
No. Blue depends on the observation conditions, and heavy depends on sample size and comparison. Measured properties and additional evidence are needed.
Compare the scientific meaning and reasoning after an independent attempt. A different accurate supported response can be correct; human review is required.
Problem 5
Answer and explanation
Two A atoms and two B atoms are represented on each side. Their combinations change from A₂ and B₂ to two AB particles. The closed-system mass remains 60 g. This model directly states a rearrangement; a color change alone would not supply the same evidence of substance identity.
Compare the scientific meaning and reasoning after an independent attempt. A different accurate supported response can be correct; human review is required.
Problem 6
Answer and explanation
No. Oxygen from the surroundings enters the material. Account for both the metal and the added oxygen rather than treating the exposed object as a closed system.
Compare the scientific meaning and reasoning after an independent attempt. A different accurate supported response can be correct; human review is required.
Problem 7
Answer and explanation
After removing 3 g sand, 105 g of salt solution remains in the idealized model. After water enters the air, 5 g salt remains. The water is transferred, not destroyed. Real recovery may be incomplete; these assumptions do not show that the liquid was safe to drink.
Compare the scientific meaning and reasoning after an independent attempt. A different accurate supported response can be correct; human review is required.
Problem 8
Answer and explanation
No. Filter pore size and other treatment properties matter. Visible clarity alone cannot establish removal of dissolved substances or drinking safety.
Compare the scientific meaning and reasoning after an independent attempt. A different accurate supported response can be correct; human review is required.
Problem 9
Answer and explanation
Ice density is 0.9 g/mL. The hull displaces 60 g of modeled water, consistent with buoyancy balancing its weight. A hollow hull can float because of its overall geometry and average density; a compact piece of the same material may behave differently. Neither result is a rule for every temperature or liquid.
Compare the scientific meaning and reasoning after an independent attempt. A different accurate supported response can be correct; human review is required.
Problem 10
Answer and explanation
Ask for mass, volume or displacement, shape, enclosed air, and the liquid’s properties. Size alone does not determine average density or buoyancy.
Compare the scientific meaning and reasoning after an independent attempt. A different accurate supported response can be correct; human review is required.