Properties of Matter — Chapter Practice, Grade 8

electron

Electrons carry negative charge and are described by an electron cloud, not literal planet-like orbits.

In the nucleus

Protons and neutrons are packed together in the central nucleus.

12

Neutrons = mass number − atomic number = 23 − 11 = 12.

24

Mass number = protons + neutrons = 12 + 12 = 24.

A neutral atom has the same number of protons (positive) as electrons (negative), so the charges cancel out and the atom has no overall charge.

20

Neutrons = mass number − atomic number = 37 − 17 = 20.

An oxygen-17 ion with charge −2.

Atomic number 8 identifies oxygen.

Mass number is 8+9=17; 8 positive and 10 negative charges give a net charge of −2.

Response criteria

• Use the stated conditions.
• Show the evidence or calculation.
• Explain the conclusion and any stated limitation.

Check that changing neutron count leaves both proton count and charge unchanged.

Mass number=13+14=27; there are 13 electrons because neutrality requires equal positive and negative charges.

State what the supplied observations or conditions establish, and separate that from any assumption.

No. Isotopes share proton count. These have different atomic numbers and are different elements.

Use the illustrated study titled Molecules and extended structures. Keep the observations, model and conditions separate.

Two water molecules contain four H and two O atoms. A salt patch represents part of an extended arrangement; its bulk Na:Cl ratio is one-to-one rather than a claim that the crystal is a collection of separate NaCl molecules. Retain element/ion labels and note that colors, bonding lines and distances are conventions.

Two water molecules contain four H and two O atoms. A salt patch represents part of an extended arrangement; its bulk Na:Cl ratio is one-to-one rather than a claim that the crystal is a collection of separate NaCl molecules. Retain element/ion labels and note that colors, bonding lines and distances are conventions.

Response criteria

• Distinguish molecule from extended arrangement.
• Count water atoms correctly.
• Explain formula ratio and a model limit.

It is oxygen, because it still has 8 protons. It has two more electrons than protons and therefore charge −2 in elementary-charge units.

Read p⁺ as positively charged protons, n as neutral neutrons and e⁻ as negatively charged electrons. Element identity and net charge answer different questions. Changing only electrons cannot turn oxygen into carbon.

It is oxygen, because it still has 8 protons. It has two more electrons than protons and therefore charge −2 in elementary-charge units.

Response criteria

• Use the named labels, directions or supplied quantities accurately.
• Connect the visual evidence to the scientific mechanism or calculation.
• State the relevant model conditions or limitation.

Atomic number 12; mass number 25; charge +2 in elementary-charge units. Electron loss changes charge, while the 12 protons still determine the element.

Read proton count for atomic number.

Add 12+13 for mass number; compare 12 positive with 10 negative charges.

Atomic number 12; mass number 25; charge +2 in elementary-charge units. Electron loss changes charge, while the 12 protons still determine the element.

Response criteria

• Use the named labels, directions or supplied quantities accurately.
• Connect the visual evidence to the scientific mechanism or calculation.
• State the relevant model conditions or limitation.

flammability

Flammability describes how a substance reacts (burns), which is a chemical property.

tearing paper

Tearing paper changes its shape and size, but it's still the same paper — no new substance forms.

3 g/cm³

Density = mass ÷ volume = 45 ÷ 15 = 3 g/cm³.

the formation of a new substance with new properties

The clearest sign of a chemical change is a genuinely new substance forming, with properties different from what you started with.

The salt doesn't become a new substance — it's still salt, just spread out among the water molecules. If you evaporated the water, you'd get the same salt back.

7.9 g/cm³

Density = mass ÷ volume = 158 ÷ 20 = 7.9 g/cm³.

Tearing is physical; burning is chemical. New substances, rather than size alone, distinguish the changes.

Tearing changes shape while the paper remains the same material.

Burning rearranges atoms into substances such as gases and ash.

Response criteria

• Use the stated conditions.
• Show the evidence or calculation.
• Explain the conclusion and any stated limitation.

Ask whether a new substance was demonstrated, rather than whether the sample looked different.

Physical: shape changes while the substance remains copper. No new composition is described.

State what the supplied observations or conditions establish, and separate that from any assumption.

Color is evidence to investigate, not proof alone. Compare composition or characteristic properties before and after and rule out physical causes.

Use the illustrated study titled Mixtures and physical separation. Keep the observations, model and conditions separate.

The separate-component tests and recovery of sand and salt are consistent with a physical combination. The illustrative recovered masses are 4.8+2.9=7.7 g versus 8.0 g originally added, leaving 0.3 g unaccounted for in that collection. Transfer residues, remaining dissolved salt or reading uncertainty can explain a shortfall. Better rinsing, complete collection and repeated matched tests can improve the account; the data do not establish a new compound.

The separate-component tests and recovery of sand and salt are consistent with a physical combination. The illustrative recovered masses are 4.8+2.9=7.7 g versus 8.0 g originally added, leaving 0.3 g unaccounted for in that collection. Transfer residues, remaining dissolved salt or reading uncertainty can explain a shortfall. Better rinsing, complete collection and repeated matched tests can improve the account; the data do not establish a new compound.

Response criteria

• Specify components and a property-based method.
• Use recovered identity and mass evidence.
• Explain a limitation or controlled improvement.

Predicted mass is 12×3 = 36 g. Density alone is insufficient because different substances can share similar values; compare additional characteristic properties.

The slope of this mass-versus-volume graph is density, not speed. Both axes name measured properties. Use several properties, with conditions stated, before deciding whether a substance’s identity has changed.

Predicted mass is 12×3 = 36 g. Density alone is insufficient because different substances can share similar values; compare additional characteristic properties.

Response criteria

• Use the named labels, directions or supplied quantities accurately.
• Connect the visual evidence to the scientific mechanism or calculation.
• State the relevant model conditions or limitation.

No. Their distinct melting behavior provides another characteristic-property comparison. One matching density is not unique identification.

Compare multiple characteristic properties under stated conditions.

Separate an observed property from an inference of identity.

No. Their distinct melting behavior provides another characteristic-property comparison. One matching density is not unique identification.

Response criteria

• Use the named labels, directions or supplied quantities accurately.
• Connect the visual evidence to the scientific mechanism or calculation.
• State the relevant model conditions or limitation.

3 g/cm³

24 ÷ 8 = 3.

Volume

Density = mass ÷ volume.

Its hull encloses air, reducing its average density below that of water.

2.7 g/mL; the solid sinks if fully wetted, without trapped air or other support.

Displaced volume is 40−20=20 mL.

Density is 54/20=2.7 g/mL, greater than water's density.

Response criteria

• Use the stated conditions.
• Show the evidence or calculation.
• Explain the conclusion and any stated limitation.

Multiply density by volume: 3×8=24 and 3×24=72, recovering the masses.

The solid volume is 14 mL; density=35/14=2.5 g/mL.

State what the supplied observations or conditions establish, and separate that from any assumption.

90/120=0.75 g/mL, below water's density. It can float under the model, provided it retains its enclosed air and does not take in water.

Its mass and volume both double, so 48/16 equals 24/8, or 3 g/cm³.

Its mass and volume both double, so 48/16 equals 24/8, or 3 g/cm³.

Response criteria

• Use the supplied conditions and relevant evidence.
• Explain the scientific mechanism or calculation.
• State a meaningful limitation or testable next step.

A and B each have density 3 g/cm³; the hollow object averages 0.6 g/cm³. A and B tend to sink if no other supporting effect dominates; the hollow object can float if it keeps that effective enclosed volume and displaces sufficient water. Larger mass does not imply larger density, and a leak can change the hollow object’s average density.

A and B each have density 3 g/cm³; the hollow object averages 0.6 g/cm³. A and B tend to sink if no other supporting effect dominates; the hollow object can float if it keeps that effective enclosed volume and displaces sufficient water. Larger mass does not imply larger density, and a leak can change the hollow object’s average density.

Response criteria

• Use the supplied conditions and relevant evidence.
• Explain the scientific mechanism or calculation.
• State a meaningful limitation or testable next step.

Use the illustrated study titled Density as identification evidence. Keep the observations, model and conditions separate.

Density is 54/20=2.7 g/cm³. A and C remain consistent while B is inconsistent under matched conditions and adequate measurement precision. Density alone cannot distinguish A from C; choose a safe characteristic-property comparison with supplied reference data, such as solubility for approved samples. Name uncertainty and whether the object is homogeneous.

Density is 54/20=2.7 g/cm³. A and C remain consistent while B is inconsistent under matched conditions and adequate measurement precision. Density alone cannot distinguish A from C; choose a safe characteristic-property comparison with supplied reference data, such as solubility for approved samples. Name uncertainty and whether the object is homogeneous.

Response criteria

• Calculate with correct units.
• Distinguish support from unique identification.
• Choose relevant further evidence and controls.

Volume = 58−40 = 18 cm³; density = 54/18 = 3 g/cm³, subject to the supplied complete-submersion assumptions.

First find the change in liquid reading; it is the volume of the submerged object. Then combine that volume with a separately measured mass. A water displacement of 18 mL does not mean the object itself has a mass of 18 g.

Volume = 58−40 = 18 cm³; density = 54/18 = 3 g/cm³, subject to the supplied complete-submersion assumptions.

Response criteria

• Use the named labels, directions or supplied quantities accurately.
• Connect the visual evidence to the scientific mechanism or calculation.
• State the relevant model conditions or limitation.

Volume 12 cm³; density 30/12 = 2.5 g/cm³. Assume no trapped bubbles, spill or incomplete submersion. Density alone does not uniquely identify the substance.

Use the change 47−35, not the final reading.

Divide the independently measured mass by that volume.

Volume 12 cm³; density 30/12 = 2.5 g/cm³. Assume no trapped bubbles, spill or incomplete submersion. Density alone does not uniquely identify the substance.

Response criteria

• Use the named labels, directions or supplied quantities accurately.
• Connect the visual evidence to the scientific mechanism or calculation.
• State the relevant model conditions or limitation.

numbers of valence electrons

Elements in a group share the same number of valence electrons, which drives similar chemical behavior.

6

For main-group elements, the group number (16) tells you the valence electron count: 16 − 10 = 6.

Metals (left and center)

Metals, found on the left and center of the table, are typically shiny, malleable, and conduct electricity well.

They have the same number of valence electrons, and valence electrons mostly determine how an atom bonds and reacts with other atoms.

decreases

Atomic size generally decreases across a period because added protons pull the electron cloud inward.

2

Group 2 elements (alkaline earth metals) all have 2 valence electrons.

They are different elements with some similar chemical properties; the table alone does not give identical reaction rates.

Their proton counts are 12 and 20, so they are different elements.

Their shared main group suggests similar outer-electron patterns and some similar reactions.

Response criteria

• Use the stated conditions.
• Show the evidence or calculation.
• Explain the conclusion and any stated limitation.

Check the group labels; neighboring position alone is not the comparison rule.

No. Atomic number specifies 16 protons. Neutron count requires isotope mass number or other evidence.

State what the supplied observations or conditions establish, and separate that from any assumption.

No. Similar valence patterns suggest some related chemical behavior; they do not require identical physical properties such as density.

It has 11 protons and 11 electrons. Na-23 has 12 neutrons because its specified mass number is 23. The decimal table value is an isotope-weighted average, not an isotope definition.

The unit u is an atomic-mass unit used for masses at the atomic scale. Mass number is a count, whereas the listed average mass is a measured average for an isotope mixture. Keep both distinct from atomic number.

It has 11 protons and 11 electrons. Na-23 has 12 neutrons because its specified mass number is 23. The decimal table value is an isotope-weighted average, not an isotope definition.

Response criteria

• Use the named labels, directions or supplied quantities accurately.
• Connect the visual evidence to the scientific mechanism or calculation.
• State the relevant model conditions or limitation.

Chlorine-37 has 17 protons and 37−17 = 20 neutrons. The decimal 35.45 is an isotope-average mass, not one atom’s particle count.

Use the specified isotope mass number, not the decimal average.

Subtract atomic number from mass number.

Chlorine-37 has 17 protons and 37−17 = 20 neutrons. The decimal 35.45 is an isotope-average mass, not one atom’s particle count.

Response criteria

• Use the named labels, directions or supplied quantities accurately.
• Connect the visual evidence to the scientific mechanism or calculation.
• State the relevant model conditions or limitation.

Electrical conductivity

Charge must move through the wire.

Safety depends on the specific material, exposure, and use, not only its origin.

It can be reused because it is durable; its durability can also make discarded waste persist.

B uses 5 g/trip versus A's 10 g/trip in this model; the model alone does not settle total environmental impact.

Divide each bag's material by its number of trips: 20/2 and 100/20.

Compare 10 g/trip with 5 g/trip, while noting that energy, cleaning and disposal are not included.

Response criteria

• Use the stated conditions.
• Show the evidence or calculation.
• Explain the conclusion and any stated limitation.

Equivalent trips must be compared; weight of one unused bag is insufficient.

240/120=2 g/use. This omits production energy, cleaning and disposal and assumes all 120 uses occur.

State what the supplied observations or conditions establish, and separate that from any assumption.

Compare required load capacity and durability under salt exposure, then cost and other constraints. Strength alone cannot decide suitability.

One use-performance feature. It does not measure extraction, manufacturing energy, lifetime reuse or disposal impacts.

One use-performance feature. It does not measure extraction, manufacturing energy, lifetime reuse or disposal impacts.

Response criteria

• Use the supplied conditions and relevant evidence.
• Explain the scientific mechanism or calculation.
• State a meaningful limitation or testable next step.

For a leak-sensitive one-time model test, A’s observed leak record is favorable. For repeated use, B’s larger tested lifetime matters, but one observed leak requires improvement or further testing. Cost and full environmental impacts remain unknown. A recommendation must identify the intended criterion and hard constraints rather than combine unlike measures without a rule.

For a leak-sensitive one-time model test, A’s observed leak record is favorable. For repeated use, B’s larger tested lifetime matters, but one observed leak requires improvement or further testing. Cost and full environmental impacts remain unknown. A recommendation must identify the intended criterion and hard constraints rather than combine unlike measures without a rule.

Response criteria

• Use the supplied conditions and relevant evidence.
• Explain the scientific mechanism or calculation.
• State a meaningful limitation or testable next step.

Use the illustrated study titled Natural resources and different plastics. Keep the observations, model and conditions separate.

The reuse record supports longer tested service for the rigid container, while the disposal record warns that acceptance depends on product and local processing. Resource origin alone does not establish total impacts. Compare the same task, include cleaning and replacement, and obtain specific production/disposal information. Describe chains and processing as a plausible property explanation rather than proof from the label plastic.

The reuse record supports longer tested service for the rigid container, while the disposal record warns that acceptance depends on product and local processing. Resource origin alone does not establish total impacts. Compare the same task, include cleaning and replacement, and obtain specific production/disposal information. Describe chains and processing as a plausible property explanation rather than proof from the label plastic.

Response criteria

• Connect a synthetic material to a resource and processing.
• Use two different source records.
• Assess benefits, concerns and missing evidence fairly.

Radiation or conduction through contacts can remain, and moving air may contribute. Compare matched temperature-versus-time records and other design criteria instead of claiming all energy transfer stops.

Structure describes how materials are arranged; a property describes their measured response under conditions. Processing and assembly can change the arrangement. A useful design explanation connects the feature to a particular mechanism and a measurable outcome.

Radiation or conduction through contacts can remain, and moving air may contribute. Compare matched temperature-versus-time records and other design criteria instead of claiming all energy transfer stops.

Response criteria

• Use the named labels, directions or supplied quantities accurately.
• Connect the visual evidence to the scientific mechanism or calculation.
• State the relevant model conditions or limitation.

A uses 8 initial material-units; B uses 40 for 20 cups. Cleaning energy/water, production, transport, disposal and actual lifetime are omitted, so initial material alone is not a complete impact ranking.

Compare the same amount of service.

Keep the measured raw-material comparison separate from the full life cycle.

A uses 8 initial material-units; B uses 40 for 20 cups. Cleaning energy/water, production, transport, disposal and actual lifetime are omitted, so initial material alone is not a complete impact ranking.

Response criteria

• Use the named labels, directions or supplied quantities accurately.
• Connect the visual evidence to the scientific mechanism or calculation.
• State the relevant model conditions or limitation.