Worked example 1: solution

Step 1 · Count the kinds of particles

Every particle in the tank is the same CO₂ molecule. One kind of particle: a pure substance.

Worked example 1: solution

A pure substance, so count elements. The formula holds carbon and oxygen: two elements.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

A common first attempt

Stir in more salt: still clear, still salt water. The ratio changed with no chemical reaction. A compound's composition cannot change without one.

Worked example 2: solution

Water molecules and dissolved salt: two kinds. A mixture, and boiling, a physical change, separates them.

Worked example 2: solution

A mixture, so count phases. One phase, uniform throughout:

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 · answer: B

A: same properties throughout shows one phase, not one substance; the composition can still vary. C: uniform appearance does not show bonding; the acid and water separate by distillation, a physical change. D: several substances can share one phase; classification follows the phase count, not the substance count.

Worked example 3: kinds of particles

Step 1 · Count the kinds of particles

Worked example 3: kinds of particles

Step 1 · Count the kinds of particles

Each sample holds more than one kind of particle: three mixtures, in three physical states.

Worked example 3: phases and the class

Step 2 · Count the elements or the phases

Worked example 3: phases and the class

Step 3 · Name the class

Worked example 3: phases and the class

Step 3 · Name the class

Worked example 3: the route on the map

Practice 2 · answer: C

Step 1: the same composition in every sample means one kind of particle, a pure substance. Step 2: heating turns it into carbon and water, new substances: a chemical change. A pure substance that a reaction breaks down holds more than one element: a compound.

A: a mixture varies in composition and separates by a physical change; charring is chemical. B: fixed composition marks any pure substance, and no reaction breaks an element down. D: counts the phases of the products, not of the sample.

Worked example 1: solution

Step 1 · Name the substance before and after

Melting: solid butter becomes liquid butter. Cool the pan and the same butter hardens back. Same substance, new form:

Worked example 1: solution

Browning: the butter becomes brown solids, smoke, and a sharp smell. Cooling leaves them brown. The butter is gone.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Name the substance before and after Step 2 · Check the evidence

Sawing leaves the same metal in two pieces: a physical change. No new color, no gas, no new solid, no temperature change; nothing signals a reaction.

Worked example 2: solution

Step 1 · Name the substance before and after Step 2 · Check the evidence
Step 3 · Halve the sample mentally

Worked example 2: solution

Step 1 · Name the substance before and after Step 2 · Check the evidence
Step 3 · Halve the sample mentally

Worked example 2: solution

Step 1 · Name the substance before and after Step 2 · Check the evidence
Step 3 · Halve the sample mentally

Practice 1 · answer: D

A: state is observed by looking; the liquid stays gasoline. B: floating compares two densities, and both liquids survive the comparison. C: evaporation trades liquid for vapor, still gasoline; chill the vapor and it condenses right back.

Practice 2 · answer: B

Melted copper is still copper, so (4) is physical; the green coating is a new substance, so (3) is chemical. The halving test sorts the numbers: mass halves, density and melting point hold.

A: melting changes the form, never the identity. C: density held when the wire was halved; mass and volume shrink together. D: "melts at 1085 °C" is a number on a label, not an event on the bench. Properties describe; changes happen.

Practice · answer: C

A accepts the claim; particles keep their size in every state, and the gaps between them grew. B describes a liquid; touching particles hold their own volume, and the vapor took the flask's. D breaks the identity rule; the vapor is the same bromine, molecule for molecule, and cooling returns the liquid.

Worked example 1: solution

Step 1 · Compute the range Step 2 · Compute the average

Worked example 1: solution

Step 1 · Compute the range Step 2 · Compute the average

A spread of 0.12% on a 50-g measurement: the trials agree. Precise.

Worked example 1: solution

Step 1 · Compute the range Step 2 · Compute the average

Worked example 1: solution

Step 1 · Compute the range Step 2 · Compute the average

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Compute the range Step 2 · Compute the average

Worked example 2: solution

Step 1 · Compute the range Step 2 · Compute the average

Not precise. Precision compares the trials with one another; the true value is not part of that comparison.

Worked example 2: solution

Step 1 · Compute the range Step 2 · Compute the average

Accurate. Nearness to the true value is accuracy's comparison, the one the first answer called precision.

Worked example 2: solution

Step 1 · Compute the range Step 2 · Compute the average

Worked example 2: the route on the map

Practice 1 · answer: A

C swapped the definitions: the 0.06-mL agreement is precision, and the 1.31-mL miss is inaccuracy. B judged the set by one trial: 26.28 − 25.00 = 1.28 mL compares a single reading with the true value, which is accuracy's comparison. D counted digits: significant figures report how finely the pipette is read, and every reading is still about 1.3 mL high.

Practice 1 · answer: A

Practice 2 · answer: D

A treated the 2% bar as 2 grams: 0.18 g is small only next to a large mass, and on a 5-g weight it is 3.6%. B swapped the definitions: relative range judges precision, percent error judges accuracy. C let one verdict answer both: high and low readings cancel in the average.

Worked example 1: solution

The small number moves right (negative n); the large number moves left (positive n).

Worked example 1: solution

The small number moves right (negative n); the large number moves left (positive n).

Worked example 1: the route on the map

Worked example 2: solution

A common first attempt

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 · answer: A

A: a number below 1 always takes a negative exponent, and the decimal moves right to build 4.2. B flips the sign, C confuses zeros with places, D leaves M below 1.

Practice 2 · answer: C

A keyed the −6 as +6: 2.0 × 10⁶ × 4.5 × 10³ = 9.0 × 10⁹. B keyed × 10 before EE, an extra ten: 2.0 × 10 × 10⁻⁶ × 4.5 × 10³ = 9.0 × 10⁻². D read the display back with the sign dropped: E−3 is 10⁻³, not 10³.

Worked example 1: solution

Step 1 · Nonzero digits always count

The 4 and the 3 count.

Worked example 1: solution

The zero between 4 and 3 counts.

Worked example 1: solution

The two zeros in front only locate the decimal point.

Worked example 1: solution

The number has a decimal point, so the final zero counts.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Nonzero digits always count

The 1 and the 2 count.

Worked example 2: solution

No zero sits between nonzero digits, and none leads. Neither rule applies here.

Worked example 2: solution

There is no decimal point. The two zeros are placeholders.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 · answer: B

The 2 and 6 count, the captive zero between them counts, and the decimal point makes the final zero count. A dropped the trailing zero: 3. C started counting at the first zero after the decimal point: 5. D counted every written digit: 6.

Worked example 3: solution

Count each factor's sig figs

8.5 carries two; 4.27 and 1.36 carry three each. The fewest is two.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Worked example 4: solution

A common first attempt

Rounding to three sig figs threw away the tenths digit the cylinder measured. Addition and subtraction do not count sig figs.

Worked example 4: solution

The cylinder is estimated in the tenths, the burette in the hundredths. The sum ends where the least precise input ends: the tenths.

Worked example 4: solution

Worked example 4: the route on the map

Worked example 5: solution

Subtract, and mark the entitlement

Do not round yet: carry 0.362 unrounded into the division.

Worked example 5: solution

Worked example 5: solution

Worked example 5: the route on the map

Practice 2 · answer: C

A rounded the sum to 22.0 before dividing: 22.0 ÷ 150.0 = 0.1467 → 0.147. B used one rule throughout: 1.9 has 2 sig figs, so 0.15. D stopped at the sum: 22.0 g is the salt, not the salt per milliliter. E kept every calculator digit.

Practice 1 · answer: A

B is the working unit of the bench, not the base unit; 1 kg = 1000 g ties every gram reading to the kilogram. C measures volume, the space a sample fills. D measures length.

Practice 2 · answer: D

A has the right unit and an absurd size: 1 kg is a bag of sugar, and a paper clip is about a gram. B misnames the kelvin; it is one of the five base units. C reports a length; a volume takes mL or L.

Worked example 1: solution

One conversion factor is needed.

A common first attempt

No unit cancels, and the answer is not in meters.

Worked example 1: solution

Both factors come from the same equality, and both equal 1. Only one cancels the given unit:

Worked example 1: solution

One factor completes the plan: cm cancels, m survives.

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Two conversion factors are needed.

Step 2 · Pick the factor that cancels its unit

The prefix equality gives the first factor, with µm in the denominator:

Worked example 2: solution

The result is in meters, not the wanted millimeters. The second factor cancels m and leaves mm:

Worked example 2: solution

Worked example 2: the route on the map

Worked example 3: solution

One conversion factor is needed, cubed.

A common first attempt

m³ is m × m × m. One linear factor cancels one m; two survive, and the answer is not in cm³.

Worked example 3: solution

Cancelling all three copies of m takes the whole equality cubed, number and unit:

Worked example 3: solution

One cubed factor completes the plan: m³ cancels, cm³ survives.

Worked example 3: solution

Worked example 3: the route on the map

Practice 1 · answer: C

A cubed the unit without cubing the number, dividing by only 100: 7.5 × 10⁵ / 100 = 7500. B pointed the cubed factor the wrong way, so nothing cancels: 7.5 × 10⁵ × 10⁶ = 7.5 × 10¹¹. D used the linear factor, inverted: 7.5 × 10⁵ × 100 = 7.5 × 10⁷.

Practice 2 · answer: B

Four sig figs survive: 2500. carries four, and the English–metric factor 3.785 counts too, so nothing limits below four. A stopped mid-plan: 2500. mL is 2.500 L, and liters are not gallons. C flipped the gallon factor: 2500./1000 × 3.785 = 9.463. D treated milliliters as liters: 2500./3.785 = 660.5.

Worked example 4: solution

Four conversion factors are needed, one per arrow of the plan.

Step 2 · Pick the factor that cancels its unit

The first arrow removes mL. The prefix equality gives the factor, with mL in the denominator:

Worked example 4: solution

Each factor cancels the unit left by the one before. The chain ends when km survives:

Worked example 4: solution

Worked example 4: the route on the map

Practice 3 · answer: D

The flow rate is a measured equality, 12.0 L = 1 min, and it converts liters to minutes. A flipped the gallon factor: 5.00 / 3.785 / 12.0 × 60 = 6.60. B skipped the gallon hop and treated gallons as liters: 5.00 / 12.0 × 60 = 25.0. C stopped at minutes: 5.00 × 3.785 / 12.0 = 1.58, a count of min, not s.

Worked example 1: solution

Step 2 · Pick the linking equation

Celsius to Kelvin is one shift. No stretching, because a kelvin and a Celsius degree are already the same size.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

A common first attempt

The 32 is already in Fahrenheit-sized degrees. Only the reading gets stretched.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 · answer: D

A dropped the minus sign: 78 + 273.15 = 351.15. C subtracted the shift instead of adding it: −78 − 273.15 = −351.15, and negative kelvins do not exist. B built Fahrenheit instead of Kelvin: 1.8(−78) + 32 = −108.4.

Practice 2 · answer: A

B stopped at Celsius: −50.15, never stretched or shifted into °F. C divided by 1.8, the coming-back move: −50.15 / 1.8 + 32 = 4.14. D dropped the minus sign: 1.8(50.15) + 32 = 122.27. E fed kelvins straight into the Fahrenheit equation: 1.8(223) + 32 = 433.4.

Worked example 1: solution

One conversion factor is needed.

Step 2 · Start with the given

The given is 15.0 mL, so the fraction must cancel milliliters.

Worked example 1: solution

Milliliters go in the denominator: 13.6 g over 1 mL.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

A common first attempt

Nothing cancels, and no quantity carries g²/mL. The fraction is upside down.

Worked example 2: solution

The given unit is grams, so grams belong in the denominator. Flip the fraction: 1 mL over 19.3 g.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 1 · answer: B

A divided the density by the mass: 10.5 ÷ 170. = 0.0618, and its units are not milliliters. C slipped a decimal: 170. ÷ 10.5 = 16.2, not 1.62. D flipped the fraction: 170. × 10.5 = 1.79 × 10³, with units of g²/mL.

Worked example 3: solution

Build the fraction from the data

Mass on top, volume underneath, units in place. The surviving unit, g/mL, is a density.

Worked example 3: solution

Of the candidates, only iron matches 7.87 g/mL. The pellets are iron.

Worked example 3: solution

Worked example 3: the route on the map

Worked example 4: solution

Three conversion factors are needed.

Step 2 · Start with the given Step 3 · Write the density fraction

The metric factor converts kilograms to grams. The density fraction, written 1 mL over 0.789 g, then cancels grams.

Worked example 4: solution

Worked example 4: solution

Worked example 4: the route on the map

Practice 2 · answer: D

Practice 2 · answer: D