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 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

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

Practice 2 — answer: C

A: no fixed ratio exists — every grade holds a different amount of carbon. B: same properties throughout shows one phase; the composition can still vary. D: two substances can share one phase; the carbon atoms sit dissolved among the iron atoms.

Worked example 1 — solution

Step 1 · Compute the range

Worked example 1 — solution

A spread of 0.06 g on a 50-g measurement: the trials agree. Precise.

Worked example 1 — solution

Worked example 1 — solution

Worked example 1 — solution

Worked example 2 — solution

Step 1 · Compute the range

Worked example 2 — solution

The trials disagree by up to 4.2 °C. Not precise. Precision compares the trials with one another; the true value is not part of that comparison.

Worked example 2 — solution

Worked example 2 — solution

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

Worked example 2 — solution

Practice 1 — answer: C

A 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: C

Worked example 3 — solution

Step 1 · Compute the range

Worked example 3 — solution

The trials spread across 0.86 g on a 10-g mass. Not precise. No single trial, however close to 10.00 g, can change a verdict about the spread.

Worked example 3 — solution

Worked example 3 — solution

The average equals the true mass. Accurate because the whole set averages onto 10.00 g, not because trial 4 landed close.

Worked example 3 — solution

Practice 2 — answer: B

A judged the set by one trial: one reading inside a 1.1-cm scatter, and precision comes from the whole set's range. C swapped the definitions: the on-target average is accuracy; the 1.1-cm spread is imprecision. D counted digits: the decimal place shows how finely the tape reads, not whether the trials repeat.

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 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

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 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

Practice 2 — answer: B

The 5.8 mL is estimated in the tenths, so the sum ends at the tenths. A kept every calculator digit: 42.783. C applied the multiplication rule, two sig figs from 5.8: 43. D rounded to the hundredths, a place the graduated cylinder never measured: 42.78.

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 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

Practice 1 — answer: B

A stopped mid-plan: 2500 mL is 2.50 L, and liters are not gallons. C flipped the gallon factor: 2500/1000 × 3.7854 = 9.46. D treated milliliters as liters: 2500/3.7854 = 660.

Worked example 3 — 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 3 — solution

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

Worked example 3 — solution

Practice 2 — answer: C

A stopped at miles: 2000/1000/3.7854 × 44 = 23.2, a count of mi, not km. B treated milliliters as liters: 1000 times too large, 3.74 × 10⁴. D flipped the gallon factor: 2000/1000 × 3.7854 × 44 / 0.62137 = 536.

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 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

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 = 1790, 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 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

Practice 2 — answer: C

A flipped the density: 1250 ÷ 13.6 ÷ 1000 = 0.0919. B slipped a decimal: 17,000 g is 17.0 kg, not 1.70. D applied g/mL to liters without converting: 1.25 × 13.6 = 17.0 g, which is 0.0170 kg. The volume must become milliliters before the density applies.