10/27 Practice Problems - Answer Key

Problem 1

A smartphone screen is 6.2 inches long and 2.9 inches wide. What is the area of the screen in square centimeters? (1 inch = 2.54 cm)

Solution:

Step 1: Calculate area in square inches

\[\text{Area} = \text{length} \times \text{width} = 6.2 \text{ in} \times 2.9 \text{ in} = 17.98 \text{ in}^2\]

Step 2: Convert square inches to square centimeters

Remember: When converting area, you must square the conversion factor!

\[17.98 \cancel{\text{ in}^2} \times \left(\frac{2.54 \text{ cm}}{1 \cancel{\text{ in}}}\right)^2 = 17.98 \times 6.4516 \text{ cm}^2 = 116.0 \text{ cm}^2\]

Problem 2

A pharmaceutical company produces pills that contain 325 mg of acetaminophen each. If a bottle contains 250 pills, what is the total mass of acetaminophen in the bottle in ounces? (1 oz = 28.35 g)

Solution:

Step 1: Find total mass in milligrams

\[325 \text{ mg/pill} \times 250 \text{ pills} = 81,250 \text{ mg}\]

Step 2: Convert mg → g

\[81,250 \cancel{\text{ mg}} \times \frac{1 \text{ g}}{1000 \cancel{\text{ mg}}} = 81.25 \text{ g}\]

Step 3: Convert g → oz

\[81.25 \cancel{\text{ g}} \times \frac{1 \text{ oz}}{28.35 \cancel{\text{ g}}} = 2.866 \text{ oz}\]

Problem 3

How many moles of water (H₂O) are in 54.0 grams of water?

Solution:

Step 1: Calculate molar mass of H₂O

H: 2 × 1.01 g/mol = 2.02 g/mol
O: 1 × 16.00 g/mol = 16.00 g/mol
Total: 18.02 g/mol

Step 2: Convert grams to moles

\[54.0 \cancel{\text{ g H}_2\text{O}} \times \frac{1 \text{ mol H}_2\text{O}}{18.02 \cancel{\text{ g H}_2\text{O}}} = 2.996 \text{ mol}\]

Problem 4

A sample of sodium chloride (NaCl) contains \(3.01 \times 10^{23}\) formula units. How many moles of NaCl is this?

Solution:

Step 1: Use Avogadro's number to convert formula units to moles

\[3.01 \times 10^{23} \cancel{\text{ formula units}} \times \frac{1 \text{ mol}}{6.022 \times 10^{23} \cancel{\text{ formula units}}} = 0.500 \text{ mol}\]

Problem 5

How many atoms of carbon are present in 11.0 grams of propane, C₃H₈?

Solution:

Step 1: Calculate molar mass of C₃H₈

C: 3 × 12.01 g/mol = 36.03 g/mol
H: 8 × 1.01 g/mol = 8.08 g/mol
Total: 44.11 g/mol

Step 2: Convert grams of C₃H₈ to moles of C₃H₈

\[11.0 \cancel{\text{ g C}_3\text{H}_8} \times \frac{1 \text{ mol C}_3\text{H}_8}{44.11 \cancel{\text{ g C}_3\text{H}_8}} = 0.2493 \text{ mol C}_3\text{H}_8\]

Step 3: Convert moles of C₃H₈ to moles of C atoms

Each molecule of C₃H₈ contains 3 carbon atoms, so:

\[0.2493 \cancel{\text{ mol C}_3\text{H}_8} \times \frac{3 \text{ mol C}}{1 \cancel{\text{ mol C}_3\text{H}_8}} = 0.7479 \text{ mol C}\]

Step 4: Convert moles of C to atoms of C

\[0.7479 \cancel{\text{ mol C}} \times \frac{6.022 \times 10^{23} \text{ atoms}}{1 \cancel{\text{ mol}}} = 4.50 \times 10^{23} \text{ C atoms}\]

Problem 6

A brass key has a mass of 12.5 grams and is 65% copper. How many grams of copper are in the key?

Solution:

Step 1: Convert percentage to decimal

\[65\% = \frac{65}{100} = 0.65\]

Step 2: Calculate mass of copper

\[\text{Mass of Cu} = 0.65 \times 12.5 \text{ g} = 8.125 \text{ g}\]

Problem 7

The air in a diving tank contains 21% oxygen by volume. If a diver uses 45 liters of air from the tank during a dive, how many liters of oxygen did the diver breathe?

Solution:

Step 1: Convert percentage to decimal

\[21\% = \frac{21}{100} = 0.21\]

Step 2: Calculate volume of oxygen

\[\text{Volume of O}_2 = 0.21 \times 45 \text{ L} = 9.45 \text{ L}\]

Problem 8

The human body is approximately 0.25% potassium by mass. If a person weighs 68 kg, how many grams of potassium does their body contain? Then, calculate how many moles of potassium atoms this represents. (Molar mass of K = 39.1 g/mol)

Solution:

Step 1: Convert percentage to decimal

\[0.25\% = \frac{0.25}{100} = 0.0025\]

Step 2: Calculate mass of potassium in kg

\[\text{Mass of K} = 0.0025 \times 68 \text{ kg} = 0.17 \text{ kg}\]

Step 3: Convert to grams

\[0.17 \text{ kg} \times \frac{1000 \text{ g}}{1 \text{ kg}} = 170 \text{ g K}\]

Step 4: Convert grams to moles

\[170 \cancel{\text{ g K}} \times \frac{1 \text{ mol K}}{39.1 \cancel{\text{ g K}}} = 4.35 \text{ mol K}\]

Problem 9

How much heat energy is required to raise the temperature of 250 g of water from 20°C to 85°C? (\(c_{\text{water}} = 4.18\) J/g°C)

Solution:

Step 1: Identify given values

m = 250 g
c = 4.18 J/g°C
ΔT = 85°C - 20°C = 65°C

Step 2: Use the heat transfer equation

\[q = mc\Delta T\]

\[q = (250 \text{ g})(4.18 \text{ J/g°C})(65°C)\]

\[q = 67,925 \text{ J}\]

Step 3: Convert to kilojoules (optional but cleaner)

\[67,925 \text{ J} \times \frac{1 \text{ kJ}}{1000 \text{ J}} = 67.925 \text{ kJ}\]

Problem 10

A 75.0 g piece of copper at 95°C is placed in 200 g of water at 22°C. What will be the final temperature when thermal equilibrium is reached? (\(c_{\text{copper}} = 0.385\) J/g°C, \(c_{\text{water}} = 4.18\) J/g°C)

Solution:

Step 1: Set up the thermal equilibrium equation

Heat lost by copper = Heat gained by water

\[q_{\text{copper}} = -q_{\text{water}}\]

\[m_{\text{Cu}}c_{\text{Cu}}\Delta T_{\text{Cu}} = -m_{\text{H}_2\text{O}}c_{\text{H}_2\text{O}}\Delta T_{\text{H}_2\text{O}}\]

Step 2: Express ΔT in terms of final temperature \(T_f\)

For copper: \(\Delta T_{\text{Cu}} = T_f - 95\) (negative, cooling down)
For water: \(\Delta T_{\text{H}_2\text{O}} = T_f - 22\) (positive, warming up)

Step 3: Substitute values

\[(75.0)(0.385)(T_f - 95) = -(200)(4.18)(T_f - 22)\]

\[28.875(T_f - 95) = -836(T_f - 22)\]

Step 4: Expand and solve for \(T_f\)

\[28.875T_f - 2743.125 = -836T_f + 18,392\]

\[28.875T_f + 836T_f = 18,392 + 2743.125\]

\[864.875T_f = 21,135.125\]

\[T_f = \frac{21,135.125}{864.875} = 24.43°C\]

Check: The final temperature is between the initial temperatures (22°C and 95°C), and much closer to the water temperature because water has much higher heat capacity. ✓