✨ Morning Practice - Answer Key

Problem 1

Round 3458.23 to 2 significant figures.

Solution:

Step 1: Identify the first 2 significant figures

The first two significant figures are 3 and 4.

These represent: 3458.23 (the 3 is in the thousands place, the 4 is in the hundreds place)

Step 2: Look at the next digit to decide on rounding

The next digit is 5, which means we round UP.

Step 3: Round and replace remaining digits with zeros

34 becomes 35 (rounded up)

Since the 4 was in the hundreds place, we need: 3500

Step 4: Express in scientific notation to show exactly 2 sig figs

\[3.5 \times 10^3\]

Key point: When rounding large numbers to fewer sig figs, scientific notation helps show exactly how many sig figs you have!

Problem 2

Convert 2.5 kg to grams.

Solution:

Step 1: Use the conversion factor

1 kg = 1000 g

Step 2: Set up the conversion

\[2.5 \cancel{\text{ kg}} \times \frac{1000 \text{ g}}{1 \cancel{\text{ kg}}} = 2500 \text{ g}\]

Problem 3

Convert 45,000 cm to km.

Solution:

Step 1: Plan the conversion path

cm → m → km

Step 2: Convert cm to m

100 cm = 1 m

\[45,000 \cancel{\text{ cm}} \times \frac{1 \text{ m}}{100 \cancel{\text{ cm}}} = 450 \text{ m}\]

Step 3: Convert m to km

1000 m = 1 km

\[450 \cancel{\text{ m}} \times \frac{1 \text{ km}}{1000 \cancel{\text{ m}}} = 0.45 \text{ km}\]

Alternative: Do it in one step!

1 km = 100,000 cm (or 10⁵ cm)

\[45,000 \cancel{\text{ cm}} \times \frac{1 \text{ km}}{100,000 \cancel{\text{ cm}}} = 0.45 \text{ km}\]

Problem 4

A car travels 120 miles in 2.0 hours. Convert this speed to kilometers per minute. (1 mile = 1.609 km)

Solution:

Step 1: Calculate the speed in miles per hour

\[\text{Speed} = \frac{120 \text{ miles}}{2.0 \text{ hours}} = 60 \text{ miles/hour}\]

Step 2: Convert miles to kilometers

\[60 \frac{\cancel{\text{miles}}}{\text{hour}} \times \frac{1.609 \text{ km}}{1 \cancel{\text{ mile}}} = 96.54 \text{ km/hour}\]

Step 3: Convert hours to minutes

1 hour = 60 minutes

\[96.54 \frac{\text{km}}{\cancel{\text{hour}}} \times \frac{1 \cancel{\text{ hour}}}{60 \text{ min}} = 1.609 \text{ km/min}\]

Step 4: Apply significant figures

2.0 hours has 2 sig figs, so: 1.6 km/min

Key point: When converting rates/speeds, you must convert BOTH the numerator (distance) AND denominator (time)!

Problem 5

A fruit smoothie contains 15% strawberries by volume. If you make a 500 mL smoothie, how many mL of strawberries are in it?

Solution:

Step 1: Convert percentage to decimal

15% = 0.15

Step 2: Multiply by the whole amount

Part = Decimal × Whole

Volume of strawberries = 0.15 × 500 mL = 75 mL

Quick check: Does 75 mL seem reasonable for 15% of 500 mL? Yes! 10% would be 50 mL, so 15% should be a bit more.

Problem 6

A brass doorknob has a mass of 280 g and is 70% copper. How many ounces of copper does the doorknob contain? (1 oz = 28.35 g)

Solution:

Step 1: Convert percentage to decimal

70% = 0.70

Step 2: Calculate mass of copper in grams

Mass of copper = 0.70 × 280 g = 196 g

Step 3: Convert grams to ounces

\[196 \cancel{\text{ g}} \times \frac{1 \text{ oz}}{28.35 \cancel{\text{ g}}} = 6.914 \text{ oz}\]

Step 4: Apply significant figures

280 has 2 sig figs (limiting), so: 6.9 oz

Key concepts: This problem combines percentage calculation (finding part of a whole) with unit conversion (g → oz)!

Problem 7

Calculate the molar mass of sodium hydroxide (NaOH). (Na = 22.99 g/mol, O = 16.00 g/mol, H = 1.01 g/mol)

Solution:

Step 1: Identify atoms in the formula

NaOH contains: 1 Na, 1 O, 1 H

Step 2: Calculate contribution from each element

Na: 1 × 22.99 g/mol = 22.99 g/mol
O: 1 × 16.00 g/mol = 16.00 g/mol
H: 1 × 1.01 g/mol = 1.01 g/mol

Step 3: Add them together

Molar mass = 22.99 + 16.00 + 1.01 = 40.00 g/mol

Problem 8

Calculate the molar mass of table sugar, sucrose (C₁₂H₂₂O₁₁). (C = 12.01 g/mol, H = 1.01 g/mol, O = 16.00 g/mol)

Solution:

Step 1: Identify atoms in the formula

C₁₂H₂₂O₁₁ contains: 12 C, 22 H, 11 O

Step 2: Calculate contribution from each element

C: 12 × 12.01 g/mol = 144.12 g/mol
H: 22 × 1.01 g/mol = 22.22 g/mol
O: 11 × 16.00 g/mol = 176.00 g/mol

Step 3: Add them together

Molar mass = 144.12 + 22.22 + 176.00 = 342.34 g/mol

Fun fact: This is why sugar is so energy-dense - look at all those carbon and hydrogen atoms that can be burned for energy!

Problem 9

A 50.0 g piece of aluminum is heated from 25.0°C to 75.0°C. How much heat energy was absorbed? (Specific heat of aluminum = 0.900 J/g°C)

Solution:

Step 1: Write the heat equation

\[q = mc\Delta T\]

Step 2: Identify the given values

m = 50.0 g
c = 0.900 J/g°C
ΔT = 75.0°C - 25.0°C = 50.0°C

Step 3: Substitute and calculate

\[q = (50.0 \text{ g})(0.900 \text{ J/g°C})(50.0°C)\]

\[q = 2250 \text{ J}\]

Step 4: Convert to kJ (optional but cleaner)

\[2250 \text{ J} \times \frac{1 \text{ kJ}}{1000 \text{ J}} = 2.25 \text{ kJ}\]

Problem 10

A student has 25.0 grams of table salt (NaCl) and needs to calculate how many moles this represents. The molar mass of NaCl is 58.44 g/mol. How many moles of NaCl does the student have?

Solution:

Step 1: Write the conversion relationship

\[\text{moles} = \frac{\text{grams}}{\text{molar mass}}\]

Step 2: Set up the calculation with units

\[25.0 \cancel{\text{ g NaCl}} \times \frac{1 \text{ mol NaCl}}{58.44 \cancel{\text{ g NaCl}}}\]

Step 3: Calculate

\[\frac{25.0}{58.44} = 0.4277 \text{ mol}\]

Step 4: Apply significant figures

25.0 has 3 sig figs, so: 0.428 mol

Real-world connection: This is about how much salt you might use to season a large pot of pasta water!

Problem 11

Multi-part metal identification problem using thermal equilibrium and density analysis.

Solution:

Part (a): Calculate experimental specific heat of the metal

Step 1: Calculate heat gained by water

Use: \(q = mc\Delta T\)

For water:

• m = 100.0 g
• c = 4.18 J/g°C
• ΔT = 28.5°C - 22.0°C = 6.5°C

\[q_{\text{water}} = (100.0 \text{ g})(4.18 \text{ J/g°C})(6.5°C) = 2717 \text{ J}\]

Step 2: Calculate specific heat of metal

Heat lost by metal = Heat gained by water

For metal: ΔT = 28.5°C - 95.0°C = -66.5°C

Using \(q = mc\Delta T\), solve for c:

\[c_{\text{metal}} = \frac{q}{m \Delta T} = \frac{-2717 \text{ J}}{(45.0 \text{ g})(-66.5°C)} = 0.908 \text{ J/g°C}\]

With sig figs: 0.91 J/g°C

Part (b): Calculate experimental density of the metal

Step 1: Find volume using water displacement

Volume of metal = Final reading - Initial reading

Volume = 56.7 mL - 50.0 mL = 6.7 mL = 6.7 cm³

Step 2: Calculate density

\[\text{Density} = \frac{\text{mass}}{\text{volume}} = \frac{45.0 \text{ g}}{6.7 \text{ cm}^3} = 6.716 \text{ g/cm}^3\]

With sig figs (2 from 6.7): 6.7 g/cm³

Part (c): Identify the metal

Compare experimental values to reference table:

Metal	Specific Heat	Our Value	Density	Our Value
Aluminum	0.900	0.91 ✓✓	2.70	6.7 ✗
Copper	0.385	0.91 ✗	8.96	6.7 ~
Iron	0.85	0.91 ✓	7.87	6.7 ✓✓

Analysis:

• Our density (6.7 g/cm³) is closest to iron (7.87 g/cm³) - only 15% error
• Our specific heat (0.91 J/g°C) is very close to both aluminum (0.900) and iron (0.85)
• Aluminum's specific heat matches slightly better, BUT its density (2.70) is way off from our value
• Remember: Density is more reliable than specific heat for metal identification
• Based on density being the most reliable measurement, the metal is iron

Part (d): Calculate percent error

Percent error formula:

\[\text{Percent error} = \left| \frac{\text{experimental} - \text{accepted}}{\text{accepted}} \right| \times 100\%\]

Percent error for specific heat (using iron as accepted):

\[\text{Percent error} = \left| \frac{0.91 - 0.85}{0.85} \right| \times 100\% = \left| \frac{0.06}{0.85} \right| \times 100\% = 7.1\%\]

Percent error for density (using iron as accepted):

\[\text{Percent error} = \left| \frac{6.7 - 7.87}{7.87} \right| \times 100\% = \left| \frac{-1.17}{7.87} \right| \times 100\% = 14.9\% \approx 15\%\]

Understanding the results:

Both measurements have reasonable error for a student lab experiment:

• Specific heat error (7.1%): Excellent! This is quite accurate for a simple calorimeter.
• Density error (15%): Good! Could be due to air bubbles during displacement or meniscus reading error

Why we trust density more: Even though our specific heat is more accurate (7.1% vs 15% error), density is a more reliable identifier because:

• The specific heat of aluminum (0.900) is almost identical to our value (0.91), BUT aluminum's density (2.70) doesn't match at all
• Iron's density (7.87) is much closer to our value (6.7) than aluminum's or copper's
• In general, density measurements are less affected by experimental technique than heat measurements

Real-world application: This problem shows how scientists must weigh competing evidence. When one property (specific heat) points toward aluminum but another property (density) strongly points toward iron, we choose based on which measurement is more reliable and which has a clearer match overall.