GOB Chemistry Practice Worksheet Guidance

Q1. How many significant figures are in each number?

Background

Topic: Significant Figures

This question tests your understanding of how to count significant figures in various types of numbers, including decimals, whole numbers, and numbers with trailing zeros.

Key Terms:

• Significant figures: Digits in a number that are meaningful in terms of accuracy.

• Rules: Leading zeros are not significant; trailing zeros after a decimal are significant; all nonzero digits are significant.

Step-by-Step Guidance

1. Examine each number and identify if there are leading zeros, trailing zeros, or decimal points.

2. Apply the rules: Leading zeros (before the first nonzero digit) are not counted; zeros between nonzero digits are always significant; trailing zeros in a decimal number are significant.

3. For numbers without a decimal, consider if trailing zeros are ambiguous or not significant.

4. Count the significant digits for each number, but stop before stating the final count for each.

Try solving on your own before revealing the answer!

Final Answer:

• 0.04500: 4 significant figures

• 0.00245: 3 significant figures

• 65: 2 significant figures

• 166000: 3 significant figures (unless specified otherwise)

• 1.00001: 6 significant figures

• 1.00000: 6 significant figures

• 3.00×1025: 3 significant figures

• 3.0×1025: 2 significant figures

Each answer follows the rules for counting significant figures based on decimal placement and zeros.

Q2. Write the following numbers in scientific notation:

Background

Topic: Scientific Notation

This question tests your ability to convert standard numbers into scientific notation, paying attention to significant figures.

Key Terms and Formula:

• Scientific notation: A way to express numbers as $a \times 10^n$ where $1 \leq a < 10$.

• Significant figures: Only the specified number of digits should be shown.

Step-by-Step Guidance

1. Identify the number and the required significant figures.

2. Move the decimal point so that only one nonzero digit remains to the left of the decimal.

3. Count how many places the decimal was moved; this becomes the exponent for 10.

4. Round or truncate the number to the correct number of significant figures, but do not write the final scientific notation yet.

Try solving on your own before revealing the answer!

Final Answer:

• 10000000: $1.0 \times 10^7$

• 23456789 (3 sig figs): $2.35 \times 10^7$

• 0.0000001 (1 sig fig): $1 \times 10^{-7}$

• 0.006254 (3 sig figs): $6.25 \times 10^{-3}$

Each answer is rounded to the correct number of significant figures and written in scientific notation.

Q3. Calculate: $\frac{3.66 \times 0.2547}{1.2}$

Background

Topic: Multiplication and Division with Significant Figures

This question tests your ability to perform calculations and report the answer with the correct number of significant figures.

Key Terms and Formula:

• Multiplication/division: The answer should have the same number of significant figures as the value with the fewest significant figures.

• Perform the calculation: $\frac{3.66 \times 0.2547}{1.2}$

Step-by-Step Guidance

1. Identify the number of significant figures in each value: 3.66 (3), 0.2547 (4), 1.2 (2).

2. Multiply 3.66 and 0.2547.

3. Divide the result by 1.2.

4. Round the final answer to the correct number of significant figures (the lowest, which is 2), but do not state the final value.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{3.66 \times 0.2547}{1.2} = 0.78$ (rounded to 2 significant figures)

The answer is rounded to match the value with the fewest significant figures.

Q4. Convert 180.1 m to kilometers, millimeters, and nanometers.

Background

Topic: Unit Conversion (Metric System)

This question tests your ability to convert between metric units using conversion factors.

Key Terms and Conversion Factors:

• 1 km = 1000 m

• 1 m = 1000 mm

• 1 m = $1 \times 10^9$ nm

Step-by-Step Guidance

1. Set up the conversion for kilometers: $180.1 \text{ m} \times \frac{1 \text{ km}}{1000 \text{ m}}$

2. Set up the conversion for millimeters: $180.1 \text{ m} \times \frac{1000 \text{ mm}}{1 \text{ m}}$

3. Set up the conversion for nanometers: $180.1 \text{ m} \times \frac{1 \times 10^9 \text{ nm}}{1 \text{ m}}$

4. Calculate each value, but stop before stating the final converted numbers.

Try solving on your own before revealing the answer!

Final Answer:

• 180.1 m = 0.1801 km

• 180.1 m = 180,100 mm

• 180.1 m = $1.801 \times 10^{11}$ nm

Each conversion uses the correct metric factor.

Q5. A dose of aspirin of 250 µg per kilogram of body weight has been prescribed to reduce the fever of an infant weighing 4.2 kg. The number of milligrams of aspirin that should be administered is ____________.

Background

Topic: Dosage Calculation and Unit Conversion

This question tests your ability to calculate a total dose based on body weight and convert micrograms to milligrams.

Key Terms and Conversion Factors:

• 1 mg = 1000 µg

• Total dose = dose per kg × weight in kg

Step-by-Step Guidance

1. Multiply the prescribed dose (250 µg/kg) by the infant's weight (4.2 kg) to get the total dose in µg.

2. Convert the total dose from µg to mg using the conversion factor.

3. Set up the calculation, but do not state the final value.

Try solving on your own before revealing the answer!

Final Answer:

Total dose: $250 \text{ µg/kg} \times 4.2 \text{ kg} = 1050 \text{ µg}$

Convert to mg: $\frac{1050 \text{ µg}}{1000} = 1.05 \text{ mg}$

The infant should be given 1.05 mg of aspirin.

Q6. A patient needs to receive 0.125 g of ampicillin. The ampicillin solution in vial contains 200 mg/5.0 mL. How many milliliters of the solution are required?

Background

Topic: Dosage Calculation and Unit Conversion

This question tests your ability to convert grams to milligrams and use concentration to find the required volume.

Key Terms and Formula:

• 1 g = 1000 mg

• Concentration: $\frac{200 \text{ mg}}{5.0 \text{ mL}}$

• Volume required: $\text{Volume} = \frac{\text{Dose}}{\text{Concentration}}$

Step-by-Step Guidance

1. Convert 0.125 g to mg: $0.125 \text{ g} \times 1000 = 125 \text{ mg}$

2. Set up the proportion: $\frac{200 \text{ mg}}{5.0 \text{ mL}}$

3. Calculate the volume needed to deliver 125 mg, but do not state the final value.

Try solving on your own before revealing the answer!

Final Answer:

$\text{Volume} = \frac{125 \text{ mg}}{\frac{200 \text{ mg}}{5.0 \text{ mL}}} = 3.1 \text{ mL}$

The patient needs 3.1 mL of the solution.

Q7. A gold crown was put in a water tank and the volume of water increased from 50.0 mL to 85.0 mL. The mass of the crown is 555 g. Is this crown made by pure gold? (density of gold = 19.3 g/mL)

Background

Topic: Density and Identification of Substances

This question tests your ability to calculate density and compare it to a known value to determine purity.

Key Terms and Formula:

• Density: $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$

• Volume displaced: $85.0 \text{ mL} - 50.0 \text{ mL} = 35.0 \text{ mL}$

Step-by-Step Guidance

1. Calculate the volume of the crown by subtracting the initial from the final water volume.

2. Calculate the density using the mass and the volume you just found.

3. Compare the calculated density to the density of pure gold (19.3 g/mL), but do not state the conclusion.

Try solving on your own before revealing the answer!

Final Answer:

Volume of crown: $35.0 \text{ mL}$

Density: $\frac{555 \text{ g}}{35.0 \text{ mL}} = 15.9 \text{ g/mL}$

Since 15.9 g/mL is less than 19.3 g/mL, the crown is not made of pure gold.

Q8. Mercury has a specific gravity of 13.6. How many milliliters of mercury have a mass of 0.65 kg?

Background

Topic: Density and Specific Gravity

This question tests your ability to use specific gravity to find volume from mass.

Key Terms and Formula:

• Specific gravity = density of substance / density of water

• Density of water = 1.00 g/mL

• Density of mercury = 13.6 g/mL

• Volume = mass / density

Step-by-Step Guidance

1. Convert mass from kg to g: $0.65 \text{ kg} \times 1000 = 650 \text{ g}$

2. Use the density (13.6 g/mL) to set up the calculation for volume: $\text{Volume} = \frac{650 \text{ g}}{13.6 \text{ g/mL}}$

3. Calculate the volume, but do not state the final value.

Try solving on your own before revealing the answer!

Final Answer:

$\text{Volume} = \frac{650 \text{ g}}{13.6 \text{ g/mL}} = 47.8 \text{ mL}$

47.8 mL of mercury has a mass of 0.65 kg.

Q9. Classify each substance:

Background

Topic: Classification of Matter

This question tests your ability to distinguish between elements, compounds, homogeneous mixtures, and heterogeneous mixtures.

Key Terms:

• Element: Pure substance made of one type of atom.

• Compound: Pure substance made of two or more elements chemically combined.

• Homogeneous mixture: Uniform composition throughout.

• Heterogeneous mixture: Non-uniform composition.

Step-by-Step Guidance

1. Consider the nature of each substance: hydrogen, carbon dioxide, coke, blueberry muffin.

2. Determine if each is an element, compound, homogeneous mixture, or heterogeneous mixture based on its composition.

3. Match each to the correct classification, but do not state the final answers.

Try solving on your own before revealing the answer!

Final Answer:

• Hydrogen: C) element

• Carbon dioxide: A) compound

• Coke: D) homogeneous mixture

• Blueberry muffin: B) heterogeneous mixture

Each classification is based on the definition of matter types.

Q10. Which of the following is a physical property of both liquids and gases?

Background

Topic: Properties of States of Matter

This question tests your understanding of physical properties shared by liquids and gases.

Key Terms:

• Physical property: Characteristic that can be observed without changing the substance.

• States of matter: Solid, liquid, gas.

Step-by-Step Guidance

1. Review the properties listed: definite shape, definite volume, strong interactions, random arrangement, large spaces.

2. Recall which properties are shared by liquids and gases.

3. Identify the property that applies to both, but do not state the final answer.

Try solving on your own before revealing the answer!

Final Answer:

D) has randomly arranged particles

Both liquids and gases have randomly arranged particles, unlike solids.

Q11. Which of the following would NOT be a physical change?

Background

Topic: Physical vs Chemical Changes

This question tests your ability to distinguish between physical and chemical changes.

Key Terms:

• Physical change: Change in state or appearance without changing composition.

• Chemical change: Change that produces a new substance.

Step-by-Step Guidance

1. Review each option and determine if the process changes the substance's chemical identity.

2. Identify which option involves a chemical reaction, but do not state the final answer.

Try solving on your own before revealing the answer!

Final Answer:

D) burning gasoline in a lawnmower

Burning gasoline is a chemical change because it produces new substances.

Q12. Convert temperatures: 104°F to °C, 196°C to K, 298 K to °C

Background

Topic: Temperature Conversion

This question tests your ability to convert between Fahrenheit, Celsius, and Kelvin.

Key Formulas:

• $\text{°C} = \frac{5}{9}(\text{°F} - 32)$

• $\text{K} = \text{°C} + 273$

• $\text{°C} = \text{K} - 273$

Step-by-Step Guidance

1. For 104°F to °C: Plug 104 into the formula $\text{°C} = \frac{5}{9}(104 - 32)$

2. For 196°C to K: Add 273 to 196.

3. For 298 K to °C: Subtract 273 from 298.

4. Calculate each value, but do not state the final answers.

Try solving on your own before revealing the answer!

Final Answer:

• 104°F = 40°C

• 196°C = 469 K

• 298 K = 25°C

Each conversion uses the correct formula.

Q13. 650 J is the same amount of energy as:

Background

Topic: Energy Units Conversion

This question tests your ability to convert between joules and calories.

Key Terms and Formula:

• 1 cal = 4.184 J

• To convert J to cal: $\text{cal} = \frac{\text{J}}{4.184}$

Step-by-Step Guidance

1. Set up the conversion: $\text{cal} = \frac{650}{4.184}$

2. Calculate the value and match it to the closest answer choice, but do not state the final answer.

Try solving on your own before revealing the answer!

Final Answer:

650 J = 155 cal (A)

650 divided by 4.184 gives approximately 155 calories.

Q14. How many calories are required to raise the temperature of a 25.0 g sample of iron from 35°C to 55°C? Iron has a specific heat of 0.108 cal/g°C.

Background

Topic: Heat Calculation (Specific Heat)

This question tests your ability to use the specific heat formula to calculate heat required for a temperature change.

Key Formula:

• $q = m \times C \times \Delta T$

• Where $q$ = heat (cal), $m$ = mass (g), $C$ = specific heat (cal/g°C), $\Delta T$ = change in temperature (°C)

Step-by-Step Guidance

1. Calculate $\Delta T$: $55°C - 35°C = 20°C$

2. Plug values into the formula: $q = 25.0 \text{ g} \times 0.108 \text{ cal/g°C} \times 20°C$

3. Multiply the values, but do not state the final answer.

Try solving on your own before revealing the answer!

Final Answer:

$q = 25.0 \times 0.108 \times 20 = 54.0 \text{ cal}$

54.0 calories are required to raise the temperature as specified.

Q15. When 20.0 g, 100°C H2O(g) is converted into 0°C H2O(s), how many kJ of heat will be released? (For water, S.H. = 4.184 J/g°C, heat of vaporization = 2260 J/g, heat of fusion = 334 J/g)

Background

Topic: Phase Changes and Heat Calculations

This question tests your ability to calculate total heat released during condensation, cooling, and freezing.

Key Formulas:

• Condensation: $q_1 = m \times \text{heat of vaporization}$

• Cooling liquid: $q_2 = m \times C \times \Delta T$

• Freezing: $q_3 = m \times \text{heat of fusion}$

Step-by-Step Guidance

1. Calculate heat released during condensation: $q_1 = 20.0 \text{ g} \times 2260 \text{ J/g}$

2. Calculate heat released during cooling from 100°C to 0°C: $q_2 = 20.0 \text{ g} \times 4.184 \text{ J/g°C} \times 100°C$

3. Calculate heat released during freezing: $q_3 = 20.0 \text{ g} \times 334 \text{ J/g}$

4. Add all three values and convert to kJ, but do not state the final answer.

Try solving on your own before revealing the answer!

Final Answer:

Condensation: $20.0 \times 2260 = 45,200 \text{ J}$

Cooling: $20.0 \times 4.184 \times 100 = 8,368 \text{ J}$

Freezing: $20.0 \times 334 = 6,680 \text{ J}$

Total: $45,200 + 8,368 + 6,680 = 60,248 \text{ J} = 60.2 \text{ kJ}$

60.2 kJ of heat will be released.