Earth Science Quiz: Scientific Notation And Units

What this quiz covers

This quiz focuses on Scientific Notation And Units, giving you a quick way to practice the rules, question types, and explanations that matter most for Earth Science.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. 0.041 cm
2. 0.41 cm (correct answer)
3. 2.41 cm
4. 4.14 cm

Explanation: The correct answer is 0.41 cm. First, calculate the total mass of ice lost over a decade: 150 Gt/year * 10 years = 1,500 Gt. Second, calculate the total sea-level rise in millimeters using the given conversion factor: Rise (mm) = Total Mass Lost / (Mass per mm rise) = 1,500 Gt / (362 Gt/mm) ≈ 4.14 mm. Finally, convert the rise from millimeters to centimeters: 4.14 mm / (10 mm/cm) = 0.414 cm.

Question 2

1. 5.0 x 10⁻³ g/cm³
2. 5.0 x 10⁻¹ g/cm³
3. 5.0 x 10¹ g/cm³
4. 5.0 x 10⁰ g/cm³ (correct answer)

Explanation: When you encounter mineral density problems, you're working with the fundamental relationship: density = mass ÷ volume. The key challenge is usually unit conversion, which is exactly what this question tests. To find the density, start by calculating the volume of the cubic crystal. Since each side is 2000 μm, the volume is $2000^3 = 8.0 \times 10^9 \text{ μm}^3$. Now convert this to cm³: since 1 cm = 10,000 μm (or $10^4$ μm), then 1 cm³ = $(10^4)^3 = 10^{12}$ μm³. Therefore: $8.0 \times 10^9 \text{ μm}^3 \times \frac{1 \text{ cm}^3}{10^{12} \text{ μm}^3} = 8.0 \times 10^{-3} \text{ cm}^3$ With mass = $4.0 \times 10^{-2}$ g and volume = $8.0 \times 10^{-3}$ cm³, the density is: $\frac{4.0 \times 10^{-2}}{8.0 \times 10^{-3}} = 5.0 \times 10^0 = 5.0 \text{ g/cm}^3$ Choice A ($5.0 \times 10^{-3}$) results from incorrectly converting micrometers to centimeters by dividing by 1000 instead of 10,000. Choice B ($5.0 \times 10^{-1}$) comes from using the linear dimension (2000 μm) as the volume instead of cubing it. Choice C ($5.0 \times 10^1$) occurs if you forget to convert units entirely and mix grams with μm³. Remember: always convert all units before calculating density, and when converting volume units, cube the linear conversion factor. Unit errors are the most common trap in density problems.

Question 3

1. 9.6 Ma
2. 96 Ma (correct answer)
3. 192 Ma
4. 1920 Ma

Explanation: The correct answer is 96 Ma. First, determine the total distance the plates have spread from the ridge. Since the 4,800 km is the total distance between the two plates, each plate has moved half that distance from the ridge: 4,800 km / 2 = 2,400 km. Second, convert units to be consistent. Convert the distance to cm: 2,400 km * (100,000 cm/km) = 2.4 x 10^8 cm. Third, calculate the time: Time = Distance / Rate = (2.4 x $10^8$ cm) / (2.5 cm/year) = 9.6 x 10^7 years. Finally, convert years to mega-annum (Ma): (9.6 x $10^7$ years) / ($10^6$ years/Ma) = 96 Ma.

Question 4

1. 1.94 x 10³ tons (correct answer)
2. 1.94 x 10⁶ tons
3. 1.62 x 10⁸ tons
4. 1.62 x 10¹¹ tons

Explanation: The correct answer is 1.94 x 10³ tons. First, calculate the volumetric discharge rate (Q) of the water in m³/s: Q = width × depth × velocity = 25 m × 3.0 m × 1.5 m/s = 112.5 m³/s. Second, calculate the total volume of water discharged in one day: Volume/day = (112.5 m³/s) × (3600 s/hr) × (24 hr/day) = 9.72 x 10⁶ m³/day. Third, convert the water volume to liters: (9.72 x 10⁶ m³/day) × (1 L / 10⁻³ m³) = 9.72 x 10⁹ L/day. Fourth, calculate the mass of sediment in milligrams: (9.72 x 10⁹ L/day) × (200 mg/L) = 1.944 x 10¹² mg/day. Finally, convert mg to metric tons: (1.944 x 10¹² mg) × (1 g / 10³ mg) × (1 kg / 10³ g) × (1 ton / 10³ kg) = 1.944 x 10³ tons.

Question 5

1. 63.2 PJ
2. 1,995 PJ (correct answer)
3. 3,990 PJ
4. 63,080 PJ

Explanation: The correct answer is 1,995 PJ. A magnitude 9.0 earthquake is two whole numbers greater than a magnitude 7.0. The energy released increases by a factor of 31.6 for each whole number. Therefore, the total increase in energy is 31.6 (for 7.0 to 8.0) × 31.6 (for 8.0 to 9.0), which is approximately 998.56. The energy of the magnitude 9.0 earthquake is (2.0 x 10¹⁵ J) × 998.56 ≈ 1.997 x 10¹⁸ J. The question asks for how much more energy is released, which is the energy of the M9 minus the energy of the M7: (1.997 x 10¹⁸ J) - (2.0 x 10¹⁵ J) = 1.995 x 10¹⁸ J. Finally, convert this to petajoules: (1.995 x 10¹⁸ J) / (10¹⁵ J/PJ) = 1,995 PJ.

Question 6

1. 6.25 x 10⁸ years
2. 1.25 x 10⁹ years
3. 2.50 x 10⁹ years (correct answer)
4. 5.00 x 10⁹ years

Explanation: The correct answer is 2.50 x 10⁹ years. If 25% of the original Potassium-40 remains, two half-lives have passed (100% -> 50% is one half-life; 50% -> 25% is the second half-life). The age of the rock is the number of half-lives multiplied by the duration of one half-life. Age = 2 half-lives × (1.25 billion years/half-life) = 2.50 billion years. Expressed in scientific notation, this is 2.50 x 10⁹ years.

Question 7

1. 1.3 g/cm³
2. 7.9 g/cm³
3. 12.7 g/cm³ (correct answer)
4. 102.0 g/cm³

Explanation: The correct answer is 12.7 g/cm³. First, convert the radius from km to cm: 1,220 km * (10⁵ cm/km) = 1.22 x 10⁸ cm. Second, calculate the volume in cm³: V = (4/3) * π * (1.22 x 10⁸ cm)³ ≈ 7.61 x 10²⁴ cm³. Third, convert the mass from kg to g: 9.7 x 10²² kg * (10³ g/kg) = 9.7 x 10²⁵ g. Finally, calculate density: ρ = mass/volume = (9.7 x 10²⁵ g) / (7.61 x 10²⁴ cm³) ≈ 12.7 g/cm³. Distractor A results from making a unit conversion error (e.g., using 1 kg = 100 g). Distractor D results from using the diameter instead of the radius in the volume calculation.

Question 8

1. 7.0 x 10⁻³ Tg
2. 7.0 x 10⁰ Tg
3. 7.0 x 10³ Tg (correct answer)
4. 7.0 x 10⁶ Tg

Explanation: The correct answer is 7.0 x 10³ Tg. First, convert the volume of tephra from km³ to cm³. Since 1 km = 10⁵ cm, it follows that 1 km³ = (10⁵)³ cm³ = 10¹⁵ cm³. The volume of tephra is 5.0 km³ × (10¹⁵ cm³/km³) = 5.0 x 10¹⁵ cm³. Second, calculate the total mass in grams using the density formula (Mass = Density × Volume): Mass = (1.4 g/cm³) × (5.0 x 10¹⁵ cm³) = 7.0 x 10¹⁵ g. Finally, convert the mass from grams to teragrams. Given that 1 Tg = 10¹² g, we divide the mass in grams by 10¹²: (7.0 x 10¹⁵ g) / (10¹² g/Tg) = 7.0 x 10³ Tg.

Question 9

1. 5.0 x 10³ m³
2. 1.0 x 10⁶ m³
3. 1.0 x 10⁵ m³
4. 5.0 x 10⁴ m³ (correct answer)

Explanation: When you encounter soil erosion problems, you're working with volume calculations that require careful unit conversions and systematic application of the formula: Volume = Area × Depth. To solve this problem, start by identifying your given values: erosion rate of 0.5 mm/year, area of 2.0 km², and time period of 50 years. First, calculate the total depth of soil lost over 50 years: $0.5 \text{ mm/year} \times 50 \text{ years} = 25 \text{ mm}$ Next, convert all measurements to consistent units (meters): 25 mm = 0.025 m, and 2.0 km² = 2.0 × 10⁶ m². Now apply the volume formula: $\text{Volume} = 2.0 \times 10^6 \text{ m}^2 \times 0.025 \text{ m} = 5.0 \times 10^4 \text{ m}^3$ Answer D (5.0 × 10⁴ m³) is correct. Answer A (5.0 × 10³ m³) represents an error where you might have forgotten to convert km² to m², using 2,000 m² instead of 2,000,000 m². Answer B (1.0 × 10⁶ m³) likely results from incorrectly using the area value (2.0 km²) as if it were already in the final calculation without proper unit conversion. Answer C (1.0 × 10⁵ m³) suggests a calculation error, possibly multiplying by an incorrect conversion factor. For erosion problems, always convert to consistent units first, then multiply area by total depth lost. Double-check your unit conversions—km² to m² requires multiplying by 10⁶, not 10³.

Question 10

1. 1.74 x 10⁻¹ EJ
2. 6.26 x 10⁻¹ EJ
3. 1.74 x 10² EJ
4. 6.26 x 10² EJ (correct answer)

Explanation: The correct answer is 6.26 x 10² EJ. First, calculate the cross-sectional area of the Earth that intercepts solar radiation: Area = πr² = π * (6.37 x 10⁶ m)² ≈ 1.275 x 10¹⁴ m². Second, calculate the total power received by this area in watts (J/s): Power = (1,361 W/m²) * (1.275 x 10¹⁴ m²) ≈ 1.735 x 10¹⁷ W or 1.735 x 10¹⁷ J/s. Third, calculate the total energy received in one hour: Energy = Power × Time = (1.735 x 10¹⁷ J/s) * (3600 s/hr) ≈ 6.246 x 10²⁰ J. Finally, convert this energy from joules to exajoules: (6.246 x 10²⁰ J) / (10¹⁸ J/EJ) ≈ 6.25 x 10² EJ.

Question 11

1. 0.032 km
2. 0.316 km
3. 3.16 km (correct answer)
4. 31.6 km

Explanation: The correct answer is 3.16 km. First, calculate the groundwater velocity in m/s using Darcy's Law: Velocity = Conductivity × Gradient = (2.0 x 10⁻⁴ m/s) × 0.005 = 1.0 x 10⁻⁶ m/s. Second, calculate the total number of seconds in one century: Time = 100 years × 365.25 days/year × 24 hours/day × 3600 s/hour ≈ 3.156 x 10⁹ s. Third, calculate the total distance traveled in meters: Distance = Velocity × Time = (1.0 x 10⁻⁶ m/s) × (3.156 x 10⁹ s) = 3156 m. Finally, convert the distance from meters to kilometers: 3156 m / (1000 m/km) = 3.156 km, which is approximately 3.16 km.

Question 12

1. 1.1%
2. 11.2% (correct answer)
3. 12.6%
4. 88.8%

Explanation: The correct answer is 11.2%. First, express the Earth's age in scientific notation: 4.54 billion years = 4.54 x 10⁹ years. Second, calculate the time that passed before the oldest rocks formed by subtracting the rock age from the Earth's age: (4.54 x 10⁹ years) - (4.03 x 10⁹ years) = 0.51 x 10⁹ years. Third, calculate the percentage this period represents of Earth's total history: Percentage = (Time before rocks / Total age of Earth) × 100 = ((0.51 x 10⁹) / (4.54 x 10⁹)) × 100 ≈ 0.1123 × 100 = 11.23%.

Question 13

1. The surface is 0.09 m higher. (correct answer)
2. The surface is 0.9 m higher.
3. The surface is 0.09 m lower.
4. The surface elevation change cannot be determined.

Explanation: The correct answer is A. This question is about elevation, not horizontal retreat. The horizontal retreat of 50 m/year is extraneous information. The key is the vertical isostatic uplift. First, calculate the total uplift over a decade: Total Uplift = (9 mm/year) * 10 years = 90 mm. Second, convert this uplift from millimeters to meters: 90 mm / (1000 mm/m) = 0.09 m. Since the ice thickness at the terminus is assumed constant, the surface of the glacier rises along with the bedrock. Therefore, the net change is an increase in elevation of 0.09 m.

Question 14

1. 5.28 x 10⁵ m³ (correct answer)
2. 2.40 x 10⁶ m³
3. 5.28 x 10⁸ m³
4. 2.11 x 10¹⁰ m³

Explanation: The correct answer is 5.28 x 10⁵ m³. First, calculate the total volume of the sandstone layer in km³. Area = 8.0 km * 12.0 km = 96 km². Total Volume = Area * Thickness = 96 km² * (2.5 x 10⁻⁵ km) = 0.0024 km³. Second, convert the total volume to m³: 1 km³ = 10⁹ m³, so Total Volume = 0.0024 km³ * (10⁹ m³/km³) = 2.4 x 10⁶ m³. Third, calculate the volume of pore space (which holds the water) by multiplying the total volume by the porosity: Water Volume = (2.4 x 10⁶ m³) * 0.22 = 528,000 m³ = 5.28 x 10⁵ m³.

Question 15

1. 0.03 centuries
2. 0.3 centuries
3. 3.0 centuries (correct answer)
4. 30 centuries

Explanation: The correct answer is 3.0 centuries. First, ensure the units for the rise are consistent. Convert the total rise from meters to millimeters: 1.05 m * 1000 mm/m = 1050 mm. Second, calculate the number of years required for this rise: Time (years) = Total Rise / Rate = 1050 mm / 3.5 mm/year = 300 years. Finally, convert years to centuries: 300 years / 100 years/century = 3.0 centuries.

Question 16

1. 7.0 x 10⁻³ Tg
2. 7.0 x 10⁰ Tg
3. 7.0 x 10³ Tg (correct answer)
4. 7.0 x 10⁶ Tg

Explanation: The correct answer is 7.0 x 10³ Tg. First, convert the volume of tephra from km³ to cm³. Since 1 km = 10⁵ cm, it follows that 1 km³ = (10⁵)³ cm³ = 10¹⁵ cm³. The volume of tephra is 5.0 km³ × (10¹⁵ cm³/km³) = 5.0 x 10¹⁵ cm³. Second, calculate the total mass in grams using the density formula (Mass = Density × Volume): Mass = (1.4 g/cm³) × (5.0 x 10¹⁵ cm³) = 7.0 x 10¹⁵ g. Finally, convert the mass from grams to teragrams. Given that 1 Tg = 10¹² g, we divide the mass in grams by 10¹²: (7.0 x 10¹⁵ g) / (10¹² g/Tg) = 7.0 x 10³ Tg.

Question 17

1. 0.032 km
2. 0.316 km
3. 3.16 km (correct answer)
4. 31.6 km

Explanation: The correct answer is 3.16 km. First, calculate the groundwater velocity in m/s using Darcy's Law: Velocity = Conductivity × Gradient = (2.0 x 10⁻⁴ m/s) × 0.005 = 1.0 x 10⁻⁶ m/s. Second, calculate the total number of seconds in one century: Time = 100 years × 365.25 days/year × 24 hours/day × 3600 s/hour ≈ 3.156 x 10⁹ s. Third, calculate the total distance traveled in meters: Distance = Velocity × Time = (1.0 x 10⁻⁶ m/s) × (3.156 x 10⁹ s) = 3156 m. Finally, convert the distance from meters to kilometers: 3156 m / (1000 m/km) = 3.156 km, which is approximately 3.16 km.

Question 18

1. 5.0 x 10⁻³ g/cm³
2. 5.0 x 10⁻¹ g/cm³
3. 5.0 x 10¹ g/cm³
4. 5.0 x 10⁰ g/cm³ (correct answer)

Explanation: When you encounter mineral density problems, you're working with the fundamental relationship: density = mass ÷ volume. The key challenge is usually unit conversion, which is exactly what this question tests. To find the density, start by calculating the volume of the cubic crystal. Since each side is 2000 μm, the volume is $2000^3 = 8.0 \times 10^9 \text{ μm}^3$. Now convert this to cm³: since 1 cm = 10,000 μm (or $10^4$ μm), then 1 cm³ = $(10^4)^3 = 10^{12}$ μm³. Therefore: $8.0 \times 10^9 \text{ μm}^3 \times \frac{1 \text{ cm}^3}{10^{12} \text{ μm}^3} = 8.0 \times 10^{-3} \text{ cm}^3$ With mass = $4.0 \times 10^{-2}$ g and volume = $8.0 \times 10^{-3}$ cm³, the density is: $\frac{4.0 \times 10^{-2}}{8.0 \times 10^{-3}} = 5.0 \times 10^0 = 5.0 \text{ g/cm}^3$ Choice A ($5.0 \times 10^{-3}$) results from incorrectly converting micrometers to centimeters by dividing by 1000 instead of 10,000. Choice B ($5.0 \times 10^{-1}$) comes from using the linear dimension (2000 μm) as the volume instead of cubing it. Choice C ($5.0 \times 10^1$) occurs if you forget to convert units entirely and mix grams with μm³. Remember: always convert all units before calculating density, and when converting volume units, cube the linear conversion factor. Unit errors are the most common trap in density problems.

Question 19

1. 5.28 x 10⁵ m³ (correct answer)
2. 2.40 x 10⁶ m³
3. 5.28 x 10⁸ m³
4. 2.11 x 10¹⁰ m³

Explanation: The correct answer is 5.28 x 10⁵ m³. First, calculate the total volume of the sandstone layer in km³. Area = 8.0 km * 12.0 km = 96 km². Total Volume = Area * Thickness = 96 km² * (2.5 x 10⁻⁵ km) = 0.0024 km³. Second, convert the total volume to m³: 1 km³ = 10⁹ m³, so Total Volume = 0.0024 km³ * (10⁹ m³/km³) = 2.4 x 10⁶ m³. Third, calculate the volume of pore space (which holds the water) by multiplying the total volume by the porosity: Water Volume = (2.4 x 10⁶ m³) * 0.22 = 528,000 m³ = 5.28 x 10⁵ m³.

Question 20

1. The surface is 0.09 m higher. (correct answer)
2. The surface is 0.9 m higher.
3. The surface is 0.09 m lower.
4. The surface elevation change cannot be determined.

Explanation: The correct answer is A. This question is about elevation, not horizontal retreat. The horizontal retreat of 50 m/year is extraneous information. The key is the vertical isostatic uplift. First, calculate the total uplift over a decade: Total Uplift = (9 mm/year) * 10 years = 90 mm. Second, convert this uplift from millimeters to meters: 90 mm / (1000 mm/m) = 0.09 m. Since the ice thickness at the terminus is assumed constant, the surface of the glacier rises along with the bedrock. Therefore, the net change is an increase in elevation of 0.09 m.