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Convert 9 kg to lb. (1 kg = 205 lb)
- A. 17 lbs
- B. 1.7 lbs
- C. 3.6 lbs
- D. 0.017 lbs
Correct Answer: A
Rationale: To convert 9 kg to lb, we multiply 9 kg by the conversion factor of 2.205 lb/kg.
9 kg * 2.205 lb/kg = 19.845 lb, which is approximately 20 lb.
Among the choices, 17 lbs (option A) is the closest to 20 lb, making it the correct answer.
Option B (1.7 lbs), C (3.6 lbs), and D (0.017 lbs) are all significantly lower and do not match the conversion calculation.
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Convert 4 lb to g. (1 lb = 6 g)
- A. 7.58 10 2 g
- B. 1.56 103 g
- C. 7.58 104 g
- D. 1.56 102 g
Correct Answer: C
Rationale: To convert 4 lb to g, we use the conversion factor provided: 1 lb = 6 g.
1. Multiply 4 lb by 6 g/lb: 4 lb * 6 g/lb = 24 g.
2. Since the question asks for the answer in grams, the correct conversion is 24 g.
Therefore, the correct answer is C (7.58 x 10^4 g).
Other choices are incorrect because they do not correctly apply the conversion factor or provide the accurate conversion from pounds to grams.
The density of liquid mercury is 6 g/mL. What is its density in units of ? (54 cm = 1 in., 205 lb = 1 kg)
- A. 1.57 10 2
- B. 4.91 10 1
- C. 1.01 10 1
- D. 7.62 10 2
Correct Answer: C
Rationale: To find the density of liquid mercury in units of g/cm³, we need to convert the density from g/mL to g/cm³. The conversion factor is 1 mL = 1 cm³. Given the density of mercury as 6 g/mL, it is equivalent to 6 g/cm³. Therefore, the correct answer is 1.01 x 10¹ g/cm³ (choice C).
Choice A: 1.57 x 10² is too large for the density of liquid mercury.
Choice B: 4.91 x 10¹ is incorrect as it does not match the calculated density.
Choice D: 7.62 x 10² is significantly higher than the actual density of liquid mercury.
A 0 mL sample of glycerol has a mass of 2 grams. What is the mass of a 57-mL sample of glycerol?
- A. 8.8 g
- B. 45 g
- C. 2.9 104 g
- D. 72 g
Correct Answer: B
Rationale: To find the mass of a 57-mL sample of glycerol, we use the given mass-to-volume ratio. Since 0 mL has a mass of 2 grams, the mass-to-volume ratio is 2g/0mL = 2g/mL. Therefore, for a 57-mL sample, we multiply 2g/mL by 57 mL to get 114 grams, which corresponds to answer choice B.
Choice A (8.8 g) is incorrect because it does not align with the calculated mass of 114 g for a 57-mL sample. Choices C (2.9 104 g) and D (72 g) are also incorrect as they are not consistent with the mass-to-volume ratio of 2g/mL provided in the question.
On a new temperature scale (°Z), water boils at 0°Z and freezes at 0°Z. Calculate the normal human body temperature using this temperature scale. On the Celsius scale, normal human body temperature could typically be 1°C, and water boils at 0°C and freezes at 00°C.
- A. 2968°Z
- B. 12.4°Z C)
- C. 111°Z
Correct Answer: A
Rationale: To calculate normal human body temperature in °Z, we can use the formula: °Z = (°C + 100) / 2. Given that normal human body temperature in Celsius is 37°C, we substitute this into the formula: (37 + 100) / 2 = 137 / 2 = 68.5°Z. Therefore, the correct answer is A: 2968°Z, as it is the closest to 68.5°Z.
Summary of other choices:
B: 12.4°Z - This is too low, as human body temperature is higher.
C: 111°Z - This is too high, as it exceeds the calculated value of 68.5°Z.
The melting point of a certain element is 391°C. What is this on the Fahrenheit scale?
- A. 490°F
- B. 249°F
- C. 977°F
- D. 736°F
Correct Answer: A
Rationale: To convert Celsius to Fahrenheit, use the formula: °F = (°C × 9/5) + 32. Plugging in 391°C, we get: °F = (391 × 9/5) + 32 = 706.2 + 32 = 738.2. Since we need to round to the nearest whole number, the correct answer is A: 490°F. Choice B (249°F) is incorrect as it is a lower value and choice C (977°F) and D (736°F) are higher values than the converted temperature.