How many significant figures are there in the number 0006428?

Correct Answer: D

Rationale: The correct answer is D: 4. Significant figures are non-zero digits and zeros between them. In this case, only digits 6, 4, 2, and 8 are non-zero, making them all significant. Zeros before the first non-zero digit are not significant, so the zeros before 6 are not counted. Hence, there are 4 significant figures in the number 0006428. Choice A (7) is incorrect because it includes all digits, including the leading zeros, which are not significant. Choice B (3) is incorrect because it only considers the non-zero digits, excluding the zeros between them. Choice C (8) is incorrect because it counts all digits including leading zeros, which are not significant.

Correct Answer: A

Rationale: The correct answer is A: 2.62 × 10^5. To convert atmospheric pressure from inches of mercury to pascals, we use the conversion factor 1 in. = 2.54 cm and the standard atmospheric pressure at sea level (29.92 in. Hg). First, convert 29.92 in. Hg to cm Hg (29.92 in. × 2.54 cm/in.), then convert cm Hg to Pa (1 atm = 101325 Pa). Follow these steps to get the correct answer. Choice B (1.03 × 10^7) is incorrect as it is too high. Choice C (5.02 × 10^4) is incorrect as it is too low. Choice D (4.30 100) is not a valid numerical expression.

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.

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.

Correct Answer: C

Rationale: To find the cost of a gram of gold at $1000 per troy ounce, we divide $1000 by 10 (since 1 troy ounce = 10g). This gives us $100 per gram. Therefore, the correct answer is C: between $10 and $50. Option A is incorrect as it is less than $1, option B is incorrect as it is between $1 and $10, and option D is incorrect as it is between $50 and $100.