409 Kelvin equals

Correct Answer: D

Rationale: To convert Kelvin to Celsius, subtract 273. So, 409K - 273 = 136°C. Choice D is correct. Choice A (136°F) is incorrect because it's a Fahrenheit conversion. Choice B (273°F) is incorrect, as it doesn't convert Kelvin to Fahrenheit. Choice C (682°F) is incorrect, as it's double the incorrect Fahrenheit conversion.

Correct Answer: A

Rationale: The correct order is A: nano- < milli- < centi- < kilo-. The rationale is as follows: 1. Start from the smallest prefix, which is nano- representing 10^-9. 2. Moving up, milli- comes next representing 10^-3, which is larger than nano-. 3. Following milli- is centi- representing 10^-2, larger than milli-. 4. Lastly, kilo- is the largest prefix representing 10^3, making it the largest of the four options. Therefore, the correct order is nano- < milli- < centi- < kilo-. The other choices are incorrect because they do not follow the correct increasing order of magnitude for the metric prefixes.

Correct Answer: D

Rationale: To convert Kelvin to Celsius, subtract 273. So, 409K - 273 = 136°C. Choice D is correct. Choice A (136°F) is incorrect because it's a Fahrenheit conversion. Choice B (273°F) is incorrect, as it doesn't convert Kelvin to Fahrenheit. Choice C (682°F) is incorrect, as it's double the incorrect Fahrenheit conversion.

Correct Answer: D

Rationale: The correct answer is D because when alpha particles are beamed at a thin metal foil, some pass directly through due to their small size and high energy (option A), while others are reflected by direct contact with nuclei in the metal foil (option C). This is based on the Rutherford scattering experiment which showed that alpha particles can be deflected by the positive nuclei in the metal foil. Option B is incorrect as alpha particles are not diverted by attraction to electrons in the foil. Option D combines the correct explanations for the behavior of alpha particles when beamed at thin metal foil.

Correct Answer: B

Rationale: To convert 4338 mL to qt, first convert mL to L by dividing by 1000 (4338 mL / 1000 = 4.338 L). Then, convert L to qt using the conversion factor 1 L = 0.946 qt. So, 4.338 L * 0.946 qt/L = 4.092 qt (Answer B). Choice A (4598 qt) is incorrect as it does not follow the correct conversion process. Choice C (4.092 10 3 qt) is incorrect due to incorrect formatting and notation. Choice D (4092 qt) is incorrect as it uses the wrong conversion factor.

Correct Answer: A

Rationale: The correct answer is A: By either adding or subtracting protons from the atom. An ion is formed when an atom gains or loses electrons, leading to an unequal number of protons and electrons. Protons determine the atom's identity, so changing the number of protons would result in a different element. This process results in the formation of ions with a positive charge if protons are added or a negative charge if protons are subtracted. Choices B and C are incorrect because adding or subtracting electrons or neutrons does not change the fundamental identity of the atom, only its charge or mass. Choice D is incorrect as not all options are true for ion formation.

Correct Answer: C

Rationale: The correct answer is C: 0.045. When dealing with significant figures, trailing zeros after the decimal point are considered significant. In this case, the number 0.045006700 should be rounded to 0.045 since it has four significant figures. Choice A has too many significant figures, B rounds incorrectly, and D removes significant figures after the decimal point. Thus, C is the correct choice.

Correct Answer: B

Rationale: The correct answer is B: 1 gram = 10^3 kilograms. This is incorrect because 1 kilogram is equal to 1000 grams, not 100 grams. A is correct as 1 microliter is indeed 10^-6 liters. C is correct as 10^3 milliliters equals 1 liter. D is correct as 1 gram is equal to 10^2 centigrams. The incorrect relationship in choice B violates the metric system conversion factor of 1 kilogram being equal to 1000 grams.

Correct Answer: B

Rationale: To determine the physical state of indium at 323°F, we convert the temperature to Celsius (323°F = 161.7°C). Since the melting point of indium is 2°C, which is lower than 161.7°C, indium would be in a liquid state at 323°F. The correct choice is B: Liquid. A: Solid - Incorrect because indium is in a liquid state at 323°F. C: Gas - Incorrect, as indium would be in a liquid state at 323°F. D: Not enough information - Incorrect since the temperature provided allows us to determine the physical state of indium.