A species with 12 protons and 10 electrons is

Correct Answer: C

Rationale: The correct answer is C: Mg2+. A species with 12 protons must have an atomic number of 12, which corresponds to magnesium (Mg). The 10 electrons indicate that it has a 2+ charge, making it Mg2+. Choice A (Ne2+) is incorrect because neon (Ne) has an atomic number of 10, not 12. Choice B (Ti2+) is incorrect because titanium (Ti) has an atomic number of 22, not 12. Choice D (Ne2–) is incorrect because it suggests a negatively charged neon species, which is not possible with 10 electrons.

Correct Answer: C

Rationale: To calculate the mass of water in the water bed, first convert all dimensions to the same unit (centimeters). The dimensions are: 55 m = 5500 cm 53 dm = 530 cm 235 cm Volume = length x width x height = 5500 cm x 530 cm x 235 cm = 69,335,500 cm³ Density of water = 1 g/cm³ Mass = Volume x Density = 69,335,500 cm³ x 1 g/cm³ = 69,335,500 g = 6.93 x 10^7 g Therefore, the correct answer is C: 1.52 x 10^5 g. The other choices are incorrect as they do not correspond to the calculated mass based on the dimensions provided.

Correct Answer: A

Rationale: To convert 0494 to L, we first convert it to inches: 494 cm ÷ 54 cm/in = 9.148 in. Then, we convert inches to L: 9.148 in × 1 L/54 in = 0.169 L. The correct answer is A: 1.40 101 L, which is the correct conversion of 0.169 L to scientific notation. Choices B, C, and D are incorrect due to incorrect conversions or not being in scientific notation.

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: C

Rationale: The correct answer is C: Mg2+. A species with 12 protons must have an atomic number of 12, which corresponds to magnesium (Mg). The 10 electrons indicate that it has a 2+ charge, making it Mg2+. Choice A (Ne2+) is incorrect because neon (Ne) has an atomic number of 10, not 12. Choice B (Ti2+) is incorrect because titanium (Ti) has an atomic number of 22, not 12. Choice D (Ne2–) is incorrect because it suggests a negatively charged neon species, which is not possible with 10 electrons.

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: 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: B

Rationale: The correct answer is B (5 significant figures) because leading zeros are not considered significant. In the number 04560700, the zeros before the 4 are placeholders and do not add to the precision of the number. Therefore, only the digits 4, 5, 6, 0, and 7 are significant. Choice A (4) is incorrect because it does not account for the zeros after the 6. Choice C (7) is incorrect as it counts all the digits including the leading zeros. Choice D (8) is incorrect as it includes all the digits, including non-significant zeros.

Correct Answer: B

Rationale: The correct answer is B: Alpha particles greatly deflected by the metal foil. In the Thomson model, the atom was thought to be a uniform, positively charged sphere. If this model had been correct, alpha particles would have passed through the foil with little or no deflection. However, Rutherford observed significant deflections, indicating a concentrated positive charge at the center (nucleus). This aligns with the concept of the Rutherford model of the atom. Choices A, C, and D are incorrect because they do not correspond to the experimental observations made by Rutherford.