Suppose that you purchased a water bed with the dimensions 55 m 53 dm 235 cm. What mass of water does this bed contain?

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: Step-by-step rationale: 1. J.J. Thomson discovered the electron in 1897 through his cathode ray experiment. 2. Electrons are negatively charged particles emitted by atoms. 3. This groundbreaking discovery proved atoms emit negative particles. 4. Therefore, J.J. Thomson is the correct answer. Summary: - Lord Kelvin focused on thermodynamics, not subatomic particles. - Ernest Rutherford discovered the nucleus, not electrons. - William Thomson, also known as Lord Kelvin, did not directly contribute to the discovery of negative particles emitted by atoms.

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: 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.

Correct Answer: A

Rationale: The correct answer is A because Bromine-79 has an atomic number of 35, indicating the number of protons. Neutrons are calculated by subtracting the atomic number from the mass number, so Bromine-79 with a mass number of 79 has 44 neutrons. Electrons are equal to the number of protons in a neutral atom, so there are 35 electrons in an atom of Bromine-79. Choice B is incorrect because it does not account for the correct number of neutrons present in Bromine-79. Choice C is incorrect as it has an incorrect number of protons and electrons for Bromine-79. Choice D is incorrect as it incorrectly assigns 79 neutrons to Bromine-79, which is the total mass number, not the number of neutrons.