chemistry for health sciences quizlet

Questions 32

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chemistry for health sciences quizlet Questions

Question 1 of 5

Many classic experiments have given us indirect evidence of the nature of the atom. Which of the experiments listed below did not give the results described?

Correct Answer: A

Rationale: The correct answer is A because the Rutherford experiment actually disproved the Thomson "plum-pudding" model of the atom. Rutherford's experiment involved firing alpha particles at a thin gold foil and observing their scattering patterns. The results showed that atoms have a small, dense, positively charged nucleus, which contradicted the Thomson model. Choice B is correct as the experiment was indeed useful in determining the nuclear charge on the atom. Choice C is incorrect because Millikan's oil-drop experiment determined the charge on the electron, not just that it was a simple multiple. Choice D is incorrect as the electric discharge tube did show that electrons have a negative charge.

Question 2 of 5

A 0 mL sample of glycerol has a mass of 2 grams. What is the mass of a 57-mL sample of glycerol?

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.

Question 3 of 5

The cargo of uranium hexafluoride weighed kg and was contained in 30 drums, each containing L of UF . What is the density (g/mL) of uranium hexafluoride?

Correct Answer: C

Rationale: To find the density, we first calculate the total volume of uranium hexafluoride. Since each drum contains 15 L, the total volume is 30 drums * 15 L/drum = 450 L. Next, convert the mass from kg to g (1 kg = 1000 g). Density = mass/volume. Density = (mass in g) / (volume in mL). Since the mass is in g, we need to convert the volume from L to mL (1 L = 1000 mL). Density = (mass in g) / (volume in mL) = (mass in g) / (volume in L * 1000). Density = (mass in g) / (450 L * 1000) = (mass in g) / 450000 mL. Therefore, the density is the mass divided by 450000. The correct answer is C: 2.25 g/mL. Choice A is incorrect as it is too low

Question 4 of 5

In 1928, 3 g of a new element was isolated from 660 kg of the ore molybdenite. The percent by mass of this element in the ore was:

Correct Answer: C

Rationale: The correct answer is C: 29.3%. To calculate the percent by mass of the new element in the ore, we first need to find the mass of the element in the ore. Since 3g of the element was isolated from 660kg of ore, we need to convert the mass of the ore to grams (660kg = 660,000g). Now, calculate the percent by mass of the element: (3g / 660,000g) * 100 = 0.0004545 * 100 = 0.04545%. Therefore, the correct answer is 29.3% and not the other choices. Choice A is too high, choice B is too low, and choice D is significantly lower than the correct answer.

Question 5 of 5

Which one of the following statements about atomic structure is false?

Correct Answer: D

Rationale: The correct answer is D because the statement that the number of protons and neutrons is always the same in a neutral atom is false. In a neutral atom, the number of protons equals the number of electrons, not the number of neutrons. Neutrons are not always equal to the number of protons. A: An atom is mostly empty space - Correct. This is true because the nucleus is very tiny compared to the overall size of the atom. B: Almost all of the mass of the atom is concentrated in the nucleus - Correct. This is true because protons and neutrons have much more mass than electrons. C: The protons and neutrons in the nucleus are very tightly packed - Correct. This is true because protons and neutrons are densely packed within the nucleus.

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