Nokea
The element rhenium (Re) exists as two stable isotopes and 18 unstable isotopes. Rhenium-185 has in its nucleus
- A. 75 protons, 75 neutrons
- B. 75 protons, 130 neutrons
- C. 130 protons, 75 neutrons
- D. 75 protons, 110 neutrons
Correct Answer: A
Rationale: The correct answer A is determined by knowing that the atomic number of rhenium is 75. Since the element is rhenium-185, the sum of protons and neutrons must equal 185. As the atomic number is 75, the number of protons is 75. Therefore, the number of neutrons is 185 - 75 = 110. Thus, rhenium-185 has 75 protons and 110 neutrons. Choices B, C, and D are incorrect as they do not adhere to the correct number of protons and neutrons in the nucleus of rhenium-185.
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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?
- A. The Rutherford experiment proved the Thomson "plum- pudding" model of the atom to be essentially correct.
- B. The Rutherford experiment was useful in determining the nuclear charge on the atom.
- C. Millikan's oil-drop experiment showed that the charge on any particle was a simple multiple of the charge on the electron.
- D. The electric discharge tube proved that electrons have a negative charge.
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.
A 0 mL sample of glycerol has a mass of 2 grams. What is the mass of a 57-mL sample of glycerol?
- A. 8.8 g
- B. 45 g
- C. 2.9 104 g
- D. 72 g
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.
The density of liquid mercury is 6 g/mL. What is its density in units of ? (54 cm = 1 in., 205 lb = 1 kg)
- A. 1.57 10 2
- B. 4.91 10 1
- C. 1.01 10 1
- D. 7.62 10 2
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.
A metric unit for length is
- A. gram
- B. milliliter
- C. yard
- D. kilometer
Correct Answer: D
Rationale: The correct answer is D: kilometer. A kilometer is a metric unit for length because it is used to measure long distances. In the metric system, length is typically measured in meters, and a kilometer is equal to 1000 meters. This makes it a suitable unit for measuring larger distances efficiently.
Explanation for incorrect choices:
A: Gram is a unit of mass, not length.
B: Milliliter is a unit of volume, not length.
C: Yard is a unit of length, but it is not a metric unit. The metric system uses meters and its derivatives for length measurements.
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:
- A. 44 %
- B. 6.6 %
- C. 29.3 %
- D. 0.0044 %
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.