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The first scientist to show that atoms emit any negative particles was
- A. J. J. Thomson
- B. Lord Kelvin
- C. Ernest Rutherford
- D. William Thomson
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.
You may also like to solve these questions
Which of the following metric relationships is incorrect?
- A. 1 microliter = 10–6 liters
- B. 1 gram = 103 kilograms
- C. 103 milliliters = 1 liter
- D. 1 gram = 102 centigrams
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.
Which of the following statements is true?
- A. Ions are formed by adding or removing protons or electrons.
- B. Scientists believe that solids are mostly open space.
- C. Heating water with a Bunsen burner results in a 2:1 mixture of hydrogen and oxygen gases.
- D. At least two of the above statements (A-C) are true
Correct Answer: A
Rationale: The correct answer is A: Ions are formed by adding or removing protons or electrons. This statement is true because ions are formed when an atom gains or loses one or more electrons, resulting in a positive or negative charge. This process does not involve changing the number of protons in the nucleus.
Choice B is incorrect because solids are composed of closely packed atoms or molecules, not open space. Choice C is incorrect because heating water with a Bunsen burner does not result in a 2:1 mixture of hydrogen and oxygen gases, but rather breaks down water into hydrogen and oxygen molecules. Choice D is incorrect as only statement A is true.
A 0 mL sample of glycerol has a mass of 2 grams. What is the density of glycerol in ounces/quart? (00 ounce = 4 grams, and 00 liter = 06 quarts)
- A. 41.9 oz/qt
- B. 4.19 oz/qt
- C. 837 oz/qt
- D. 47.0 oz/qt
Correct Answer: A
Rationale: To find the density in ounces/quart, we first need to convert the mass of glycerol to ounces. 2 grams is equal to 0.5 ounces (2 grams / 4 grams per ounce). Next, convert the volume from mL to quarts. Since 1000 mL is 1 liter and 1 liter is 0.26 quarts (1 liter / 3.78 liters per gallon / 4 quarts per gallon), 0 mL is equal to 0 quarts. Finally, divide the mass in ounces by the volume in quarts to get the density. Therefore, the density of glycerol is 0.5 oz/qt, which is equivalent to 41.9 oz/qt. Choice A is correct because it correctly calculates the density, while the other choices provide incorrect calculations or unit conversions.
If the Thomson model of the atom had been correct, Rutherford would have observed:
- A. Alpha particles going through the foil with little or no deflection.
- B. Alpha particles greatly deflected by the metal foil.
- C. Alpha particles bouncing off the foil.
- D. Positive particles formed in the 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.
A scientist obtains the number 0.045006700 on a calculator. If this number
actually has four significant figures, how should it be written?
- A. 0.4567
- B. 0.4501
- C. 0.045
- D. 0.045
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.