ATI RN
chemistry for health sciences Questions
Question 1 of 9
Order the four metric prefixes from smallest to largest.
Correct Answer: A
Rationale: The correct order is A: nano- < milli- < centi- < kilo-. The rationale is as follows: 1. Start from the smallest prefix, which is nano- representing 10^-9. 2. Moving up, milli- comes next representing 10^-3, which is larger than nano-. 3. Following milli- is centi- representing 10^-2, larger than milli-. 4. Lastly, kilo- is the largest prefix representing 10^3, making it the largest of the four options. Therefore, the correct order is nano- < milli- < centi- < kilo-. The other choices are incorrect because they do not follow the correct increasing order of magnitude for the metric prefixes.
Question 2 of 9
Convert 4338 mL to qt. (1 L = 06 qt)
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.
Question 3 of 9
The agreement of a particular value with the true value is called
Correct Answer: A
Rationale: The correct answer is A: accuracy. Accuracy refers to how close a measured value is to the true value. It indicates the absence of systematic errors. Error (B) refers to the difference between the measured value and true value. Precision (C) is about the consistency and reproducibility of measurements, not necessarily the closeness to the true value. Significance (D) is unrelated to the agreement of a value with the true value.
Question 4 of 9
Convert 4 mi to km. (1 m = 094 yd, 1 mi = 1760 yd)
Correct Answer: B
Rationale: To convert 4 mi to km, first convert miles to yards: 4 mi * 1760 yd/mi = 7040 yd. Then convert yards to meters: 7040 yd * 0.9144 m/yd = 6437.216 m. Finally, convert meters to kilometers: 6437.216 m/1000 m/km = 6.437216 km. Rounding to three significant figures, the answer is 3.69 x 10^1 km (B). Other choices are incorrect because they do not correctly convert miles to kilometers or have calculation errors.
Question 5 of 9
If the Thomson model of the atom had been correct, Rutherford would have observed:
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.
Question 6 of 9
Manganese makes up 3 10–4 percent by mass of the elements found in a normal healthy body. How many grams of manganese would be found in the body of a person weighing 206 lb? (205 lb = 1 kg)
Correct Answer: A
Rationale: To calculate the grams of manganese in a 206 lb body, first convert pounds to kg (206 lb = 93.44 kg). Then, use the given percentage (3 10–4%) to find the amount of manganese in the body (93.44 kg * 0.0003 = 0.028032 g). Since 205 lb = 1 kg, the correct answer is 0.028032 g or approximately 0.59 g (rounding). Summary of other choices: B: Incorrect, this does not consider the correct conversion and percentage calculation. C: Incorrect, this is a much larger value and does not align with the given percentage. D: Incorrect, this is also a much larger value and does not align with the given percentage.
Question 7 of 9
Bromine exists naturally as a mixture of bromine-79 and bromine-81 isotopes. An atom of bromine-79 contains
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.
Question 8 of 9
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)
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.
Question 9 of 9
A scientist obtains the number 0.045006700 on a calculator. If this number actually has four significant figures, how should it be written?
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.