CART - Chemistry Acceleration Readiness Measurement and Math Skills Questions and Answers

Question 1

A student performs a calculation to determine the volume of a rectangular block. The measured dimensions are 12.20 cm, 3.45 cm, and 1.9 cm. What is the volume of the block reported to the correct number of significant figures?

Accepted Answer

80 cm³

Answer

The volume is calculated by multiplying the length, width, and height: 12.20 cm × 3.45 cm × 1.9 cm = 79.959 cm³. When multiplying or dividing measured values, the result must be rounded to the same number of significant figures as the measurement with the fewest significant figures. The measurements have four (12.20), three (3.45), and two (1.9) significant figures, respectively. Therefore, the answer must be rounded to two significant figures, which is 80 cm³.

Question 2

A chemical reaction must be performed at a temperature of 398 K. What is this temperature in degrees Celsius (°C)?

Accepted Answer

125°C

Answer

The formula to convert from Kelvin (K) to degrees Celsius (°C) is °C = K - 273.15. Using this formula: °C = 398 - 273.15 = 124.85°C. When reporting the answer, it is common practice to maintain the precision of the original measurement. Since 398 K is given to the nearest whole number, the result is rounded to the nearest whole number, which is 125°C.

Question 3

Which of the following correctly expresses the number 0.0004075 in scientific notation?

Accepted Answer

4.075 x 10⁻⁴

Answer

Scientific notation requires a single non-zero digit to the left of the decimal point. To convert 0.0004075, the decimal point must be moved four places to the right to get 4.075. Since the original number is less than one, the exponent of 10 is negative. Therefore, the correct scientific notation is 4.075 x 10⁻⁴.

Question 4

A student measures the volume of a liquid using a 100 mL graduated cylinder with markings every 2 mL. The bottom of the meniscus is clearly above the 74 mL mark but below the 76 mL mark, appearing to be halfway between them. Which of the following is the most appropriate recording of this volume?

Accepted Answer

75 mL

Answer

When reading a graduated instrument, you record all certain digits plus one estimated digit. The markings are every 2 mL (74, 76, 78...). The volume is definitely greater than 74 mL and less than 76 mL. The student can estimate the position between the marks. The halfway point is 75 mL. Since the markings are at the ones place (74, 76), the estimated digit is also in the ones place. Therefore, 75 mL is the correct reading. Reporting 75.0 mL would imply certainty in the ones place and an estimated tenths place, which is not possible with this instrument.

Question 5

The density of mercury is 13.6 g/cm³. What is the mass in kilograms (kg) of 2.50 liters (L) of mercury? (Note: 1 L = 1000 cm³)

Accepted Answer

34.0 kg

Answer

First, convert the volume from liters to cm³: 2.50 L × (1000 cm³/L) = 2500 cm³. Next, use the density formula (mass = density × volume) to find the mass in grams: mass = 13.6 g/cm³ × 2500 cm³ = 34,000 g. Finally, convert the mass from grams to kilograms: 34,000 g × (1 kg / 1000 g) = 34.0 kg. The final answer is reported to three significant figures, consistent with the initial measurements.

CART - Chemistry Acceleration Readiness Test Practice Test

CART - Chemistry Acceleration Readiness Test Practice Test

CART - Chemistry Acceleration Readiness Measurement and Math Skills Questions and Answers