Question

1. what are the base units for the following?

a. mass \t\t\tkilograms (kg)
b. length \t\t\tmeters (m)
c. volume \t\t\tliter (l)
d. density \t\t\tgram/ cm³
convert the following:
a. 1 cm = 10 mm \t\t0.02 cm=.....0.2.....mm (multiply by 10)
b. 1 m = 1000 mm \t\t0.09 m =....90... mm (multiply by 1,000)
c. 1 cm = 1.0 10⁷ nm \t\t2.0 cm =.....2.0 10⁷... nm (multiply by 1.0 * 10⁷
d. 100 cm = 1m \t\t0.02 cm=.....0.0002... m (divide by 100)

1. aluminum has a density of 2.40 g/cm³. what would be the mass of a sample whose volume is 10.0 cm³?

mass= density * volume
mass = 2.4 *10.0 = 24 grams

1. how many significant figures are in each of the following values?

a. 346572 \t\tall of these numbers are significant (1, 2, 3, 4, 5, 6, 7, 8, and 9 are always significant)
b. 9.2746419 \t\tall of these numbers are significant (1, 2, 3, 4, 5, 6, 7, 8, and 9 are always significant)
c. 0.00001 \t\tonly 1 is significant. if the number is smaller than 1, zeros between the decimal point and a whole number are not significant and zero before the decimal point is not significant.
d. 0.00826 \t\tonly 8.2 and 6 are significant

Explanation:

Step1: Identify the formula for mass

The formula relating mass, density, and volume is \( \text{mass} = \text{density} \times \text{volume} \).

Step2: Substitute the given values

We know the density of aluminum is \( 2.40 \, \text{g/cm}^3 \) and the volume of the sample is \( 10.0 \, \text{cm}^3 \). Substituting these values into the formula: \( \text{mass} = 2.40 \, \text{g/cm}^3 \times 10.0 \, \text{cm}^3 \).

Step3: Perform the multiplication

When we multiply \( 2.40 \) by \( 10.0 \), we get \( 2.40 \times 10.0 = 24.0 \) (we consider significant figures here; since \( 2.40 \) has three significant figures and \( 10.0 \) has three, the result should have three significant figures, but in the given solution it's written as 24 grams, maybe a simplification for basic calculation).

Answer:

The mass of the aluminum sample is 24 grams (or 24.0 grams considering significant figures).