Question

1. what effect does an increase in temperature have on the density of air? explain your answer.
2. the mineral sample shown is galena. the chart lists the densities of several different minerals.

mineral sample
mass = 210 grams
mineral density (g/cm³) mineral density (g/cm³)
gypsum 2.3 hornblende 3.2
orthoclase 2.6 chalcopyrite 4.2
quartz 2.7 pyrite 5.0
calcite 2.7 magnetite 5.2
dolomite 2.9 galena?
fluorite 3.2 copper 8.9
a. what instrument can be used to measure the mass galena?
b. 50 ml of water is added to a 100 ml graduated cylinder. the galena is added to the graduated cylinder and the volume increases to 78ml. what is the volume of the piece of galena?
c. find the density of galena if the mass is 210 grams. show your work below and explain your choice and round to the nearest tenth.
d. a different mineral has a mass of 40.0 grams and a volume of 8.0 ml. this sample is most likely the mineral ____. show your work below and explain your choice

Explanation:

Step1: Answer for 21

When temperature increases, air molecules gain kinetic energy and move more vigorously, spreading out and occupying more space. Since density $
ho=\frac{m}{V}$ (mass per unit volume), with mass remaining constant and volume increasing, the density of air decreases.

Step2: Answer for 22 - a

A balance can be used to measure the mass of Galena. A balance compares the mass of an object to known masses to determine its mass value.

Step3: Answer for 22 - b

The volume of Galena is found by the displacement method. The initial volume of water is $V_1 = 50$ mL and the final volume is $V_2=78$ mL. So the volume of Galena $V = V_2 - V_1=78 - 50=28$ mL.

Step4: Answer for 22 - c

The density formula is $
ho=\frac{m}{V}$. Given $m = 210$ grams and $V = 28$ mL. Then $
ho=\frac{210}{28}=7.5$ g/mL. We use this formula because density is defined as mass per unit volume.

Step5: Answer for 22 - d

First, calculate the density of the new - mineral using $
ho=\frac{m}{V}$. Given $m = 40.0$ grams and $V = 8.0$ mL, so $
ho=\frac{40.0}{8.0}=5.0$ g/mL. Looking at the table, the mineral with a density closest to 5.0 g/mL is Pyrite.

Answer:

1. An increase in temperature decreases the density of air because air molecules spread out (occupy more volume) while mass remains constant.

22 - a. A balance.
22 - b. 28 mL
22 - c. 7.5 g/mL
22 - d. Pyrite