# CHEM125 – CHAPTER 05 – REGULAR HOMEWORK

## CHAPTER 5 REGULAR HOMEWORK

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### Question 1

The pressure on a balloon holding 105 mL of an ideal gas is increased from 212 torr to 1.00 atm. What is the new volume of the balloon (in mL) at constant temperature? ### Explanation

Remember to use 3 Sig. Fig. 1atm =760 torr P1V1=P2V2 212(105)=760V V=29.28947

### Question 2

A sample of hydrogen gas (H2) is in a 2.11 L container at 694 torr and 25 oC. Express the pressure of hydrogen in atm after the volume has changed to 1.09 L and the temperature is raised to 166 oC. ### Explanation

Remember to use KPV=nRT(694/760)(2.11)=n(0.08206)(25+273.15)n= 0.08009528P(1.09)=(0.08009528)(0.08206)(166+273.15)P=2.648

### Question 3

A sample of methane gas (CH4) at 17.4 oC is either heated or cooled. The pressure increases from 1.54 atm to 3.95 atm, and the volume of the sample changes from 3.73 L to 1.99 L. To what temperature, in oC, was the methane heated or cooled? ### Explanation

Last reminder–use K in calculation, then convert to C. PV/T= PV/T (1.54)(3.73)/(17.4+273.15)=(3.95)(1.99)/T T (K)=397.59 T (C) = 297.59-273.15 T=124.4455

### Question 4

A 32 L cylinder containing helium gas at a pressure of 45.1 atm is used to fill a weather balloon in order to lift equipment into the stratosphere. To what maximum pressure (in atm) could a 131 L balloon be filled. Hint is given in the feedback. ### Explanation

Hint: This is easier than it looks–just use P1V1 = PfVf. Careful! What is Vf? 32(45.1) = (131+32)PP=8.9

### Question 5

A 40.0 L cylinder containing helium gas at a pressure of 25.7 atm is used to fill a weather balloon in order to lift equipment into the stratosphere. What is the final pressure (in atm) in the cylinder after a 391 L balloon is filled to a pressure of 1.47 atm. Hint is given in the feedback. ### Explanation

Method 1, new hint, find the pressure of the gas in the balloon, if it were all in the small cylinder. Next use Dalton’s law of partial pressure to get the final pressure in the small cylinder. Method 2, Old hint(works but harder), the moles in the cylinder at the beginning = moles in cylinder + balloon at end. Simplest Way: One Step 40(25.7)=(391*1.47)+(40*X)X=11.33

### Question 6

The following procedure provides a crude method of determining the molar mass of a volatile liquid. A liquid of mass 0.0157 g is introduced into a syringe and the end is capped (sealed). The syringe is transferred to a temperature bath maintained at 63.1 oC, and the liquid vaporizes. As the liquid vaporizes the plunger is pushed out. At equilibrium, the plunger reads 4.00 mL of gas. Atmospheric pressure is 740. mmHg. What is the approximate molar mass of the compound (in g/mol)? Hint is given in the feedback. ### Explanation

Hint, use ideal gas law to find the moles of gas. PV=nRT (740/760)(0.004)=n(0.08206)(63.1+273.15) n=0.00014115 0.0157g/ X = n X=111.22

### Question 7

The reaction of magnesium metal with HCl yields hydrogen gas and magnesium chloride. What is the volume, in liters, of the gas formed at 755 torr and 17 oC from 2.69 g of Mg in excess HCl? (Hint, first write the balanced equation.) ### Explanation

Mg + 2HCl —> H2 + MgCl2 PV=nRT (755/760)V=(2.69/24.3)(0.08206)(17+273.15) (.993421050)V=(0.11069959)(0.08206)(290.15) V=2.6531

### Question 8

What mass of water (in grams) forms from the reaction of 2.93 L of hydrogen gas and 1.96 L of oxygen gas? Both gases are at 701 torr and 27.7 oC. Remember hydrogen and oxygen are a diatomic molecules. Don’t forget to balance the equation. ### Explanation

PV=nRT (2.93)(701/760)=n(0.08206)(27.7+273.15) (2.93)(0.92236842)=n(0.08206)(300.85) n=0.10946884 (1.96)(701/760)=n(0.08206)(27.7+273.15) (1.96)(0.92236842)=n(0.08206)(300.85) n=0.0732283 2H2 + O2 –>2 H2O X g/ 18 = 0.10946884 mol H2O 1.97 = X

### Question 9

What is the partial pressure of argon, in torr, in a container that also contains neon at 139 torr and is at a total pressure of 569 torr? ### Explanation

Wouldn’t life be boring if all problems were this easy? 139 + X = 569 X= 430

### Question 10

A 2.10 L sample of neon at 11.54 atm is added to a 12.0 L cylinder that contains argon. If the pressure in the cylinder is 5.47 atm after the neon is added, what was the original pressure (in atm) of argon in the cylinder? Hint is given in the feedback. ### Explanation

Hint, what pressure would the neon have in the second container if no argon were present? 2.10 (11.54) = 12 X X = 2.0195 5.47-2.0195= 3.4505

### Question 11

Calculate the total pressure, in atm, in a 8 L flask that contains 4.40 g of Ne and 6.54 g of Ar. The temperature of the gases is 25 oC. ### Explanation

4.40 g Ne = 0.2180 mol Ne 6.54 g Ar= 0.16358179 mol Ar PV=nRT P(8)= (.164+.218)(0.08206)(298.15) P=1.168

### Question 12

Calculate vrms, the root mean square velocity, in m/s of SO2 molecules at 132 oC. Hint is given in the feedback. ### Explanation

Hint, units must be self-consistent. J is kg m2/s2. So, what unit to you need for MW? (3RT/M)^(0.5) ((3 *8.314) * (132+273.15))/ (.064066))^0.5

### Question 13

An effusion container is filled with 4 L of an unknown gas. It takes 124 s for the gas to effuse into a vacuum. From the same container under the same conditions–same temperature and initial pressure, it takes 42 s for 4.00 L of O2 gas to effuse. Calculate the molar mass (in grams/mol) of the unknown gas. ### Explanation

Rate of effusion 1/Rate of effusion 2 = (MM2/MM1)^0.5 (4/124)/(4/42) =( 32/x)^0.5 X=278.929

### Question 14

Three bulbs are connected by tubing, and the tubing is evacuated. The volume of the tubing is 48.0 mL. The first bulb has a volume of 36.0 mL and contains 8.83 atm of argon, the second bulb has a volume of 250 mL and contains 1.43 atm of neon, and the third bulb has a volume of 32.0 mL and contains 8.26 atm of hydrogen. If the stopcocks (valves) that isolate all three bulbs are opened, what is the final pressure of the whole system in atm? ### Explanation

8.83 + 1.43 + 8.26 = 18.52 PV=nRT (8.83)(36)/(0.08206)(298.15)=12.99= 317.88 (1.43)(250)/(0.08206)(298.15)=14.16=357.5 (8.26)(32)/(0.08206)(298.15)=10.80=264.32 48 + 36 + 250 + 32 = 366 P(366)= (37.95) (0.08206)(298.15) P=2.536 P(366)= 939.7 P=2.5674

### Question 15

Use the van der Waals equation to calculate the pressure, in atm, of 45.80 mol of hydrogen at 216 oC in a 2.51 L container. ### Explanation (P+(0.244)(45.8/2.51)^2))( 2.51-(45.8*0.0266))=45.8*0.08206(216+273.15) P=1341.97 Honestly look do this math on mathway unless you’re a lover of math like me