chapter 5 hw solutions - chem 1, g-chem...

CHAPTER 5 HW – SOLUTIONS

GAS VARIABLES

1.) The atmospheric pressure at the summit of Mt. Everest (29,029 ft) is roughly 250 mmHg. Convert this pressure into torr, atmospheres, and pounds per square inch.

250 mmHg = 250 torr

250 mmHg × !#$%

&'(%%)* = 0.33 atm 0.33 atm ×

!+.'-./0!#$%

= 4.8 psi

2.) Use the Kinetic Molecular Theory of gases (i.e. a discussion of the gas particle’s interaction with the container walls) to explain the following observations:

a. When the plunger on a sealed syringe is extended, the pressure decreases inside the syringe. (Note: if the syringe were not sealed, the low pressure would withdraw material into the syringe; this is the normal mechanism of a syringe. This is also the mechanism of a bicycle pump.)

When the volume increases, the trapped air molecules have farther to travel before they contact the container walls. This will result in fewer collisions with the container walls, which means a lower pressure.

b. If a ping pong ball becomes dented, it can be restored to its original shape by placing it into boiling water. After a short amount of time in the hot water, the dent will be removed. Note: this only works if the ping pong ball has not become cracked (why?).

As the temperature increases inside the ping pong ball, the trapped air molecules move faster and contact the container walls more often (increasing the pressure). The high pressure inside the ball eventually creates enough force to push out the dent and increase the volume.

Note: this would no longer work if the ball were cracked as air could escape and would never build up pressure.

10 15 20 5 10 15 20air

plug

3.) A bag of potato chips is packed and sealed in Los Angeles (at sea level), then shipped to Lake Tahoe (in the mountains). The volume of the bag increases upon arrival at Lake Tahoe. What gas conditions most likely caused the volume increase? Use the Kinetic Molecular Theory of gases to explain.

If the effect was because of a difference in temperature (↓T in the mountains), that would have caused the gas volume to decrease, which didn’t happen. Since the gas volume increased, this could be explained by a lower pressure in Tahoe due to its high elevation. The chip bag was packaged under L.A. atmospheric pressure. When the bag is taken to Tahoe and the atmospheric pressure is lowered, the L.A. pressure inside the bag is greater and pushes the bag outwards. The bag volume increases until the pressure inside and outside the bag is equal, at the new Tahoe pressure.

4.) If you fill up your car’s tires at a gas station, you increase both the pressure and volume of the tires. Why does this not contradict Boyle’s law, which says that pressure and volume are inversely proportional?

Boyle’s law only works when all other gas variables are kept constant (moles, temperature). When you fill your tires with air, you increase the moles of gas.

5.) The following graph was generated by measuring the volume and Celsius temperature of a sample of gas (with constant moles and pressure).

a. What are the units of the slope and y-intercept?

Slope = ∆2∆3

= 𝑳˚𝑪

Y-intercept = L

b. Use the provided slope and y-intercept to write the equation of the linear graph, using variables (not x,y).

Volume = 0.003700 T + 0.9989

c. Use the graph to calculate the experimental temperature of absolute zero in Celsius.

Since ↓T will ↓V, at the lowest T (absolute zero), the volume is zero.

T = 789:%;<(.--=-

(.((>&(( = (<(.--=-(.((>&((

= –270.0 ˚C

6.) Use the Kinetic Molecular Theory of gases (i.e. a discussion of the gas particle’s interaction with the container walls) to explain why a latex balloon gets bigger when air is blow into it.

When air is blown into the balloon, the moles of gas are increased. This leads to a higher pressure at first as the larger number of gas particles will contact the container walls more often. Since the balloon is deformable, the higher pressure inside the balloon creates an outward force and the balloon increases its volume to compensate.

m = 0.003700 b = 0.9989

7.) Which graph (A-D) correctly shows the relationship between each set of variables? (Assume in each that other variables are kept constant.) Briefly explain your logic.

a. Pressure vs. Volume

If volume decreases, pressure increases (smaller space, more collisions with container walls and thus more pressure). Therefore P should be high at low V (either graphs B or D). The relationship is non-linear (as shown in class), so it’s Graph D.

b. Pressure vs. Inverse Volume

Boyle’s law says PV= constant, which means P = constant (1/V).

P versus 1/V should be linear (Graph A).

c. Pressure vs. Temperature

When temperature increases, pressure increases (gas particles move faster and so collide more often with the container walls). The relationship is linear, thus Graph A. Eventually in lecture you’ll see how PV = nRT summarizes the relationships; if variables are on opposite sides of the equality their relationship is linear (e.g. V vs. n, V vs. T, P vs. n, P vs. T).

d. Volume vs. Temperature

When temperature increases, pressure increases. But if pressure is constant the system will adjust, pushing outwards to a higher volume until pressures equalize. Therefore, as T↑, V↑ (Graph A).

8.) A glass vessel contains 28 g of nitrogen gas. Which of the processes listed below would double the pressure in the container? Explain your answers.

a. Adding 28 g of oxygen gas.

The vessel contains 28 g of N2 gas, which is equal to 1 mole of N2 gas, MMN2 = 2(14.01). If another mole of gas were added, that would double the pressure, as P ∝ n (PV = nRT). If 28 g of O2 gas were added, that’s not quite 1 mole of O2 gas (28 g / 32 g/mol = only 0.875 mol), so the pressure increases but doesn't double.

b. Adding 32 g of oxygen gas.

If 32 g of O2 gas were added, the pressure would double, as this is doubling the moles of gas in the container. P ∝ n.

c. Raising the temperature from –73 ˚C to 127 ˚C.

If the temperature is raised from –73˚C to 127 ˚C, the pressure would double. P ∝ T (PV = nRT), and is mathematically proportional when the temperature is in Kelvin. In Kelvin, the temperature is raised from 200 K to 400 K, so the temperature is doubled. As T doubles in Kelvin, P doubles.

d. Raising the temperature from 35 ˚C to 70 ˚C.

Raising the temperature from 35 ˚C to 70 ˚C would not double the pressure. Pressure would increase (P ∝ T), but not by double. The absolute temperature increases here from only 308 K to 343 K, so the pressure will increase only a little.

e. Adding enough liquid mercury to fill one-half the container.

Adding Hg to fill one-half the container takes up space, so in effect decrease the volume of gas by half. This would double the pressure, as P ∝ 1/V, or PV = constant. If V is decreased two-fold, pressure has to increase two-fold to keep PV constant.

GAS CALCULATIONS

9.) A steel gas cylinder containing argon gas has a volume of 43 L. At 22.0 ˚C the pressure of argon gas is 135 atm. What would be the gas pressure if the temperature is increased to 44.0 ˚C?

T1 = 22.0 ˚C + 273.15 = 295.2 K T2 = 44.0 ˚C + 273.15 = 317.2 K

@A7ABA

= @C7CBC

constant V so remove V; @ABA

= @CBC

P2 = @ABCBA

= (!>E#$%)(>!&.GH)

(G-E.GH) = 145 atm

Note: temperature is doubled in ˚C, but pressure doesn’t double.

10.) A bicycle tire has an internal volume of 1.52 L and contains 0.406 mol of air. The tire will burst if its internal pressure reaches 7.25 atm. What temperature (in ˚C) does the air in the tire need to be to cause a blow-out?

PV = nRT T = @7IJ

= (&.GE#$%)(!.EGK)

((.+('%89)L(.(=G('MNOPPQRST = 331 K

331 K – 273.15 = 58 ˚C

11.) Helium-filled weather balloons are used to carry instruments high into the atmosphere. Suppose a 4190 L balloon is launched at 22.5 ̊ C and 754 mmHg. If no helium escapes, what would be the volume of the balloon at a height of 20 miles, where the pressure is 76.0 mmHg and the temperature is –33.0 ˚C?

T1 = 22.5 ˚C + 273.15 = 295.7 K T2 = –33.0 ˚C + 273.15 = 240.2 K

@A7ABA

= @C7CBC

V2 = @A7ABCBA@C

= (&E+%%)*)(+!-(K)(G+(.GH)

(G-E.&H)(&'.(%%)*) = 33,800 L

12.) A standard commercial “Balloon Time” helium tank is labeled to contain “8.9 cubic feet of helium,” which represents the volume of helium it can produce under typical conditions (1.00 atm and 25 ˚C). The tanks are pressurized to 260 psi.

a. What is the volume of helium (in L) in the pressurized tank? Assume the tank is at 25 ˚C.

@A7ABA

= @C7CBC

constant T so P1V1 = P2V2 P2 = 1.00 atm = 14.7 psi

V1 = @C7C@A

= (!+.&./0)(=.-U$V)

(G'(./0) = 0.50 ft3

0.50 ft3× L!G0I!U$

T>× LG.E+W%

!0IT>× !%K!W%V ×

!K!(((%K

= 14 L

(Or convert 260 psi → 18 atm for P1 and 8.9 ft3 → 250 L for V2 then P1V1 = P2V2 to give 14 L)

b. It takes approximately 0.27 cubic feet (cf) of helium to fill one 9” latex balloon at 1.00 atm. How many 9” balloons could be filled with the Balloon Time helium tank?

# Balloons = =.-U$V

(.G&U$V = 33 balloons (note: the 8.9 cf tank is labeled to fill 30 balloons)

13.) Jacques Charles launched the first manned hydrogen (H2) balloon in 1783. The balloon stayed aloft for almost 45 minutes and traveled 15 miles away. When it landed in a village, the people were so terrified they tore it to shreds with pitchforks. What was the volume of the balloon used by Charles if it contained about 1300 moles of H2 at 23 ˚C and 750 mmHg?

T = 23 ˚C + 273.15 = 296 K

750 torr × !#$%&'($8XX

= 0.99 atm

PV = nRT V = IJB@

= (!>((%89)L(.(=G('MNOPPQRST(G-'H)

(.--#$% = 32,000 L

14.) A person accidentally swallows one drop of liquid oxygen (O2), which has a density of 1.149 g/mL. Assuming the drop has a volume of 0.050 mL, what volume of O2 gas (in mL) will be produced in the person’s stomach at body temperature (37 ˚C) and a pressure of 1.0 atm?

T = 37 ˚C + 273.15 = 310. K

0.050 mL O2 × !.!+-*!%K

× !%89YC>G.((*

= 0.0018 mol O2

MMO2 = 2(16.00) = 32.00 g/mol

PV = nRT V = IJB@

= ((.((!=%89)L(.(=G('MNOPPQRST(>!(.H)

!.(#$% = 0.046 L

0.046 L × !(((%K!K

= 46 mL

15.) Dry ice (solid CO2) has occasionally been used as an explosive in mining. A hole is drilled, dry ice and a small amount of gunpowder are placed in a hole, a fuse is added, and the hole plugged. When lit, exploding gunpowder rapidly vaporizes the dry ice building up an immense pressure. Assume 500.0 g of dry ice is placed in a cavity with a volume of 0.800 L and the ignited gunpowder heats the CO2 to 430 ˚C. What is the final pressure inside the hole?

T = 430 ˚C + 273.15 = 703 K (2 SF) MMCO2 = 12.01 + 2(16.00) = 44.01 g/mol

500.0 g CO2 × !%89ZYC++.(!*

= 11.36 mol CO2

PV = nRT P = IJB7

= (!!.>'%89)L(.(=G('MNOPPQRST(&(>H)

(.=((K = 820 atm

MOLAR VOLUME

16.) How many moles of hydrogen (H2) are present in 0.075 L of hydrogen gas at STP?

0.075 L × !%89GG.+!K

= 0.0033 mol H2 Note: The 22.41 L trick only works at STP.

Or PV = nRT n = @7JB

= (!#$%)((.(&EK)

L(.(=G('MNOPPQRST(G&>.!EH) = 0.0033 mol H2

17.) The electrolysis of water produces both H2 and O2 gases, which can be collected separately, as shown in the picture below. Are there a greater number of moles of H2 or O2? Briefly explain how you know.

Both tubes are collected under identical experimental conditions (temperature, pressure). H2 has the greater moles of gas as its volume is larger (V ∝ n).

18.) Many bad odors are due to sulfur compounds, for example the scent of sweaty socks, bad breath, and flatulence. These odors are commonly caused by hydrogen sulfide (H2S), dimethylsulfide (CH3SCH3), and/or methyl mercaptan (CH3SH).

How many molecules of CH3SH are present in a 1.00 mL vessel that contains CH3SH gas at STP?

1.00 mL × !K

!(((%K = 0.00100 L

PV = nRT n = @7JB

= (!#$%)((.((!((K)

L(.(=G('MNOPPQRST(G&>.!EH) = 4.46 × 10–5 mol CH3SH

4.46 × 10–5 mol × '.(GG×!(CV%89;W:9;/

!%89Z)V\) = 2.69 × 1019 molecules

Or 1.00 mL × !K

!(((%K ×

!%89GG.+!K

× '.(GG×!(CV%89;W:9;/

!%89Z)V\) = 2.69 × 1019 molecules

O2 H2

GAS DENSITY

19.) Carbon dioxide is used in some fire extinguishers as its high density allows for it to smother a fire. (Note: CO2 is an invisible gas, but comes out of the extinguisher as a white cloud. The CO2 from the tank expands and cools significantly, enough to condense moisture from the air, which is then visible).

Calculate the density of CO2 (g) at room temperature (24.2 ˚C) and 1.00 atm.

T = 24.2 ˚C + 273.15 = 297.4 K

MMCO2 = 12.01 + 2(16.00) = 44.01 g/mol

MM = ]JB@

d = @^^JB

= (!.((#$%)L++.(! _PQRT

L(.(=G('MNOPPQRST(G-&.+H) = 1.8 g/L

20.) Air, with an average molar mass of 28.96 g/mol, has a density of 1.18 g/mL at 25 ˚C and sea level. Forty miles above Earth’s surface, in the mesosphere where most meteors burn up, the temperature is –23 ˚C and the pressure is only 0.20 mmHg. What is the density of air at this altitude?

T = –23 ˚C + 273.15 = 250. K

0.20 mmHg × !#$%

&'(%%)* = 0.00026 atm

MM = ]JB@

d = @^^JB

= ((.(((G'#$%)LG=.-' _PQRT

L(.(=G('MNOPPQRST(GE(.H) = 0.00037 g/L

21.) An unknown diatomic gas has a density of 3.164 g/L at STP. What is the identity of the gas?

MM = ]JB@

= L>.!'+_MTL(.(=G('

MNOPPQRST(G&>.!EH)

(!.((#$%) = 70.92 g/mol

The gas is diatomic, so 70.92 / 2 = 35.46 g/mol is the MM of the element: it’s chlorine gas (Cl2).

22.) In each pair, which has a higher density at room temperature? Higher MM = more dense

a. N2 (g) vs. C8H18 (g) (C8H18 is similar to gasoline) c. Ar (g) vs. Ne (g)

This is why gasoline vapors settle on the ground.

b. NO2 (g) vs. CO (g) d. CO2 (s) vs. SF6 (g) Solid more dense

GAS STOICHIOMETRY

23.) A self-contained breathing apparatus uses canisters of potassium superoxide (KO2). It consumes CO2 exhaled by a person and replaces it with oxygen through the following reaction. These apparatuses are primarily used by firefighters, but also the US Navy.

4 KO2 (s) + 2 CO2 (g) → 2 K2CO3 (s) + 3 O2 (g)

What mass of KO2 is required to react with 8.90 L of CO2 at 22.0 ˚C and 767 torr?

T = 22.0 ˚C + 273.15 = 295.2 K

767 torr × !#$%&'($8XX

= 1.01 atm

PV = nRT n = @7JB

= (!.(!#$%)(=.-(K)

L(.(=G('MNOPPQRST(G-E.GH) = 0.371 mol CO2

0.371 mol CO2 × +%89HYCG%89ZYC

× &!.!(*

!%89HYC = 52.8 g KO2

MMKO2 = 39.10 + 2(16.00) = 71.10 g/mol

24.) Explosive compounds are generally ones that react quickly to produce large amounts of energy and gases. The newly generated gases then undergo a rapid expansion, destroying objects in their path. Ammonium nitrate is predominantly used in agriculture as a fertilizer, but can decompose explosively when heated via the following equation.

2 NH4NO3 (s) → 2 N2 (g) + 4 H2O (g) + O2 (g)

The most recent explosion involving ammonium nitrate occurred in China in the summer of 2015, where 800 metric tons (800,000 kg) of NH4NO3 (MM = 80.04) detonated, killing 173 people. How many liters of gas were formed during the explosion, assuming the gases were at 1.00 atm and 450. ˚C (the temperature of an explosion)?

T = 450. ˚C + 273.15 = 723 K

800,000 kg NH4NO3 × !(((*!`*

× !%89a)baYV

=(.(+* ×

&%89*#/;/G%89a)baYV

= 3.5 × 107 mol gases (N2, H2O, O2)

PV = nRT V = IJB@

= (>.E×!(c%89)L(.(=G('MNOPPQRST(&G>H)

!.((#$% = 2 × 109 L

PARTIAL PRESSURES

25.) For scuba dives below 150 feet, helium is often used to replace nitrogen in the scuba tank. If 15.2 g of He and 30.6 g of O2 are added to a previously evacuated 5.00 L tank at 22 ˚C, calculate the partial pressure of each gas as well as the total pressure in the tank.

T = 22 ˚C + 273.15 = 295 K MMO2 = 2(16.00) = 32.00 g/mol

15.2 g He × !%89);+.((*

= 3.80 mol He 30.6 g O2 × !%89);>G.((*

= 0.956 mol O2

PV = nRT PHe = IdeJB7

= (>.=(%89)L(.(=G('MNOPPQRST(G-EH)

E.((K = 18.4 atm He

PO2 = IfCJB7

= ((.-E'%89)L(.(=G('MNOPPQRST(G-EH)

E.((K = 4.63 atm O2

Ptot = PHe + PO2 = 18.4 atm + 4.63 atm = 23.0 atm

26.) Hydrazine can be used as rocket fuel, and was repurposed in the movie The Martian (with Matt Damon, 2015) to produce hydrogen gas (via the equation below). The hydrogen gas was subsequently used by Matt Damon’s character to make water (via 2 H2 + O2 → 2 H2O) for his potato farm on the planet Mars.

N2H4 (g) → N2 (g) + 2 H2 (g)

If 12.00 g of hydrazine (MM = 32.06) is decomposed, and the N2 and H2 gases collected in a 15.0 L tank at 23˚C, what is the partial pressure of each gas and total pressure in the tank?

T = 23 ˚C + 273.15 = 296 K

12.00 g N2H4 × !%89aCYb>G.('*

= 0.3743 mol N2O4 × !%89aC!%89aCYb

= 0.3743 mol N2

0.3743 mol N2O4 × G%89)C!%89aCYb

= 0.7486 mol H2

PV = nRT PN2 = IgCJB

7 = ((.>&+>%89)L(.(=G('MNOPPQRST(G-'H)

!E.(K = 0.606 atm N2

PH2 = IgCJB

7 = ((.&+='%89)L(.(=G('MNOPPQRST(G-'H)

!E.(K = 1.21 atm H2

Ptot = PN2 + PH2 = 0.606 atm + 1.21 atm = 1.82 atm

27.) A sample of O2 is collected over water at 21 ˚C and a total pressure of 742.2 mmHg. The vapor pressure of water at 21 ˚C is 18.663 mmHg. What is the partial pressure of O2 in the sample?

Ptot = PO2 + Pwater

PO2 = Ptotal – Pwater = 742.2 mmHg – 18.663 mmHg = 723.5 mmHg

28.) Reaction of 0.2273 g of iron with excess acid produces H2 through the following unbalanced equation. The H2 is collected over water at 24.2 ˚C and a total pressure of 754.7 torr. What volume (in mL) of H2 is collected? The vapor pressure of water at 24.2 ˚C is 22.648 torr.

Fe (s) + 2 HCl (aq) → FeCl2 (aq) + H2 (g)

T = 24.2 ˚C + 273.15 = 297.4 K

0.2273 g Fe × !%89h;EE.=E*

× !%89)C!%89h;

= 0.004070 mol H2

Ptot = PH2 + Pwater

PH2 = Ptotal – Pwater = 754.7 torr – 22.648 torr = 732.1 torr × !#$%&'($8XX

= 0.9632 atm

PV = nRT VH2 = IdCJB@dC

= ((.((+(&(%89)L(.(=G('MNOPPQRST(G-&.+H)

(.-'>G#$% = 0.1031 L H2

0.1031 L × !(((%K!K

= 103.1 mL H2

KINETIC ENERGY, GAS VELOCITIES

29.) Do all molecules in a 1 mol sample of CH4 have the same kinetic energy at 273 K?

No, there is a distribution of energies (bell-shaped curve). Temperature is a measure of the average kinetic energy of the particles.

30.) Consider two gas samples at 20 ˚C, one with CO at 760 torr, and another with H2 at 100 torr.

a. In which sample will the gas molecules have the greatest average velocity? Explain.

The gas sample with H2 will have the greatest average velocity as H2 has a lower molar mass than CO (H2 is “lighter”).

b. Why would it be difficult to quickly know which sample had the highest average velocity if the samples were not at the same temperature? Explain.

Temperature is related to the average kinetic energy and it’s possible that a heavier gas molecule could have a higher average velocity than a lighter gas molecule if it’s at a high enough temperature. Both temperature and mass of the gas particle affect the average velocity.

31.) Consider separate 1.0 L gaseous samples of He, Xe, Cl2, and O2, all at STP.

a. Rank the gases in order of increasing average kinetic energy. Explain your answer.

All samples are at the same temperature (STP, 0 ˚C) so they have the same average kinetic energy.

b. Rank the gases in order of increasing root-mean-square velocity. Explain your answer.

Root-mean-square velocity is closely related to average velocity: Xe < Cl2 < O2 < He. This order correlates with the gas particle’s molar mass. The particle with the highest MM (Xe, “heavier”) is the slowest on average, while the particle with the lowest MM (He, “lightest”) is the fastest on average.

[Note: molar mass trend is obvious based on the periodic table, I don’t think it has to be proven. It would make sense to show the MM calculation if they were closer.]

c. How can separate samples of O2 and He each have the same average velocity?

This could only happen if the two gas samples were are at different temperatures. Since He is “lighter” and faster, if would need to be at a lower temperature so it could slow down enough to be at the same average velocity as the “heavier” O2 particles.

32.) Calculate the root mean square velocity of CH4 (g) and N2 (g) at 273 K.

Molar masses have to be in kg to partially cancel with J (J = kgm2/s2)

MMCH4 = 12.01 + 4(1.01) = 16.05 *%89

× !`*!(((*

= 0.01605 kg/mol

MMN2 = 4(14.01) = 28.02 *%89

× !`*!(((*

= 0.02802 kg/mol

urms = i>JB^^

= j>(=.>!+E k

PQRS)(G&>H)

(.(!'(E l_PQR

= 652 m/s (CH4)

urms = i>JB^^

= j>(=.>!+E k

PQRS)(G&>H)

(.(G=(G l_PQR

= 493 m/s (N2)

33.) At what temperature (in ˚C) does the average speed of an oxygen molecule (O2) equal that of an airplane moving at 580 miles per hour?

MMO2 = 2(16.00) = 32.00 *%89

× !`*!(((*

= 0.03200 kg/mol

580 %0mX

× !.'(-`%!%0

× !(((%!`%

× !mX

'(%0I × !%0I'(/

= 259 m/s

urms = i>JB^^

(urms)2 = >JB^^

T = (:nPo)C^^

>J = LGE-Po T

C((.(>G(( l_PQR)

>(=.>!+E kPQRS)

= 86.1 K – 273.15 = –187 ˚C

GAS DIFFUSION / EFFUSION

34.) One of the main components of skunk spray is 3-methyl-1-butanethiol (C5H11SH) while one of the main components of lavender is linalool (C10H17OH). If separate containers containing gaseous samples of C5H11SH and C10H17OH are opened 10 feet away from you, which compound will you smell first? Explain your answer.

The two compound diffuse through the air at different rates. The compound with the higher molar mass (the “heavier” one) diffuses more slowly, while the compound with the lower molar mass diffuses more quickly. The compound that diffuses quickly reaches your nose quicker, and you will smell first. Therefore the skunk smell, C5H11SH, would reach you first as it has a lower MM.

35.) In each pair, which effuses more quickly at room temperature? Lower MM = faster effusion

a. N2 (g) vs. CO2 (g) b. CF4 (g) vs. F2 (g) c. 20Ne (g) vs. 22Ne (g)

36.) You have gaseous samples of ozone (O3) and ethylene (C2H4).

a. Which effuses more quickly at room temperature, O3 or C2H4?

MMO3 = 3(16.00) = 48.00 g/mol MMC2H4 = 2(12.01) + 4(1.01) = 28.06 g/mol

C2H4 has a lower MM (“lighter”) and so effuses more quickly.

b. How much faster does one sample effuse compared to the other?

XAXC

= i^^C

^^A

XpCdbXfV

= i ^^fV

^^pCdb = i+=.((*/%89G=.('*/%89

= C2H4 is 1.308 times faster

37.) What is the relative rate of effusion for F2O (g) compared to H2 (g)?

MMF2O = 2(19.00) + 16.00 = 54.00 g/mol MMH2 = 2(1.01) = 2.02 g/mol

XAXC

= i^^C

^Â

XdCXrCf

= i^^rCf

^^dC = iE+.((*/%89G.(G*/%89

= H2 is 5.17 times faster

Or XrCfXdC

= i ^^dC^^rCf

= iG.(G*/%89E+.((*/%89

= F2O is 0.193 times as fast as H2

38.) The effusion rate of an unknown gas is found to be 31.50 mL/min. Under identical conditions, the effusion rate of O2 (g) is found to be 30.50 mL/min. Which unknown gas was used: CH4, CO, NO, CO2, or NO2?

MMO2 = 2(16.00) = 32.00 g/mol, let’s make this MM2

Unknown gas effusion rate (r1) = 31.50 mL/min, O2 effusion rate (r2) = 30.50 mL/min.

XAXC

= i^^C

^Â LXA

XCTG

= ^^C

^Â MM1 =

^^C

LnAnCTC =

(>G.(( _PQR)

LVA.stPM/PuvVt.stPM/PuvT

C = 30.00 g/mol

Choices for the gas are CH4, CO, NO, CO2, and NO2

The gas that best matches this molar mass is NO: MMNO = 14.01 + 16.00 = 30.01 g/mol

39.) It takes 4.5 minutes for 1.0 L helium gas to effuse through a porous barrier. How long would it take 1.0 L of chlorine gas to effuse under identical conditions?

Let’s call He gas 1 and Cl2 gas 2 (He is lighter and will be faster).

MMCl2 = 2(35.45) = 70.90 g/mol

XAXC

= i^^C

^Â

XdeXpRC

= i&(.-(*/%89+.((*/%89 = 4.21

This means He effuses 4.21 times faster than Cl2.

If He takes 4.5 min to effuse, chlorine will effuse in 4.5 min × 4.21 = 19 min

chapter 5 hw solutions - chem 1, g-chem...

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