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BSAC Homework 1

Errors, Measurement and Gas Laws

Answer Key*************************************************

Q3: Seven BSAC students scuba dived down to the Kram Shipwreck(depth 27 m) near Pattaya and recorded the pressure increase (aboveatmospheric) at that depth. They reported their readings as follows:

a) Bearing in mind that 1 atm = 101325 Pa, how would youappropriately express the underwater pressure in Pa?

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BSAC General Chemistry I (Dr. Tom Gale) Hand in Tuesday

Homework 1: Errors, Measurements and Gas Laws 14 October 9:00

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Q1: A crucible of yellow-hot molten iron ( = 6.93 g/mL) is found toweigh 36.45 kg. Using index notation where necessary, calculate theliquid’s volume and express in terms of…...

a) mL (quote to 5 s.f.)b) m3 (3 s.f.) c) nL (2 s.f.)

d) Regardless of how we express the answer, what is the greatestnumber of significant figures we should meaningfully use? Why?

Q2: A coal-fired power station is found to release 43.4 kg of SO2 into the air every hour.

a) Assuming the plant runs continually, how many cubicmetres are emitted each week? (to 4 s.f.)b) How many moles of the gas are released every year? (3 s.f.)

c) SO2 gas can be scrubbed (removed from the powerstation’s chimney by passing the exhaust gases throughlime (calcium hydroxide). Complete the equation:

SO2(g) + Ca(OH)2 (aq) CaSO4(aq) + ?

d) How many cubic feet of SO2 can be absorbed by a tankcontaining 1 metric tonne of dissolved lime (2 s.f.)? student reading

1 257 kPa 2 2.63 atm 3 1.94 x 103 Torr 4 2.6 atm 5 2980 mmHg 6 261,431 Pa 7 2002 mmHg

b) A diver at the wreck exhaled a bubble of volume 8 mL. What volume was it when it reached the surface?

c) Suppose the same diver exhaled the same mass of air at a depth of40 m. Assuming P increases linearly with depth, what would thebubble’s initial volume now be?

Q4: You are given a gas balloon containing one gram of neon gas at ambientpressure and 36 ºC.

a) Using Avogadro’s law, what is the sample’s volume in L?b) Calculate the density () of the gas (mg/L)c) If the gas is heated to 330 ºC at constant pressure what is itsfinal volume (L)? d) If the original balloon is compressed to 310 mL volume, whatis its final pressure (atm)?

Q5: Assuming ideal gas behaviour, what volume does a 30 g sample of SF4 occupy at 85 ºC (quote answer in L to 3 s.f.)?

[Important: put on your answers: 1) hwk sheet number 2) your name 3) your ID 4) BSAC]

Q1: A crucible of yellow-hot molten iron ( = 6.93 g/mL) is found toweigh 36.45 kg. Using index notation where necessary, calculate theliquid’s volume and express in terms of…...

a) mL (quote to 5 s.f.) 5259.7 mL

b) m3 (3 s.f.) 5.26 x 10-3 m3

c) nL (2 s.f.) 5.3 x 106 nL

d) Regardless of how we express the answer, what is the greatest number of significant figures we should meaningfully use?

three (because one of the inputted datum is quoted to no more than 3 s.f.)

singular of data

Q2: A coal-fired power station is found to release 43.4 kg of SO2 intothe air every hour.

a) Assuming the plant runs continually, how many cubic metres areemitted each week? (to 4 s.f.)

b) How many moles of the gas are released every year? (3 s.f.)

1.138 x 105 mol = 2550m3

380,184 kg = 5.93 x 106 mol

Q2: A coal-fired power station is found to release 43.4 kg of SO2 intothe air every hour.

c) SO2 gas can be scrubbed (removed from the power station’s chimneyby passing the exhaust gases through lime (calcium hydroxide).Complete the equation:

SO2(g) + Ca(OH)2 (aq) CaSO4(aq) + H2O(l)

d) How many cubic feet of SO2 can be absorbed by a tank containing 1 metric tonne of dissolved lime (2 s.f.)?

1 tonne = 1000 kg = 13496 mol Ca(OH)2 1 m3 = 35.315 ft3

13496 mol SO2 = 302311 L = 302.31 m3 = 10676 ft3

= 1.1 x 104 ft3

Q3: Seven BSAC students scuba dived down to the Kram Shipwreck (depth 27 m) near Pattaya and recorded the pressure increase (above atmospheric) at that depth. They reported their readings as follows:

a) Bearing in mind that 1 atm = 101325 Pa, how would you appropriately express the underwater pressure in Pa?

mean P = 262,320 Pa (student no. 5’s reading is outlying and should be discounted). Quoted answer: ‘2.6 x 105 Pa’ (shouldn’t use more than2 s.f. because student no. 4 wasn’t more accurate than 2 s.f.)

student reading 1 257 kPa 2 2.63 atm 3 1.94 x 103 Torr 4 2.6 atm 5 2980 mmHg 6 261,431 Pa 7 2002 mmHg

Q3: Seven BSAC students scuba dived down to the Kram Shipwreck (depth 27 m) near Pattaya and recorded the pressure increase (above atmospheric) at that depth. They reported their readings as follows:

b) A diver at the wreck exhaled a bubble of volume 8 mL. What volume was it when it reached the surface?

mean P = 262,320 Pa = 2.589 atm Total pressure at wreck P28m = 2.589 + 1.00 = 3.589 atm

Volume of bubble = 3.589 / 1.00 x 8 mL = 28.7 mL

student reading 1 257 kPa 2 2.63 atm 3 1.94 x 103 Torr 4 2.6 atm 5 2980 mmHg 6 261,431 Pa 7 2002 mmHg

P at surface

Q3:c) Suppose the same diver exhaled the same mass of air at a depth of 40 m. Assuming P increases linearly with depth, what would the bubble’s initial volume now be?

P40m = [2.589 x (40/27)] + 1.00 = 4.8356 atm

Boyle’s Law: P1V1 = P2V2 3.589 x 8.0 = 4.8356V2

V2 = 5.94 mL

P at surface

(bubble is smaller, deeper down, where pressureis greater, just as we would expect…

Q4: You are given a gas balloon containing one gram of neon gas at ambient pressure and 36 ºC.

a) Using Avogadro’s law, what is the sample’s volume in L?b) Calculate the density () of the gas (mg/L)c) If the gas is heated to 330 ºC at constant pressure what is its finalvolume (L)? d) If the original balloon is compressed to 310 mL volume, what is itsfinal pressure (atm)?

A: a) 1 mol = 22.4 L 0.04955 mol 1.11L (STP) = 1.26 L (309K) (or use Ideal Gas Law:V = nRT / P = 0.04955 x 8.314 x 309 / 1010325 = 1.26 x 10-3 m3

b) mass = 1000 mg = 796 mg/Lc) (603/309) x 2.0978 L = 2.45 L (Charles’ Law)d) (1256/310) x 1.0 atm = 4.05 atm (Boyle’s Law)

Q5: Assuming ideal gas behaviour, what volume does a 30 g sampleof SF4 occupy at 85 ºC (quote answer in L to 3 s.f.)?

A: Use PV = nRTV = ?P = 1.013 x 105 Pa

Molar mass (SF4) = (1 x 32.06) + (4 x 19.00) = 108.06 g/moln = 30 x 10-6/ 108.06 = 2.776 x 10-7 mol

V = nRT/P = (2.776 x 10-7 mol x 8.314 x 358) / 101325 = 8.15 x 10-9 m3 (8.2 nL)

always use accurateperiodic table atomic masses!