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CHEM103 / CHEM 103 Module 3:

(Latest Update 2025 / 2026) General Chemistry I with Lab | Questions & Answers | Grade A | 100% Correct - Portage Learning

Question:

Show the calculation of the mole fraction and partial pressure of each gas in a mixture which consists of 3.00 moles of H2, 4.50 moles of CO2 and 6.00 moles of Ar if the total pressure of the mixture is 700 mm.Express your answer in mm.

Answer:

XHe = 3.00 / (3.00 + 4.50 + 6.00) = 0.2222

XCO2 = 4.50 / (3.00 + 4.50 + 6.00) = 0.3333

XAr = 6.00 / (3.00 + 4.50 + 6.00) = 0.4444

PHe = XHe (700 mm) = 0.2222 (700 mm) = 155.5 mm PCO2 = XCO2 (700 mm) = 0.3333 (700 mm) = 233.3 mm PAr = XAr (700 mm) = 0.4444 (700 mm) = 311.1 mm

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Question:

Show the calculation of the molecular weight of an unknown gas if the rate of effusion of oxygen gas (O2) is 1.562 times faster than that of an unknown gas.

Answer:

(rO2 /runknown)2 = MWunknown / MWO2 (1.562/1)2 = MWunknown / 32.00 MWunknown = (1.562)2 x 32.00 = 78.01

Question:

Show the calculation of the final temperature of the mixture when a 40.5 gram sample of water at 85.7C is added to a 36.8 gram sample of water at 26.3C in a coffee cup calorimeter.c (water) = 4.184 J/g C

Answer:

Heat temp change = qtemp change = m x c x ∆t-(40.5 x 4.184 x (∆t - 85.7)) = 36.8 x 4.184 x (∆t - 26.3)-(169.452 x (∆t - 85.7)) = 153.9712 x (∆t - 26.3)- (169.452∆t - 14522.0364) = 153.9712∆t - 4049.44256-169.452∆t + 14522.0364 = 153.9712∆t - 4049.4425618571.479 = 323.423∆t57.4 C = ∆t

Question:

Show the calculation of the energy involved in melting 120 grams of ice at 0oC if the Heat of Fusion for water is 0.334 kJ/g.

Answer:

qs↔l = mass x Heat of Fusion = m x ∆Hfusion120 x 0.334 = 40.08 kJ

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Question:

Sulfur undergoes combustion to yield sulfur trioxide by the following reaction equation:2 S + 3 O2 → 2 SO3 ΔH = - 792 kJIf 42.8 g of S is reacted with excess O2, what will be the amount of heat given off?

Answer:

moles = grams/molecular weightmoles (S) = 42.8/32.07 = 1.335 mols S= ΔHrx x new moles / original molesq = -792 x ( 1.335/2 )= -526.7 kJ

Question:

Hydrosulfuric acid (H2S) undergoes combustion to yield sulfur dioxide and

water by the following reaction equation:2 H2S + 3 O2 → 2 SO2 + 2

H2OWhat is the ΔH of the reaction if 26.2 g of H2S reacts with excess O2 to yield 431.8 kJ?

Answer:

moles = grams/molecular weightmoles (H2S ) = 26.2/34.086 = 0.7686 mols H2SΔHrx = q / (new moles / original moles)-431.8 / ( 0.7686/2) = -1123.6 kJ

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Question:

A sample of benzene (C6H6), weighing 7.05 g underwent combustion in a

bomb calorimeter by the following reaction:2 C6H6 (l) + 15 O2 (g) → 12 CO2

(g) + 6 H2O (l)The heat given off was absorbed by 500 g of water and caused the temperature of the water and the calorimeter to rise from 25.00 to 53.13 C.The heat capacity of water = 4.18 J/g/C and the heat capacity of the calorimeter = 10.5 kJ/C. (1) what is the ΔH of the reaction? Using the definitions at the beginning of the module describe (2) the calorimeter + contents, (3) the type of process.

Answer:

1) q water = s (specific heat of water) x mass x Δt = 4.18 J / g / K x 500 g x (53.13 C

- 25.00 C) = - 58,791 J

q calorimeter = heat capacity x Δt = (10.5 kJ/C) x (53.13 C - 25.00 C) = - 295.365 kJ x 1000 J/1 kJ= - 295,365 J q reaction = - 58,791 J + (-295,365 J) = - 354,156 J = - 354,156 J x 1 kJ / 1000 J = - 354.16 kJ

(new) moles C6H6 = 7.05 g / 78 = 0.0904 mole C6H6

ΔH = q / (new moles / original moles) ΔH = - 354.16 / (0.09048 / 2) = - 7828 kJ / mole

(2) Isolated system (bomb calorimeter)

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