Portage Learning CHEM 103 chemistry EXAM A+ GRADE ASSURED COMPLETE SOLUTIONS AND VERIFIED ANSWERS LATEST UPDATE!!!!

PORTAGE LEARNING EXAM zm zm




Exam Solution zm




Portage Learning Chemistry CHEM 103 Unit 3 2026 A+ zm zm zm zm zm zm zm zm zm




GRADE ASSURED COMPLETE SOLUTIONS AND VERIFIE zm zm zm zm zm




D ANSWERS (C3887)
zm zm




QUESTION 1 zm




Heat temp change = qtemp change = m x c x ∆t -(40.5 x 4.184 x (∆t -
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm



85.7)) = 36.8 x 4.184 x (∆t - 26.3) -(169.452 x (∆t - 85.7)) = 153.9712 x (∆t - 26.3) -
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm



(169.452∆t - 14522.0364) = 153.9712∆t - 4049.44256 - zm zm zm zm zm zm zm



169.452∆t + 14522.0364 = 153.9712∆t - zm zm zm zm zm



4049.44256 18571.479 = 323.423∆t 57.4 C = ∆t
zm zm zm zm zm zm zm zm




ANSWER

1. Show the calculation of the final temperature of the mixture when a 40.5 gram sample of water a
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm



t 85.7C is added to a 36.8 gram sample of water at 26.3C in a coffee cup calorimeter. c (water) = 4.
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm



184 J/g C zm zm




QUESTION 2 zm




qs↔l = mass x Heat of Fusion = m x ∆Hfusion 120 x 0.334 = 40.08 kJ
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm




ANSWER

2. Show the calculation of the energy involved in melting 120 grams of ice at 0oC if the Heat of Fus
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm



ion for water is 0.334 kJ/g.
zm zm zm zm zm




QUESTION 3 zm




moles = grams/molecular weight moles (S) = 42.8/32.07 = 1.335 mols S = ΔHrx x new
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm z



moles / original moles q = -792 x ( 1.335/2 )= -526.7 kJ
m zm zm zm zm zm zm zm zm zm zm zm zm




ANSWER

3. Sulfur undergoes combustion to yield sulfur trioxide by the following reaction equation: 2 S + 3 O
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm



2 → 2 SO3 ΔH = -
zm zm zm zm zm zm



792 kJ If 42.8 g of S is reacted with excess O2, what will be the amount of heat given off?
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm

, QUESTION 4 zm




moles = grams/molecular weight moles (H2S ) = 26.2/34.086 = 0.7686 mols H2S ΔHrx
zm zm zm zm zm zm zm zm zm zm zm zm zm z



= q / (new moles / original moles) -431.8 / ( 0.7686/2) = -1123.6 kJ
m zm zm zm zm zm zm zm zm zm zm zm zm zm zm




ANSWER

4. Hydrosulfuric acid (H2S) undergoes combustion to yield sulfur dioxide and water by the following
zm zm zm zm zm zm zm zm zm zm zm zm zm zm



reaction equation: 2 H2S + 3 O2 → 2 SO2 + 2 H2O What is the ΔH of the reaction if 26.2 g of H2S
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm z



reacts with excess O2 to yield 431.8 kJ?
m zm zm zm zm zm zm zm




QUESTION 5 zm




2 (2 NH3 (g) → 3 H2 (g) + 2 N2 (g) ΔH = + 91.8 kJ) 3 (2 H2 (g) + O2 (g) → 2 H2O (g)
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm



ΔH = - zm zm



483.7 kJ) 2 (N2 (g) + O2 (g) → 2 NO (g) ΔH = 180.6 kJ) 4 NH3 (g) + 5 O2 (g) → 4 NO
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm



(g) + 6 H2O (g) ΔHrxn = - 906.3 kJ ΔHrxn = 2 (+ 91.8) + 3 (-483.7) + 2 (180.6) = -
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm



906.3 kJ
zm zm




ANSWER

7. The combustion of ammonia by the following reaction yields nitric oxide and water 4 NH3 (g) +
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm



5 O2 (g) → 4 NO (g) + 6 H2O (g) Determine the heat of reaction (ΔHrxn) for this reaction by using
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm z



the following thermochemical data: N2 (g) + O2 (g) → 2 NO (g) ΔH = 180.6 kJ N2 (g) + 3 H2 (g)
m zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm



→ 2 NH3 (g) ΔH = - 91.8 kJ 2 H2 (g) + O2 (g) → 2 H2O (g) ΔH = - 483.7 kJ
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm




QUESTION 6 zm




Pi x Vi = Pf x Vf Ti Tf 680 ml/1000 = 0.680 liters = Vi 720 mm/760 = 0.947 atm = Pi
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm



820 mm/760 = 1.08 atm = Pf 28 C + 273 = 301 K = Ti 55 C + 273 = 328 K = Tf Put in
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm z



the data: (0.947) x (0.680) = (1.08) x Vf (301) (328) Solve for Vf: 0.002139402 = 0.00
m zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm



3292683 x Vf Vf = 0.650 literrs zm zm zm zm zm zm




ANSWER

9. A gas sample has an original volume of 680 ml when collected at 720 mm and 28
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm



C. What will be the volume of the gas sample if the pressure increases to 820 mm and the temperat
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm



ure increases to 55 C?
zm zm zm zm




QUESTION 7 zm




P x V = n x R x T 0.546 mole = n R = 0.0821 700 mm/760 = 0.921 atm = P 25 C + 27
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm



3 = 298 K = T (0.921) x V = (0.546) x (0.0821) x (298) (0.921) x V = 13.358 V = 13.3
zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm zm



58 = 14.5 liters 0.921
zm zm zm zm




ANSWER