Review+Chapter+5+Answers

q= 16 kJ m of brick = 938 g T1 = 19.5o C T2 = 35o C ∆T = T2 - T1 =35 – 19.5 o C= 15.5o Cq = mc∆T c = q/ m∆T c = 16kJ / (938g X 15.5o C) c = 1.1 J/g oC
 * 2)**

m of kettle = 2.0 Kg c = 0.385J/g.oC T1 = 20.0 C T2 = 80 C ∆T = 60 C q =2.0 kg X (0.385 J / g C) X 60 C q = 46.2 kJ = 0.5 kg X (4.18 J/g C) X 60 C = 125.4 kJ Total heat needed to raise the temperature of the kettle and its content 46.2 kJ + 125.4 kJ = **172 kJ
 * 4)**
 * q (kettle)** = mc∆T
 * q (water)** = mc∆T

5)** 200.0 L H2O (l) ΔT of H2O = 45°C Δ H of C3H8 = -2220 kJ/mol converts to -2.22 x 10^6 J/mol molar mass C3H8 = 44 g/mol molar mass H2O = 18 g/mol mass C3H8 = ?

V H2O --> m H2O m = 200 L X 1000 mL/L X 1 g/1 mL = 200,000 g

mol C3H8 = (m X c X ΔT)/ ΔH = (200,000 g X 4.18 J/g*°C X 45°C)/ 2.22 x 10^6 J/mol = 16.9 mol

mol C3H8 --> m C3H8 m = 16.9 mol X 44g/mol m = 743.6 g

(textbook gives 547 g, but I think it’s wrong) (I think the text is wrong too... I got 754.3g propane -Morgan) The energy gained by the water is: Quantity equals 200 kg X 4200 J/kgC X 45 C equals 3.78 X 10 E 7 Joules Now, how much energy did the propane supply? The same amount (but negative = -3.78 X 10 E 7 Joules) How much propane was needed? Well we know that each mole of propane provides 2.2 X 10 E 6 Joules ... (again negative because it is exothermic.) So how many moles of propane are necessary: -3.78 X 10 E 7 / -2.2 X 10 E 6 = 17.18 moles Each mole of propane has a mass of 44g 17.18 moles X 44g / mole = 755.92g .... so your answer is correct, but I couldn't follow your logic ... Paul.**
 * This question makes no sense as it was done: hang on ... correctly, break it down into steps:

10) //C6H12O6 + 6O2 -> 6CO2 + 6H2O + 2813 kJ// //C2H6 + 7/2O2 -> 2CO2 + 3H2O + 1369 kJ//

C6H12O6 + 6O2 -> 6CO2 + 6H2O ∆H= - 2813 kJ 3H2O + 2CO2 -> 7/2O2 + C2H6 ∆H= + 1369

C6H12O6 + 5/2O2 -> 4CO2 + 3H2O + C2H6 + 1454 kJ ∆H= - 1454 kJ

Molar mass of glucose: 6(12) + 12 (1) + 6(16) = 180 g/mol

500g(1mol/180g) = 2.78 mol 2.78mol ( -1454 kJ/mol) = **-4830.89 kJ**

∆T=81.8 C c= 4.18 kJ/kg C m= 3.77 kg ∆H= -802 kJ/mol Molar Mass of Methane= 16 g/mol
 * 11)**

Q= mc∆T Q= 3.77 kg X 4.18 kJ/kg C X 81.8 C Q= 1289 kJ

number of moles of methane Q/∆H 11 289 kJ/ 802 kJ per mol = 1.6 mol

Mass of methane needed # of moles X Molar Mass 1.6 mol X 16 g/mol = 25.7g

Therefore, the mass of methane needed is 25.7 g.

CH3COCOOH → CH3COOH + CO ∆H =? Reference Equations: a) CH3COCOOH + 5/2 O2 → 3CO2 + 2H2O ∆H =-1275 kJ/mol b) CH3COOH + 2O2 → 2CO2 + 2H2O ∆H =-875.3 kJ/mol c) CO + ½ O2 → CO2 ∆H =-282.7 kJ/mol Solving: CH3COCOOH + 5/2 O2 → 3CO2 + 2H2O ∆H = -1275 kJ/mol (remains same) 2CO2 + 2H2O → CH3COOH + 2O2 ∆H= -1(-875.3 kJ/mol) 875.3 kJ (multiply equation by -1 as equation is reversed) CO2 → CO + ½ O2 ∆H = -1(-282.7 kJ/mol) 282.7 kJ (equation is reversed so multiply by -1) CH3COCOOH → CH3COOH + CO ∆H = -117 kJ/mol
 * 12)**
 * Answer:**