ID: A Class: SptBaus Date: \' 1 IN 25B Spring 2014 Test # 3 (4,5,11) e Choice th
ID: 561491 • Letter: I
Question
ID: A Class: SptBaus Date: ' 1 IN 25B Spring 2014 Test # 3 (4,5,11) e Choice the letter of the choice that best completes the statement or answers the question Be sure to show work to get CREDIT ulas on board 7-84 +1+S(Gases) +S(Hess's Law)+S(Phase change)- 100 : BATMAN ,ete 1. When 18.0 g of an unknown metal at 79.0 C is placed in 112 g H0 at 22.2 C, the final temperature of the is the specific heat capacity of the metai? The specific heat of water is 4.184 J/g K. water is 24.9 C. What a. 0.34 J/gK Cm b. 0.72 J/g K c. 0.93 J/g K d. 1.0 J/gK e) 1.3 J/gK Calculate the amount of heat required to change 35.0 g ice at -25.0°C to steam at 125 C. (Heat of fusion- 333 J/g: heat of vaporization-2260 Js; specific heats: ice 2.09 J/g K, water 4.18 J/g K, steam-1.84 J/g K) 112 2 ip ks heat of vaperiation -2260 1a- a. 22.0 kJ b. 90.9 kJ 109 kJ 276 kJ e 3290 kJ C H 12.01x4: 1.01 xi0 8.04 10.10 58.14 =MM 3. The standard molar enthalpy ofcombustion of butane is-2877kJ What is the enthalpy change for the combustion of 15.00 g CaHo? 160°allmol 281|L=-742.5 a. 4315kJ -1114k/ -742.5 kJ 491.2 k 58-14 e. 4 Ir 132 g MgoO is combined with 100.0 mL, of 100 MHCI in a coffee cup calorimeter, the temperature of the n increases from 24.2 C to 34.4 C. Calculate the enthalpy change for the reaction per mole of MgO. Assume that the specific heat of the HCI solution is 4.18 J/g K and its density is 1.00 g/mL 409 kJ mol d 3273m 100.0 ml x 1:0009 100.09 mu T-614-24.22 10.2 4264Explanation / Answer
2)
Ti = -25.0 oC
Tf = 125.0 oC
here
Cs = 2.09 J/g.oC
Heat required to convert solid from -25.0 oC to 0.0 oC
Q1 = m*Cs*(Tf-Ti)
= 35 g * 2.09 J/g.oC *(0--25) oC
= 1828.75 J
Lf = 333.0 J/g
Heat required to convert solid to liquid at 0.0 oC
Q2 = m*Lf
= 35.0g *333.0 J/g
= 11655 J
Cl = 4.18 J/g.oC
Heat required to convert liquid from 0.0 oC to 100.0 oC
Q3 = m*Cl*(Tf-Ti)
= 35 g * 4.18 J/g.oC *(100-0) oC
= 14630 J
Lv = 2260.0 J/g
Heat required to convert liquid to gas at 100.0 oC
Q4 = m*Lv
= 35.0g *2260.0 J/g
= 79100 J
Cg = 1.84 J/g.oC
Heat required to convert vapour from 100.0 oC to 125.0 oC
Q5 = m*Cg*(Tf-Ti)
= 35 g * 1.84 J/g.oC *(125-100) oC
= 1610 J
Total heat required = Q1 + Q2 + Q3 + Q4 + Q5
= 1828.75 J + 11655 J + 14630 J + 79100 J + 1610 J
= 108823.75 J
= 109 KJ
Answer: c