.7. How much energy is needed to change the temperatue of 25.00 ml of water from
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.7. How much energy is needed to change the temperatue of 25.00 ml of water from 100 Ct95 c ID: -4.184 Jrg OC),d-1.00 g/ml.) a. 160 k b. 2880 k d. 64 8.9k What is the change in internal energy (AE) when a system is heated with 35 J of energy a. +50 b. +20J c. -20 105k 38. while it does 15 3 of work? d. 50J e +35K 39 What is the change in internal energy of a system when it loses 10 kJ of energy and 5000 J of work is done on the system? a. +15 k b.-10k d. +5010 J +4990 for this proces? 40. A pot of water is heated with 10 J of energy at constant pressure. What is the enthalpy change (AJr) a. -10 c +10 Answer following questions in details and show each step of calculation 41. (20pts) 50 g NaCI is added to 6.0 kg of H,O. 1) What is the mole fraction of NaCi? 2) What is molality? 3) What is weight percent of NaCI? 4) If density of solution is 1.2 g/ml. What is molarity of the solution? How much Cacl, is needed to 7.5kg of water to lower the freezing point of water from o°C to -20°C? Kp -1.86 °c/m. (10 pts). 42. 43. Gold crystalizes in FCC (Face Centered Cubic). rAu radius- 144 pm. Calculate 1) length of edge? 2) volume of unit cell? 3) how many Au in each unit cell? 4) density? (20pts) For NaCI solution, molality is 0.25, what is its weight percent and mole fraction ? (10 pts) 44.Explanation / Answer
37)
Given:
volume of water = 25.00 mL
Since density is 1.00 g/mL, mass of water will be 25.00 g
m = 25.00 g
C = 4.184 J/g.oC
Ti = 10 oC
Tf = 95 oC
use:
Q = m*C*(Tf-Ti)
Q = 25.00*4.184*(95.0-10.0)
Q = 8891 J
Q = 8.9 KJ
Answer: c
38)
Q = 35 J (positive because system absorbs heat)
W = -15 J (negative because system does work)
use:
delta U = Q + W
= 35 J - 15 J
= 20 J
Answer: b
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