1. Instant cold packs, commonly used by athletic and medical professionals, cons
ID: 927394 • Letter: 1
Question
1. Instant cold packs, commonly used by athletic and medical professionals, consist of a bag of water inside another bag which contains ammonium nitrate. By squeezing the outer bag, the inner bag will break allowing the ammonium nitrate to dissolve in the water. A standard instant cold pack has approximately 15.0 g of water and 10.0 g of ammonium nitrate. What will be the temperature of the cold pack after breaking the inner bag if the cold pack was initially at 25 °C? Assume the solution has a heat capacity of 4.18 J/g·°C and that the solution does not undergo a phase change. NH4NO3(s) NH4 + (aq) + NO3 – (aq) H = +25.7 kJ (a) –26 °C (b) –6 °C (c) 25 °C (d) 56 °C (e) 76 °C
2. 100. mL of a 0.200 M silver nitrate solution is mixed with 100. mL of a 0.400 M sodium chloride solution. After precipitate is formed, [Ag+ ] = 1.8×10–9 M. What is the Ksp of AgCl? (a) 3.2×10–18 (b) 3.6×10–11 (c) 1.8×10–10 (d) 1.8×10–9 (e) 4.2×10–5
3. The rate of effusion of ammonia is 3.32 times faster than the rate of effusion of an unknown gas at 350 K. What is the molar mass of the unknown gas? (a) 31.0 g/mol (b) 45.5 g/mol (c) 56.5 g/mol (d) 112 g/mol (e) 188 g/mol
4. A solution of Ba(OH)2 is formed by adding Ba(OH)2 to 500. mL of water until no more solid will dissolve. What is the concentration of OH– in this solution? Ksp of Ba(OH)2 is 5.0×10–3 . (a) 0.071 M (b) 0.11 M (c) 0.17 M (d) 0.22 M (e) 0.34 M
5. Gas A is confined in a cylinder fitted with a movable piston and it spontaneously decomposes under an external pressure of 1.00 atm to produce gases B and C. As the decomposition occurs, the system takes in 1337 J of heat from the surroundings. The volume of the cylinder changes by 50.00 mL during this process. What is the change in internal energy of the system? (a) 1332 J (b) 1337 J (c) 1342 J (d) –1337 J (e) –1332 J
Explanation / Answer
1)
m = 15 g of water
m = 10 g of amm nitrate
MW NH4NO3= 80.04 mol
mol = mass/MW = 10/80.04 = 0.1249
Tf when
Ti= 25°C
Cp = 4.18
dHrxn = 25.7
Q =m*Cp*(Tf-Ti)
Q = Hrxn*mol = -0.1249*25.7 = -3.20993 kJ
Q = -320.993 J
Then
Q =-m*Cp*(Tf-Ti)
Tf = Q/(m*Cp) + Ti = -320.993 /(15+10)/(4.184) + 25 = 21.931 °C
Tf = 21.931
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