### Chemistry Chapter 6 Answer Key

```Chapter 6 Answers
Page 195
#1. Use the tables on page 420 (heat of dissolution) and
questions.
a) HClO3(s) H+(aq) +ClO3-(aq)
ΔH=
b) HI(s)  H+(aq) + I-(aq)
ΔH=
c) KNO3(s)K+(aq) +NO3-(aq)
ΔH=
d) CuSO4(s) Cu2+(aq) + SO42-(aq) ΔH=
e) Li2CO3(s)  2 Li+(aq) + CO3 2-(aq) ΔH=
-42.0
-30.0
+36.0
-68.0
-13.0
exothermic
exothermic
endothermic
exothermic
exothermic
Page 195, #2
Step 1: From the problem we identify that the mass of
water is 20g, the ΔT is 50°C and the c is 4.184
DATA
Step 2: Find Q
Q=mwcw ΔT
Calorimeter:
=20 g x4.184 J/g°C x 50°C
mw=20g
=4184.0 J
cw=4.184 J/g°C
Step 3: Calculate the number of moles of the solute
ΔT=50°C
n=ms / Ms
=4g / 39.99g/mol
Solute:
=0.100 mol
mS = 4g
Step 4: Convert Joules to kiloJoules and determine sign
MS= 39.99g/mol
4184.0 J = 4.184 kJ
The reaction is exothermic so the sign is negative
To find:
n=
Q=
ΔH=
Step 5: Calculate ΔH
ΔH= Q / n
= -4.184 kJ / 0.100 mol
= -41.8 kJ/mol
Page 195 #3
Step 1: calculate heat...
Q
=
mwcwΔT
mw=1000g
=
1000g x 4.184 J/g°C x -2°C
Cw=4.184
=
-8368 J
ΔT=-2°C
Step 2: Determine the sign... And change to kJ
ms=24.42g
endothermic, so positive  +8.368 kJ
Ms=122.5
Step 3: Find the number of moles of KClO3
ns
=
ms / Ms
find
=
24.42g / 122.5 g.mol-1
ns=
=
0.1993 mol
Step 4: Find ΔHd =
Q / ns = 8.368 kJ / 0.199 mol
=41.9 kJ/mol
Data
Page 195 #4
• Dissolution takes place in two steps. In the
first step, solute particles are separated from
each other. This process is endothermic
(absorbs energy from the water, positive
enthalpy change). The second step the solute
particles are distributed through the solvent.
This step is exothermic (releases energy,
negative enthalpy change).
Page 196 #5
Data
mw=1000g
Cw=4.184
ΔT=5.5°C
n= 1 mol
Step 1: Calculate heat...
Q=mwcwΔT = 1000g x 4.184 J/g°C x 5.5°C
Q=
= 23 012 J or 23.012 kJ
Step 2: Determine the sign... It’s positive
Step 3: find the ΔHd...
ΔHd = Q / n (but n=1, so that’s easy)
ΔHd = +23.012
Step 4: look it up on page 420...
The only salt with a ΔHd of about +23 is
AgNO3, (silver nitrate)
Data
mw=100g
Cw=4.184
Ti=20°C
ms=8g.
Ms=159.6
ΔHd=68.0
Find
ns=
ΔT=
Tf =
Page 196 #6
Step 1: find the number of moles of CuSO4
n=
ms / Ms
= 8g / 159.6 = 0.0501 mol
Step 2: calculate the heat using the table p. 420
Q= n x ΔHd
= 0.0501 mol x 68 kJ/mol
= 3.400 kJ or 3400 J
Step 3: calculate ΔT = Q / mwcw
= 3400 J / (100g•4.18 J/g°C)
= 8.13°C
Step 4: Calculate Tf
= 20 + 8.13 = 28.13°C
Round answer to a sensible number
Data
mw=200g
Cw=4.184
Ti=22
ms=8.42g.
Ms= 42.4
ΔHd=
Find
ns=
ΔT=
Tf =
Page 196 #7
Step 1: Find number of moles of solute
ns=ms / Ms = 8.42g / 42.4 g/mol
=0.1986 mol
Step 2: Calculate the heat from dissolution
Q = ΔHd x ns = -35 kJ/mol x 0.1986 mol
= -6.951 or -6951 J
Step 3: Determine effect (Exothermic = temperature up)
Step 4: Calculate temperature change, Q=mwcwΔT, so...
ΔT= Q / mc = 6951J / (200•4.184)
= 8.31°C
Step 5: Calculate Tf...
= 22° + 8.31 ° =30.31°C
Page 196, #8,
8. Step 1:
Step 2:
ΔT
Q
= 80°C – 25°C = 55°C
= mcΔT
= 100 (4.184)(55)
= 23 012 J or 23.012 kJ
Step 3: its exothermic, so it will be negative
Step 4: n
= Q / ΔH,
= -23.012 / -55.0 = 0.4184 mol
Step 5: m
= nM
= 0.4184 x 56.1
=23.47 g
The necessary amount is 23g of NaOH
Page 195, #9
Steps:
1. Calculate ΔT
ΔT = Tf – Ti
2. Calculate energy Q=mc ΔT
3. Determine sign
endothermic
4. Calculate moles, if the salt were NaCl
n = m/M
= -7°C
=-1464.4 J
ΔHd+
= 10/58.4
= 0.1712 mol
5. Calculate ΔHd
ΔHd= Q/n
= 4.3
Since ΔHd of NaCl is +4.3 kJ/mol, and the salt used absorbs
8.56 kJ/mol, it is not the same salt
Page 195, #10
Steps:
1. Find number of moles
n= C(mol/L)V(L)
2. Calculate heat transfer
Q=n ΔHd
3. Determine effect
endothermic
4. Calculate ΔT
ΔT = Q / mc
5. Find final temperature
The final temperature is 23.5°C