Each of the following is a Lewis acid-base reaction. Which reactant is the acid, and which is the base? Explain. (a) SO 3 + H 2 O → H 2 SO 4 (b) Zn ( OH ) ( s ) +2OH - ( aq ) → [ Zn ( OH ) 4 ] 2- ( aq )
Each of the following is a Lewis acid-base reaction. Which reactant is the acid, and which is the base? Explain. (a) SO 3 + H 2 O → H 2 SO 4 (b) Zn ( OH ) ( s ) +2OH - ( aq ) → [ Zn ( OH ) 4 ] 2- ( aq )
Solution Summary: The author explains the Lewis structure of sulfurtrioxide, which is able to accept electrons from other chemical species known as Lewis base.
Each of the following is a Lewis acid-base reaction. Which reactant is the acid, and which is the base? Explain. (a)
SO
3
+
H
2
O
→
H
2
SO
4
(b)
Zn
(
OH
)
(
s
)
+2OH
-
(
aq
)
→
[
Zn
(
OH
)
4
]
2-
(
aq
)
5
What would the complete ionic reaction be if aqueous solutions of potassium sulfate and barium acetate were mixed?
ed
of
Select one:
O a
2 K SO4 + Ba2 +2 C₂H3O21
K+SO4 + Ba2+ + 2 C2H3O21
K+SO42 + Ba2 +2 C2H3O2
BaSO4 +2 K+ + 2 C2H3O
estion
Ob.
O c.
Od.
2 K SO4 +Ba2 +2 C₂H₂O₂
BaSO4 + K+ + 2 C2H3O
BaSO4 + K + 2 C2H301
→Ba² +SO42 +2 KC2H3O
s page
(28 pts.) 7. Propose a synthesis for each of the following transformations. You must include the
reagents and product(s) for each step to receive full credit. The number of steps is provided.
(OC 4)
4 steps
4 steps
OH
b.
LTS
Solid:
AT=Te-Ti
Trial 1
Trial 2
Trial 3
Average
ΔΗ
Mass water, g
24.096
23.976
23.975
Moles of solid, mol
0.01763
001767
0101781
Temp. change, °C
2.9°C
11700
2.0°C
Heat of reaction, J
-292.37J -170.473
-193.26J
AH, kJ/mole
16.58K 9.647 kJ 10.85 kr
16.58K59.64701
KJ
mol
12.35k
Minimum AS,
J/mol K
41.582
mol-k
Remember: q = mCsAT (m = mass of water, Cs=4.184J/g°C) & qsin =-qrxn &
Show your calculations for:
AH in J and then in kJ/mole for Trial 1:
qa (24.0969)(4.1845/g) (-2.9°C)=-292.37J
qsin =
qrxn =
292.35 292.37J
AH in J = 292.375 0.2923kJ
0.01763m01
=1.65×107
AH in kJ/mol =
=
16.58K
0.01763mol
mol
qrx
Minimum AS in J/mol K (Hint: use the average initial temperature of the three trials, con
Kelvin.)
AS=AHIT
(1.65×10(9.64×103) + (1.0
Jimai
Chapter 16 Solutions
General Chemistry: Principles and Modern Applications (11th Edition)
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell