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1.45g KCl ×
.250 𝑘𝑔 𝐾𝑁𝑂3
.50 𝑔
= .019mol KCl
1
×
.45 L
= .04 M
× 100% = .07%
(12M)(Vd)=(.75L)(.25M)
𝐾𝐶𝑙
.019mol
× 1𝐸6 = 166.7 𝑃𝑃𝑀
1500.250 𝑘𝑔
(750𝑔+.50𝑔)
1 mol
74.55 g
74.55𝑔
1𝑚𝑜𝑙
×
3𝑚𝑜𝑙
1𝐿
×
2.0𝐿
1
1.56x10-2L
= 447.3𝑔𝐾𝐶𝑙
KClO3
150g
190g
unsaturated
20g
NaNO3
The minimum energy (kinetic and bond energy)
required by the reactants in order to become
products.
Collision theory states that molecules must collide in
order for a reaction to occur. The molecules must
have sufficient minimum kinetic energy as well as
correct orientation.
Increased surface area allows for more collisions,
increased temperature means the collisions are more
forceful and more likely to break bonds, and Stirring
removes newly dissolved particles from the solid
surface and continually exposes the surface to fresh
solvent.
Dynamic equilibrium refers to reactions in which the
rate of the forward reaction is the same as the reverse
and the concentrations of reactants and products is
constant.
A reaction in which the product molecules collide and
break apart and then create the original reactant
molecules
Rates are equal.
Concentrations no longer change.
This is an equilibrium graph showing how the
concentration of the reactants decrease over time and
the products increase over time. The point in time
when the lines are parallel is when dynamic
equilibrium has been reached.
LeChatelier
The system (reaction) shifts (speeds up) either left or
right in order to decrease the pressure. The system
shifts to the side with fewer moles of molecules.
The system shifts in the direction in order to increase
the temperature, the side with the energy. To the
right if its exothermic and to the left if endothermic.
No change, because the forward and reverse
reactions are increased simultaneously.
⇆
The system (reaction) shifts (speeds up) in the
direction to minimized the stress.
Changing the volume affects the pressure inside a
closed system. Decrease the volume and pressure
increases.
[NO2 ]4 [H 2O]6
Keq =
[NH 3 ]4 [O2 ]7
2
Write equilibrium expression: K = [NH 3 ]
eq
[N 2 ][H 2 ]3
Insert concentrations in M into the expression
[1.2]2
K eq =
= .72
[2][1]3
29. reactants, left
30. products, right
31. a. temperature- rxn. speeds up, more collisions
b. concentration- rxn. speeds up, more
collisions
c. surface area- rxn. speeds up, more collisions
d. catalyst- no effect
32. N2(g) + 3H2(g)
2NH3(g) + energy
Stress
Equilibrium
Shift
Add N2
right
Add H 2
right
Remove N2
left
decrease pressure
left
decrease temp.
right
[N2]
¯
[H2]
[NH3]
¯
¯
¯
-
¯
¯
33. check your solution notes bottom of page 1.
34. “Like dissolves like”. Nonpolar solutes require
nonpolar solvents to dissolve, the intermolecular
bonding must be similar. In the case of nonpolar
molecules it would be London Dispersion forces
causing them to attract.
35. Gases are less soluble because at increased
temperatures they are moving faster and can more
easily go from being dissolved in the liquid state to a
gas. They “escape” the attraction of the solvent.
36. Gases are more soluble at higher pressures because
of equilibrium: H2O(l) « H2O(g). If the stress is an
increase of pressure than the system shifts to the side
that will provide a lower pressure, that is to the liquid
side.
37. Intermolecular forces, temperature, and pressure.
38. pH<7, electrolytes, turn blue litmus paper red, react
with metals to produce H2 gas.
39. pH>7, electrolytes, turn red litmus paper blue, bitter
taste.
A
B
CA
CB
40. a) H2SO4(s) + H2O(l) ↔ H3O(aq+ + HSO4– (aq)
B
b)
c)
A
CA
CH3COO–(aq)
+ H3O
A
B
NH2NH3+
+
+ H2O ↔
(aq)
CB
↔ HCH3COO(aq) + H2O(aq)
CA
H3O(aq+
CB
+ NH2NH2(aq)
41. A strong acid completely dissociates when it is in
solution. It has a very weak conjugate base. HCl,
H2SO4, HNO3.
A weak acid only partially dissociates in solution. Ex:
acetic acid HC2H3O2.
42. A strong electrolyte creates a solution that creates
many charge carriers that can conduct electricity.
They usually completely dissociate. Weak
electrolytes partially dissociate and conduct electricity
poorly.
43. a, f-h, k are bases; b-e, I are acids
H 2O
44. KOH(s) « K+(aq) + OH–(aq)
HC2H3O2(g) + H2O(l) « H3O+(aq) + C2H3O2–(aq)
HCOOH(g) + H2O(l) « H3O+(aq) + COOH–(aq)
45.
Base
I–
SO32–
PO43–
C2H3O2–
Conjugate Acid
HI_____
__HSO3–_
__HPO42-_
_ HC2H3O2_
Acid
HClO4
H2S
HCO3–
H2SO4
Conjugate Base
___ClO4–_
___HS–__
___CO32-_
___HSO4–_
46.
2HClaq) + Mg(OH)2(aq)  MgCl2(aq) + 2H2O(l)
H2CO3(aq) + 2KOH(aq)  K2CO3(aq) + 2H2O(l)
H3PO4(aq) + 3NaOH(aq)  Na3PO4(aq) + 3H2O(l)
47. Acid strength is measured using pH and it’s related to
the % dissociation of the acid. Acid concentration is
measured using molarity and it’s the # of moles per
Liter.
48. What remains of the base after it has accepted a
hydrogen ion from the acid.
49. What remains of the acid after it has donated it’s
hydrogen ion to the base.
50. H2O(l) + H2O(l) « H3O(aq) + OH–(aq)
51. Keq=1.0x10-14 = [H+] [OH–]
52. pH = -log[1x10-5]
=5
53. 1x10-14=[H+][OH-]
1x10-14=[H+][1x10-12]
solve for H+
[H+]= .01M, use pH equation
pH=-log[.01] =2
54. pH=-log[8.5 x 10-3]
= 2.07
55. You could solve like question 53 or solve for pOH
first, then use pH +pOH=14.
pOH=-log[2.8x10-12] = 11.55
pH + 11.55=14
2.45
56. 12.75=-log[H+], multiply both sides by (-1), take
inverse log of both sides. pH= 1.78
57. use: 11.63 + pOH=14
pOH=2.37, next use: pOH=-log[OH–]
2.37=-log[OH–], solve for OH– = 4.3x10–3M