Question
                A solution has  [OH−]
[
O
H
−
]
= 2.1×10−2 M
M
. Use the ion product constant of water
Kw=[H3O+][OH−]
K
w
=
[
H
3
O
+
]
[
O
H
−
]
 
to find the [H3O+]
[
H
3
O
+
]
of the solution.
            
        [
O
H
−
]
= 2.1×10−2 M
M
. Use the ion product constant of water
Kw=[H3O+][OH−]
K
w
=
[
H
3
O
+
]
[
O
H
−
]
to find the [H3O+]
[
H
3
O
+
]
of the solution.
Answers
                    Answer
            
    A 125 mL
m
L
bubble of hot gases at 213 ∘C
∘
C
and 1.87 atm
a
t
m
is emitted from an active volcano.
Part A
What is the final temperature, in degrees Celsius, of the gas in the bubble outside the volcano if the final volume of the bubble is 136 mL
m
L
and the pressure is 0.800 atm
a
t
m
and if the amount of gas does not change?
    
m
L
bubble of hot gases at 213 ∘C
∘
C
and 1.87 atm
a
t
m
is emitted from an active volcano.
Part A
What is the final temperature, in degrees Celsius, of the gas in the bubble outside the volcano if the final volume of the bubble is 136 mL
m
L
and the pressure is 0.800 atm
a
t
m
and if the amount of gas does not change?
                    Answer
            
    A sample of argon gas has a volume of 735 mL
m
L
at a pressure of 1.20 atm
a
t
m
and a temperature of 112 ∘C
∘
C
. What is the final volume of the gas, in milliliters, when the pressure and temperature of the gas sample are changed to the following, if the amount of gas does not change?
664mmHg and 274 K
    
m
L
at a pressure of 1.20 atm
a
t
m
and a temperature of 112 ∘C
∘
C
. What is the final volume of the gas, in milliliters, when the pressure and temperature of the gas sample are changed to the following, if the amount of gas does not change?
664mmHg and 274 K
                    Answered by
            Bot
            
    We know that:
Kw = [H3O+][OH−]
Plugging in the given value for [OH−]:
Kw = [H3O+](2.1×10−2 M)
Rearranging the equation to solve for [H3O+]:
[H3O+] = Kw/[OH−]
[H3O+] = (1.0×10−14)/(2.1×10−2 M)
[H3O+] = 4.76×10−13 M
Therefore, the [H3O+] of the solution is 4.76×10−13 M.
    
Kw = [H3O+][OH−]
Plugging in the given value for [OH−]:
Kw = [H3O+](2.1×10−2 M)
Rearranging the equation to solve for [H3O+]:
[H3O+] = Kw/[OH−]
[H3O+] = (1.0×10−14)/(2.1×10−2 M)
[H3O+] = 4.76×10−13 M
Therefore, the [H3O+] of the solution is 4.76×10−13 M.
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