Answer: This is a typical acid/base equilibrium problem, that involves the use of logarithms.
Explanation:We assume that both nitric acid and hydrochloric acid dissociate to give stoichiometric
H
3
O
+
.
Moles of nitric acid:
26.0
×
10
−
3
⋅
L
×
8.00
⋅
m
o
l
⋅
L
−
1
=
0.208
⋅
m
o
l
H
N
O
3
(
a
q
)
.
And, moles of hydrochloric acid:
88.0
×
10
−
3
⋅
L
×
5.00
⋅
m
o
l
⋅
L
−
1
=
0.440
⋅
m
o
l
H
C
l
(
a
q
)
.
This molar quantity is diluted to
1.00
L
. Concentration in moles/Litre =
(
0.208
+
0.440
)
⋅
m
o
l
1
L
=
0.648
⋅
m
o
l
⋅
L
−
1
.
Now we know that water undergoes autoprotolysis:
H
2
O
(
l
)
⇌
H
+
+
O
H
−
. This is another equilibrium reaction, and the ion product
[
H
+
]
[
O
H
−
]
=
K
w
. This constant,
K
w
=
10
−
14
at
298
K
.
So
[
H
+
]
=
0.648
⋅
m
o
l
⋅
L
−
1
;
[
O
H
−
]
=
K
w
[
H
+
]
=
10
−
14
0.648
=
?
?
p
H
=
−
log
10
[
H
+
]
=
−
log
10
(
0.648
)
=
?
?
Alternatively, we know further that
p
H
+
p
O
H
=
14
. Once you have
p
H
,
p
O
H
is easy to find. Take the antilogarithm of this to get
[
O
H
−
]
.
Answer link
Answer:
<u>132.15</u>
Explanation:
Molar mass N = 14.00
Molar mass H = 1.01
Molar mass H4 = 1.01 x 4 = 4.04
Molar mass NH4 = 14.00 + 4.04 = 18.04
Molar mass (NH4)2 = 18.04 x 2 = 36.08
Molar mass S = 32.07
Molar mass O = 16.00
Molar mass O4 = 16.00 x 4 = 64.00
Molar mass SO4 = 32.07 + 64.00 = 96.07
Molar mass (NH4)2SO4 = 36.08 + 96.07 = <u>132.14</u>
Heat
gained or loss in a system can be calculated by multiplying the given mass to the
specific heat capacity of the substance and the temperature difference. It is
expressed as follows:<span>
Heat = mC(T2-T1)
When two objects are in contact,
it should be that the heat lost is equal to what is gained by the other. So, the heat released by the lead is equal to the heat that is absorbed by the water.
</span>Heat = mC(T2-T1) = 50.0 mL (1.00 g/mL) (4.18 J/g °C) (20 °C - 18 °C) = 418 J<span>
</span>
Oxygen has to be involved when methanol is ignited
