Answer:
The final volume will be 24.7 cm³
Explanation:
<u>Step 1:</u> Data given:
Initial temperature = 180 °C
initial volume = 13 cm³ = 13 mL
The mixture is heated to a fina,l temperature of 587 °C
Pressure and amount = constant
<u>Step 2: </u>Calculate final volume
V1/T1 = V2/T2
with V1 = the initial volume V1 = 13 mL = 13*10^-3
with T1 = the initial temperature = 180 °C = 453 Kelvin
with V2 = the final volume = TO BE DETERMINED
with T2 = the final temperature = 587 °C = 860 Kelvin
V2 = (V1*T2)/T1
V2 = (13 mL *860 Kelvin) /453 Kelvin
V2 = 24.68 mL = 24.7 cm³
The final volume will be 24.7 cm³
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
Roman numerals are used in naming ionic compounds when the metal cation forms more than one ion. The metals that form more than one ion are the transition metals, although not all of them do this.
Answer:
1. 4
2. 2
3. 3
4. 1
Explanation: just did it on edge 2021
