Answer:
The given reaction is a combustion reaction of benzene,
C
6
H
6
. From its balanced chemical equation,
2
C
6
H
6
+
15
O
2
→
12
C
O
2
+
6
H
2
O
,
the mass of carbon dioxide
(
C
O
2
)
produced from 20 grams (g) of
C
6
H
6
is determined through the molar mass of the two compounds, given by,
M
M
C
O
2
=
44.01
g
/
m
o
l
M
M
C
6
H
6
=
78.11
g
/
m
o
l
and their mole ratio:
12
m
o
l
C
O
2
2
m
o
l
C
6
H
6
→
6
m
o
l
C
O
2
1
m
o
l
C
6
H
6
With this,
m
a
s
s
o
f
C
O
2
=
(
20
g
C
6
H
6
)
(
1
m
o
l
C
6
H
6
78.11
g
C
6
H
6
)
(
6
m
o
l
C
O
2
1
m
o
l
C
6
H
6
)
(
44.01
g
C
O
2
1
m
o
l
C
O
2
)
=
(
20
)
(
6
)
(
44.01
)
g
C
O
2
78.11
=
5281.2
g
C
O
2
78.11
m
a
s
s
o
f
C
O
2
=
67.6
g
C
O
2
Therefore, the mass in grams of
C
O
2
formed from 20 grams of
C
6
H
6
is
67.6
g
C
O
2
.
it is a problem of app
The gases that get released form bubbles in the solution
Answer:
We can't see the options so we don't know what we can put
Explanation:
of oxygen at STP would be required to react completely with 38.8g of propane.
<u>Given that :</u>
molar mass of propane = 44 g/mol
mass of propane = 38.8 g
∴ Moles present in 38.8 g of propane =
= 0.88 mole
<u>applying rule of balanced equations </u>
1 mole of propane = 5 moles of oxygen
0.88 mole of propane = 5 * 0.88 = 4.4 moles of oxygen
Note : volume of 1 mole of oxygen at STP = 
∴Total volume of oxygen required at STP = 22.4 * 4.4 = 
Hence we can conclude that the volume of oxygen at STP required to react completely 
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Answer:
Brown color of the solution decreases
Explanation:
is brown in color whereas
is colorless.
Equilibrium reaction between
and
is as follows:

As per the Le Chatelier's principle, if pressure of a equilibrium is increased, the equilibrium will shift in the direction having fewer no. of moles of gases.
In the given equilibrium,
side has more no. of moles. So on increasing pressure, equilibrium will shift towards the side of
or more formation of
will take place.
Therefore, more
will decompose that will decrease the brown color of the solution as
is colorless.