Assuming it's a perfect gas, we have PV=nRT hence if T goes down, V goes down up. The volume will decrease.
A first-order reaction is 81omplete in 264s.The half-life for this reaction (i) t 1/2 = =3.465×10 −3 s.to reach 95% Completion = 285 s.
To measure reaction rates, chemists initiate the reaction, measure the concentration of the reactant or product at different times as the reaction progresses,
For a 0-order response, the mathematical expression that may be employed to determine the half of life is: t1/2 = [R]0/2k. For a first-order reaction, the half of-existence is given by: t1/2 = zero.693/ok. For a 2d-order response, the method for the half-life of the response is: 1/okay[R]0
The 1/2-life of a response (t1/2), is the quantity of time needed for a reactant concentration to lower via half of compared to its initial awareness. Its software is used in chemistry and medicine to are expecting the awareness of a substance over time
Half of the lifestyles is the time required for exactly 1/2 of the entities to decay 50%.
Learn more about first order reaction here:-
#SPJ4
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
Every atom tends to form configuration of noble gas , with the 8 electrons in valence shell.
Answer:
D
Explanation:
This creates a gap that we call an oceanic trench
:)
