Answer:
305 litres of NO gas will be produced from 916 L of NO₂
Explanation:
Given the balanced equation of the chemical reaction as follows:
3 NO₂ (g) + H₂O( l) —— 2 HNO₃ (l) + NO (g)
Under standard conditions, 3 moles of No₂ will react with 1 mole of water to produce 1 mole of NO gas.
Molar volume of a gas at STP is 22.4 L
Number of moles of NO₂ gas present in 916 L = 916/22.4 = 40.893 moles of NO₂ gas
From the mole ratio of NO₂ to NO in the equation of reaction,
Number of moles of NO that will be produced = 1/3 × 40.893 moles = 13.631 moles of NO gas
Volume of 13.631 moles of NO gas = 13.631 × 22.4
Volume of NO gas produced = 305.334L
Therefore, Volume of NO gas produced from the reaction of 916 L of NO₂ with water = 305 L
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
Explanation:
a) HNO2(aq) = HNO3(aq) + H2O(l) +NO(g)
b) SoCl2 (l) + H2O (l) = So2(g) + 2HCl(aq)
c) CH4 (g) + 2O2(g) = Co2 (g) + 2H2O(g)
d) 3CuO(s) + 2NH3 (g) = 3Cu(s) + 3H2O (l) + N2(g)
Answer:
A chemist searches for new knowledge about chemicals and use it to improve the way we live. He or she may develop products such as synthetic fibers, drugs and cosmetics. Chemists create processes, including all refining and petrochemical processing, that reduce energy use and pollution.
Radio active decay reactions follow first order rate kinetics.
a) The half life and decay constant for radio active decay reactions are related by the equation:
Where k is the decay constant
b) Finding out the decay constant for the decay of C-14 isotope:
c) Finding the age of the sample :
35 % of the radiocarbon is present currently.
The first order rate equation is,
![[A] = [A_{0}]e^{-kt}](https://tex.z-dn.net/?f=%20%5BA%5D%20%3D%20%5BA_%7B0%7D%5De%5E%7B-kt%7D%20%20%20)
![\frac{[A]}{[A_{0}]} = e^{-kt}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BA%5D%7D%7B%5BA_%7B0%7D%5D%7D%20%3D%20e%5E%7B-kt%7D%20%20)
t = 7923 years
Therefore, age of the sample is 7923 years.
