Responder:
2H2 + O2 → 2H2O
CaO + H2O → Ca (OH) 2
Fe + S → FeS
H2SO3 → SO2 + H2O
CaCO3 → CaO + CO2
Explicación:
2H2 + O2 → 2H2O
2 moléculas de hidrógeno gaseoso reaccionan con oxigente para producir 2 moléculas de agua
CaO + H2O → Ca (OH) 2
El óxido de calcio reacciona con el agua para producir hidróxido de calcio.
Fe + S → FeS
El hierro reacciona con el azufre para producir sulfuro de hierro.
H2SO3 → SO2 + H2O
Por descomposición, el ácido sulfuroso se descompone para producir dióxido de azufre y agua.
CaCO3 → CaO + CO2
El carbonato de calcio se descompone para producir óxido de calcio y dióxido de carbono.
As far as I know NONE.....
Explanation:
Sodium peroxide can be thermolyzed to give dioxygen gas...
N
a
2
O
2
(
s
)
+
Δ
→
N
a
2
O
(
s
)
+
1
2
O
2
(
g
)
↑
⏐
⏐
⏐
But with water, we simply get an acid base reaction....
N
a
2
O
2
(
s
)
+
2
H
2
O
(
l
)
→
2
N
a
O
H
(
a
q
)
+
H
2
O
2
(
a
q
)
...
Well a solid and a liquid is completely different from a gas. For a gas you need to have an air tight seal in order for the gas to stay in, or a special type of container so the gas doesn't decompose the container. And a liquid or a solid it's self explanatory. With some liquids you need special containers to hold the liquid so it doesn't decompose the container it's in.
Explanation:
Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.
a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement a is wrong.
b. Expression for second order reaction is as follows:
![\frac{1}{[A]} =\frac{1}{[A]_0} +kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%20%2Bkt)
the above equation can be written in the form of Y = mx + C
so, the plot between 1/[A] and t is linear. So the statement b is true.
c.
Expression for half life is as follows:
![t_{1/2}=\frac{1}{k[A]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BA%5D_0%7D)
As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.
d.
Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong
