Answer:
CaCI2 + H2O = Ca(OH)2 + HCI
Answer:
Option A:
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
Explanation:
The half reactions given are:
Zn(s) → Zn^(2+)(aq) + 2e^(-)
Cu^(2+) (aq) + 2e^(-) → Cu(s)
From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).
While in the second half reaction, Cu^(2+) is reduced to Cu.
Thus, for the overall reaction, we will add both half reactions to get;
Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)
2e^(-) will cancel out to give us;
Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)
Answer:
9.63 L.
Explanation:
Hello,
In this case, the undergoing chemical reaction is:

So the consumed amounts of hydrochloric acid and bromine are the same to the beginning based on:

In such a way, the yielded moles of hydrobromic acid and chlorine are:

Thus, the volume of the sample, after the reaction is the same as no change in the total moles is evidenced, that is 9.63L.
Best regards.
<u>Answer:</u>
<u>For A:</u> The
for the given reaction is 
<u>For B:</u> The
for the given reaction is 1642.
<u>Explanation:</u>
The given chemical reaction follows:

The expression of
for the above reaction follows:

We are given:

Putting values in above equation, we get:

Hence, the
for the given reaction is 
Relation of
with
is given by the formula:

where,
= equilibrium constant in terms of partial pressure = 
= equilibrium constant in terms of concentration = ?
R = Gas constant = 
T = temperature = 500 K
= change in number of moles of gas particles = 
Putting values in above equation, we get:

Hence, the
for the given reaction is 1642.