Answer:
<h3>The answer is 2.75 × 10²⁴ molecules</h3>
Explanation:
The number of molecules of CO2 can be found by using the formula
<h3>N = n × L</h3>
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
N = 4.56 × 6.02 × 10²³
We have the final answer as
<h3>2.75 × 10²⁴ molecules</h3>
Hope this helps you
Answer:
It is required 0.3 molecules of O2 to produce 0.6 molecules of H2O
Explanation:
The water´s formation reaction is: 2H2 + O2 --> 2H2O
This reaction shows the molecular O2:H2O ratio: 1:2.
Then, if we want to know how many molecules of O2 are required to produce 0.6 of H20, it is necessary calculate as it showed next:
0.6 molecules H2O * (1 molecula O2 / 2 molecules H2O)= 0.3 molecule O2
Oxygen. argon.nitrogen.carbon dioxide
Answer:
Pb²⁺(aq) --> Ni²⁺(aq)
Explanation:
The equation of the reaction is given as;
Ni(s) + Pb(NO₃)₂(aq) --> Pb(s) + Ni(NO₃)₂(aq)
In writing the ionic equation, we break the aqueous compound into ions. The solid and liquid compounds are ignored.
We have;
Pb²⁺(aq) + NO₃²⁻(aq) --> Ni²⁺(aq) + NO₃²⁻(aq)
Canceling the spectator ions;
Pb²⁺(aq) --> Ni²⁺(aq)
<u>Answer:</u> The cell potential of the cell is +0.118 V
<u>Explanation:</u>
The half reactions for the cell is:
<u>Oxidation half reaction (anode):</u> 
<u>Reduction half reaction (cathode):</u> 
In this case, the cathode and anode both are same. So, 
will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Ni^{2+}_{diluted}]}{[Ni^{2+}_{concentrated}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BNi%5E%7B2%2B%7D_%7Bdiluted%7D%5D%7D%7B%5BNi%5E%7B2%2B%7D_%7Bconcentrated%7D%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 2
= ?
![[Ni^{2+}_{diluted}]](https://tex.z-dn.net/?f=%5BNi%5E%7B2%2B%7D_%7Bdiluted%7D%5D)
= 
![[Ni^{2+}_{concentrated}]](https://tex.z-dn.net/?f=%5BNi%5E%7B2%2B%7D_%7Bconcentrated%7D%5D)
= 1.0 M
Putting values in above equation, we get:
Hence, the cell potential of the cell is +0.118 V
