potassium belongs to group IA of the periodic table.
Answer:
A
C
A
Explanation:
Iam not sure if it's correct
Explanation:
A compound is defined as follows.
- Compounds are pure substance.
- The atoms bonded in a compound are in specific proportion.
- A compound is formed by chemical combination.
- For a compound, melting point and boiling point is defined.
A mixture is defined as follows.
- Mixtures are impure substance.
- The bonded atoms in a mixture are in any proportion.
- A mixture is formed by physical combination.
- For a mixture, melting point and boiling point is not defined.
On changing the amount of one substance will affect the formation of compound as a compound requires atoms to be bonded in a specific proportion. Whereas changing the amount of one substance will not affect the formation of mixtures as atoms can be bonded in any proportion in a mixture.
Hey There!:
2 C2H2 + 5 O2 = 4 CO2 + 2 H2O
2 moles C2H2 ----------- 5 moles O2
35.0 moles C2H2 ------- moles O2
moles O2 = 35.0 * 5 / 2
moles O2 = 87.5 moles
hope this helps!
<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
