Answer:
cerium (iii) sulfate is less soluble
Vegetable soup should be the correct answer. The reason for this is that it can easily be split, where as the rest would be very difficault if not imposible because of chemical change.
<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
0.000132 g of hydrated sodium borate (Na₂B₄O₇ · 10 H₂O)
Explanation:
First we need to find the number of moles of sodium borate (Na₂B₄O₇) in the solution:
molar concentration = number of moles / volume (L)
number of moles = molar concentration × volume (L)
number of moles of Na₂B₄O₇ = 0.1 × 0.5 = 0.05 moles
We know now that we need 0.05 moles of hydrated sodium borate (Na₂B₄O₇ · 10 H₂O) to make the solution.
Now to find the mass of hydrated sodium borate we use the following formula:
number of moles = mass / molar weight
mass = number of moles × molar weight
mass of hydrated sodium borate = 0.05 / 381 = 0.000132 g
Learn more about:
molar concentration
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