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Answer:0.58 atm
Explanation:see attached photo
<h3>
Answer:</h3>
266.325 g
<h3>
Explanation:</h3>
We are given the balanced equation;
2NaOH + H₂SO₄ → H₂O + Na₂SO₄
We are required to determine the mass of Na₂SO₄ that will be formed.
<h3>Step 1: Determine the number of moles of NaOH</h3>
Moles = Mass ÷ molar mass
Molar mass of NaOH is 40.0 g/mol
Therefore;
Moles of NaOH = 150 g ÷ 40 g/mol
= 3.75 moles
<h3>Step 2: Determine the number of moles of sodium sulfate formed</h3>
- From the equation 2 moles of NaOH reacts with sulfuric acid to form 1 mole of sodium sulfate.
- Therefore; mole ratio of NaOH : Na₂SO₄ is 2 : 1
Thus, moles of Na₂SO₄ = Moles of NaOH ÷ 2
= 3.75 moles ÷ 2
= 1.875 moles
<h3>Step 3: Determine the mass of Na₂SO₄ produced.</h3>
we know that;
Mass = Moles × Molar mass
Molar mass of Na₂SO₄ is 142.04 g/mol
Therefore;
Mass of Na₂SO₄ = 1.875 moles × 142.04 g/mol
= 266.325 g
Thus, the mass of sodium sulfate formed 266.325 g
<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
