First, we write the half equations for the reduction of the chemical species present:
Cu⁺² + 2e → Cu; E° = 0.34 V
Ni⁺² + 2e → Ni; E° = - 0.23 V
In order to determine the potential of the cell, we find the difference between the two values. For this:
E(cell) = 0.34 - (-0.23)
E(cell) = 0.57 V
The second option is correct. (The difference in values is due to different values in literature, and it is negligible)
Answer:
Explanation:When silver Ag metal is added to copper sulphate CuSO4 solution, no reaction takes place as silver is less reactive than that of copper and cannot displace copper from its solution. Hence, when a piece of silver metal is added to copper sulphate solution there will be no reaction.
<u>Answer:</u> The
for the reaction is -1052.8 kJ.
<u>Explanation:</u>
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The given chemical reaction follows:

The intermediate balanced chemical reaction are:
(1)

(2)

The expression for enthalpy of the reaction follows:
![\Delta H^o_{rxn}=[1\times \Delta H_1]+[1\times (-\Delta H_2)]](https://tex.z-dn.net/?f=%5CDelta%20H%5Eo_%7Brxn%7D%3D%5B1%5Ctimes%20%5CDelta%20H_1%5D%2B%5B1%5Ctimes%20%28-%5CDelta%20H_2%29%5D)
Putting values in above equation, we get:

Hence, the
for the reaction is -1052.8 kJ.
The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL
<h3>Balanced equation </h3>
2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 3
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of acid, H₃PO₄ (Ma) = 0.390 M
- Volume of acid, H₃PO₄ (Va) = 24.5 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(0.39 × 24.5) / (0.279 × Vb) = 2/3
9.555 / (0.279 × Vb) = 2/3
Cross multiply
2 × 0.279 × Vb = 9.555 × 3
0.558 × Vb = 28.665
Divide both side by 0.558
Vb = 28.665 / 0.558
Vb = 51.4 mL
Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL
Learn more about titration:
brainly.com/question/14356286
Answer:
hydrgen = i think it is 4
oxygen = i think it is 3
Explanation: