Answer:
+1.03 V
Explanation:
The standard emf of the voltaic cell is the value of the standard potential of it, which is calculated by the standard reduction potential (E°).
The standard reduction potential is the potential needed for the reduction reaction happen, and it's determined by the reaction with the hydrogen cell (which has E° = 0.0V). The half-reactions of reduction of Ni⁺² and Ag⁺, are:
Ni⁺²(aq) + 2e⁻ → Ni(s) E° = -0.23 V
Ag⁺(aq) + e⁻ → Ag(s) E° = +0.80 V
The value is calculated by a spontaneous reaction, in which the cell with the greater E° is reduced (gain electrons), and the other is oxidized (loses electrons). So, Ag⁺ reduces.
emf = E°reduces - E°oxides
emf = 0.80 - (-0.23)
emf = +1.03 V
Answer:
co2
Explanation:
because carbon is a gas product
Answer:
(A) 4.616 * 10⁻⁶ M
(B) 0.576 mg CuSO₄·5H₂O
Explanation:
- The molar weight of CuSO₄·5H₂O is:
63.55 + 32 + 16*4 + 5*(2+16) = 249.55 g/mol
- The molarity of the first solution is:
(0.096 gCuSO₄·5H₂O ÷ 249.55 g/mol) / (0.5 L) = 3.847 * 10⁻⁴ M
The molarity of CuSO₄·5H₂O is the same as the molarity of just CuSO₄.
- Now we use the dilution factor in order to calculate the molarity in the second solution:
(A) 3.847 * 10⁻⁴ M * 6mL/500mL = 4.616 * 10⁻⁶ M
To answer (B), we can calculate the moles of CuSO₄·5H₂O contained in 500 mL of a solution with a concentration of 4.616 * 10⁻⁶ M:
- 4.616 * 10⁻⁶ M * 500 mL = 2.308 * 10⁻³ mmol CuSO₄·5H₂O
- 2.308 * 10⁻³ mmol CuSO₄·5H₂O * 249.55 mg/mmol = 0.576 mg CuSO₄·5H₂O
Answer:
Basic solution
Explanation:
A solution containing 0.005 M potassium hydroxide has a pH of;
pOH= -log[0.005]
pOH= 2.3
But pH+pOH=14
Therefore;
pH= 14-pOH
Hence;
pH= 14-2.3= 11.7
A solution having a pH of 11.3 is clearly a basic solution because this pH indicates a substance that is basic in nature. Hence the answer given.
