It would go B. A. E. D. C.
Hope I helped!
Answer:
1.18 V
Explanation:
The given cell is:
Half reactions for the given cell follows:
Oxidation half reaction:
Reduction half reaction:
Multiply Oxidation half reaction by 2 and Reduction half reaction by 3
Net reaction:
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the
of the reaction, we use the equation:
Putting values in above equation, we get:
To calculate the EMF of the cell, we use the Nernst equation, which is:
where,
= electrode potential of the cell = ?V
= standard electrode potential of the cell = +1.21 V
n = number of electrons exchanged = 6
Putting values in above equation, we get:

Answer: the empirical formula is C3H4O3
Explanation:Please see attachment for explanation
Answer:
The answer is 465.6 mg of MgI₂ to be added.
Explanation:
We find the mole of ion I⁻ in the final solution
C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol
But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.
So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.
Hence, the weight of MgI₂ must be added is
Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg
25 g of NH₃ will produce 47.8 g of (NH₄)₂S
<u>Explanation:</u>
2 NH₃ + H₂S ----> (NH₄)₂S
Molecular weight of NH₃ = 17 g/mol
Molecular weight of (NH₄)₂S = 68 g/mol
According to the balanced reaction:
2 X 17 g of NH₃ produces 68 g of (NH₄)₂S
1 g of NH₃ will produce
g of (NH₄)₂S
25g of NH₃ will produce
of (NH₄)₂S
= 47.8 g of (NH₄)₂S
Therefore, 25 g of NH₃ will produce 47.8 g of (NH₄)₂S