Explanation:
The given data is as follows.
Current (I) = 3.50 amp, Mass deposited = 100.0 g
Molar mass of Cr = 52 g
It is known that 1 faraday of electricity will deposit 1 mole of chromium. As 1 faraday means 96500 C and 1 mole of Cr means 52 g.
Therefore, 100 g of Cr will be deposited by "z" grams of electricity.

z = 
= 185576.9 C
As we know that, Q = I × t
Hence, putting the given values into the above equation as follows.
Q = I × t
185576.9 C =
t = 53021.9 sec
Thus, we can conclude that 100 g of Cr will be deposited in 53021.9 sec.
The answer is 4.69 x 10⁻¹⁹ I hope this helped!
D. Making the reactant particles larger
Answer:
+5
Explanation:
The oxidation number of phosphorus can be obtained as follows:
H4P2O7 = 0
4(+1) + 2P + 7(—2) = 0
4 + 2P —14 = 0
Collect like terms
2P = 14 — 4
2P = 10
Divide both side by 2
P = 10/2
P = +5
The oxidation number of phosphorus is +5
Rows are periods, columns are groups