Answer:
oxidation occurs at the cathode.
Explanation:
In a voltaic cell electrons move from anode to cathode. At the anode, species give up electrons. This is an oxidation reaction depicted by the oxidation half equation. At the cathode, species accept electrons and become reduced. This is depicted by the reduction half equation. In summary; in a Voltaic cell, oxidation occurs at the anode while reduction occurs at the cathode.
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.
Answer: C
Explanation:
he never had evidence in the first place that was nearly enough.
Answer:
a. 1.21M
b. 0.119M
c. 0.00496M
Explanation:
Molarity, M, is an unit of concentration defined as the ratio between moles of solute and liters of solution:
a. 4.35 mol LiCl / 3.60L = 1.21M
b. 29.43gC6H12O6 * (1mol / 180.16g) = 0.1634moles / 1.37L = 0.119M
<em>Molar mass C6H12O6: 180.16g/mol</em>
c. 34.5mg NaCl = 0.0345g * (1mol / 58.44g) = 5.9x10⁻⁴moles / 0.1191L = 0.00496M