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melomori [17]
3 years ago
8

Please help it’s due today. I will give brainliest.

Chemistry
2 answers:
givi [52]3 years ago
8 0

Answer:

B.both conserve mass.

Tcecarenko [31]3 years ago
7 0

Answer:

both conserve mass

Explanation:

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How many amps are required to produce 75. 8 g of iron metal from a solution of aqueous iron(iii)chloride in 6. 75 hours?
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The amount of current required to produce 75. 8 g of iron metal from a solution of aqueous iron (iii)chloride in 6. 75 hours is 168.4A.

The amount of Current required to deposit a metal can be find out by using The Law of Equivalence. It states that the number of gram equivalents of each reactant and product is equal in a given reaction.

It can be found using the formula,

m = Z I t

where, m = mass of metal deposited = 75.8g

            Z = Equivalent mass / 96500 = 18.6 / 96500 = 0.0001

             I is the current passed

              t is the time taken = 75hour = 75 × 60 = 4500s

On subsituting in above formula,

75.8 = E I t / F

⇒ 75.8 = 0.0001 × I × 4500

⇒ I = 168.4 Ampere (A)

Hence, amount of current required to deposit a metal is 168.4A.

Learn more about Law of Equivalence here, brainly.com/question/13104984

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2 years ago
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