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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Answer:
I
Explanation:
The complete question can be seen in the image attached.
We need to understand what is actually going on here. In the first step that yields product A, the sodamide in liquid ammonia attacks the alkyne and abstracts the acidic hydrogen of the alkyne. The second step is a nucleophilic attack of the C6H5C≡C^- on the alkyl halide to yield product B (C6H5C≡C-CH3CH2).
Partial reduction of B using the Lindlar catalyst leads to syn addition of hydrogen to yield structure I as the product C.