Copper may be used for electroplating with a half reaction of cu2+ + 2e- > cu. if a current of 10.0a is applied to a cu2+ solution for 60.0 min, the mass of copper plated out can be calculated as
a.) 10.0 x 3600 x (1/96500) x (2.00/1) x 63.5
b.) 10.0 x 3600 x (1/96500) x (1/2.00) x 63.5
c.) 96500 x (1/10.0) x (1/2.00) x (1/3600) x 63.5
d.) 96500 x (1/10.0) x (1/2.00) x (1/3600) x (1/63.5)

The correct option is: B.) 10.0 x 3600 x (1/96500) x (1/2.00) x 63.5

Explanation:

Given reduction half-reaction: Cu²⁺ + 2e⁻ → Cu

Given: Electrons transferred: n = 2, Current: I = 10.0 A,

Time: t = 60.0 min = 60 × 60 sec = 3600 sec    (∵ 1 min = 60 sec)

Since, Electric charge (Q) = current × time

∴ Q = I × t

⇒ Q = 10.0 A × 3600 sec = (10.0 × 3600) C

Since, one faraday charge is equal to the charge of one mole electrons.

∴ One mole electron = 1 Faraday (F) = 96,500 coulombs (C)

⇒ 1 C = 1 ÷ 96,500 mole electron

∴ (10.0 × 3600) C = [(10.0 × 3600) × (1 ÷ 96,500)] mole electron

Now since 2 mole electrons reduces 1 mole Cu²⁺ to Cu.

So, 1 mole electrons reduces (1/2) mole Cu²⁺

Therefore, moles of Cu²⁺ reduced by [(10.0 × 3600) × (1 ÷ 96,500)] mole electrons = [(10.0 × 3600) × (1 ÷ 96,500)] mole electron × (1 mole Cu²⁺ ÷ 2 mole electron)

=  [(10.0 × 3600) × (1 ÷ 96,500) × (1  ÷ 2)] moles of Cu²⁺

As number of moles = mass taken ÷ molar mass

⇒ mass of copper plated = number of moles × molar mass

As the molar mass of copper = 63.5 g/ mol

∴  mass of copper plated = [(10.0 × 3600) × (1 ÷ 96,500) × (1  ÷ 2)] moles × 63.5 g/mol = [(10.0 × 3600) × (1 ÷ 96,500) × (1  ÷ 2) moles × 63.5] g

Therefore, the mass of copper plated = [(10.0 × 3600) × (1 ÷ 96,500) × (1  ÷ 2) moles × 63.5] g

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.

a saline water is fed to a reverse osmosis unit at flow rate o rotameter connected to reject (retentate) line reads 8 installed in reject and permeate line reads 15.2 milli siemens/cm and 300 micro siemens/com respectively.