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Nady [450]
3 years ago
5

Electroplating is a way to coat a complex metal object with a very thin (and hence inexpensive) layer of a precious metal, such

as silver or gold. In essence the metal object is made the cathode of an electrolytic cell in which the precious metal cations are dissolved in aqueous solution. Suppose a current of is passed through an electroplating cell with an aqueous solution of in the cathode compartment for seconds. Calculate the mass of pure silver deposited on a metal object made into the cathode of the cell. Be sure your answer has a unit symbol and the correct number of significant digits.
Chemistry
1 answer:
Lapatulllka [165]3 years ago
5 0

This is a incomplete question. The complete question is :

Electroplating is a way to coat a complex metal object with a very thin (and hence inexpensive) layer of a precious metal, such as silver or gold. In essence the metal object is made the cathode of an electrolytic cell in which the precious metal cations are dissolved in aqueous solution. Suppose a current of 0.270 A is passed through an electroplating cell with an aqueous solution of Ag_2SO_4 in the cathode compartment for 72 seconds. Calculate the mass of pure silver deposited on a metal object made into the cathode of the cell. Be sure your answer has a unit symbol and the correct number of significant digits.

Answer: 0.219 g

Explanation:

According to mole concept:

1 mole of an atom contains 6.022\times 10^{23} number of particles.

We know that:

Charge on 1 electron = 1.6\times 10^{-19}C

To calculate the charge passed, we use the equation:

I=\frac{q}{t}

where,

I = current passed = 0.272 A

q = total charge = ?

t = time required = 72.0 s

Putting values in above equation, we get:

0.272A=\frac{q}{72}\\\\q={0.272A}\times 72.0=19.6C

Ag_2SO_4\rightarrow 2Ag^+SO_4^{2-}

Ag^++e^-\rightarrow Ag

108 g of silver is deposited by 1 mole of electrons

96500 C deposits = 108 g of dilver

Thus 19.6 C deposits =\frac{108}{96500}\times 19.6=0.0219g

Thus the mass of pure silver deposited on a metal object made into the cathode of the cell is 0.0219 g

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Determine the number of atoms and the mass of zirconium, silicon and oxygen found in 0.3384 Mol of zircon ZrSiO4, a
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since ,

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0.3384 mol of zircon ZrSiO₄ ,

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from the above equation ,

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Hence ,

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The mass is calculated as -

since ,

Moles is denoted by given mass divided by the molecular mass ,  

Hence ,  

n = w / m

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m = molecular mass .

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moles of Zr = 0.3384 mol

using the above formula ,

w = n * m

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The mass of Zr = 30.79 g

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Since we know , the molecular mass of Si = 28 g/mol

moles of Si = 0.3384 mol

using the above formula ,

w = n * m

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Since we know , the molecular mass of O = 16 g/mol

moles of Si = 1.3536 mol

using the above formula ,

w = n * m

w = 1.3536 mol  * 16 g/mol = 21.65 g

hence ,  

The mass of O = 21.65 g

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