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Yakvenalex [24]
3 years ago
9

What quantity of copper is deposited by the same quantity of electricity that deposited 9g of aluminum

Chemistry
1 answer:
klio [65]3 years ago
3 0

Answer:

Mass of copper deposited = 31.75 g

Explanation:

According to Faraday's second law of electrolysis, when the same quantity of electricity is passed through different electrolytes, the relative number of moles of the elements deposited are inversely proportional to the charges on the ions of the elements.

From this law, it can be seen that the higher the charge, the lower the number of moles of a given element deposited.

Number of moles of aluminium in 9 g of aluminium = mass / molar mass

Molar mass of aluminium = 27 g

Number of moles of aluminium = 9/27 = 1/3 moles

Charge on aluminium ion = +3

3 moles of electrons will discharge 1 mole of aluminium,

1 mole of electrons will discharge 1/3 moles of aluminium

Number of moles of electrons involved = 1 mole of electrons

Charge on copper ion = +2

1 mole of electrons will discharge 1/2 moles of copper.

Mass of 1/2 moles of copper = number of moles × molar mass of copper

Molar mass of copper = 63.5 g

Mass of copper deposited = 1/2 × 63.5 = 31.75 g

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Answer:

The correct option is;

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Explanation:

The given parameters are;

The mass of the ice = 55 g

The Heat of Fusion = 80 cal/g

The Heat of Vaporization = 540 cal/g

The specific heat capacity of water = 1 cal/g

The heat required to melt a given mass of ice = The Heat of Fusion × The mass of the ice

The heat required to melt the 55 g mass of ice = 540 cal/g × 55 g = 29700 cal

The heat required to raise the temperature of a given mass ice (water) = The mass of the ice (water) × The specific heat capacity of the ice (water) × The temperature change

The heat required to raise the temperature of the ice from 0°C to 100°C = 55 × 1 × (100 - 0) = 5,500 cal

The heat required to vaporize a given mass of ice = The Heat of Vaporization × The mass of the ice

The heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal

The total heat required to boil 55 g of ice = 29700 cal + 5,500 cal + 4,400 cal = 39,600 cal

However, we note that the heat required to vaporize the 55 g mass of ice at 100°C = 80 cal/g × 55 g = 4,400 cal.

The heat required to vaporize the 55 g mass of ice at 100°C = 4,400 cal

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