A 3.50 amp power supply is used to deposit chromium from a solution of CrCl3. How long will it take to deposit 100.0 grams of ch
1 answer:
Explanation:
The given data is as follows.
Current (I) = 3.50 amp, Mass deposited = 100.0 g
Molar mass of Cr = 52 g
It is known that 1 faraday of electricity will deposit 1 mole of chromium. As 1 faraday means 96500 C and 1 mole of Cr means 52 g.
Therefore, 100 g of Cr will be deposited by "z" grams of electricity.
z =
= 185576.9 C
As we know that, Q = I × t
Hence, putting the given values into the above equation as follows.
Q = I × t
185576.9 C =
t = 53021.9 sec
Thus, we can conclude that 100 g of Cr will be deposited in 53021.9 sec.
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Answer : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ
Explanation :
First we have to calculate the moles of n-butane.
Given:
Molar mass of n-butane = 58.12 g/mole
Mass of n-butane = 58.3 g
Now put all the given values in the above expression, we get:
Now we have to calculate the energy required.
where,
Q = energy required
= enthalpy of fusion of solid n-butane = 4.66 kJ/mol
n = moles = 1.00 mol
Now put all the given values in the above expression, we get:
Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ