Answer:
THE CURRENT REQUIRED TO PRODUCE 193000 C OF ELECTRICITY IS 35.74 A.
Explanation:
Equation:
Al3+ + 3e- -------> Al
3 F of electricity is required to produce 1 mole of Al
3 F of electricity = 27 g of Al
If 18 g of aluminium was used, the quantity of electricity to be used up will be:
27 g of AL = 3 * 96500 C
18 G of Al = x C
x C = ( 3 * 96500 * 18 / 27)
x C = 193 000 C
For 18 g of Al to be produced, 193000 C of electricity is required.
To calculate the current required to produce 193 000 C quantity of electricity, we use:
Q = I t
Quantity of electricity = Current * time
193 00 = I * 1.50 * 60 * 60 seconds
I = 193 000 / 1.50 * 60 *60
I = 193 000 / 5400
I = 35.74 A
The cuurent required to produce 193,000 C of electricity by 18 g of aluminium is 35.74 A
<span>If the human body were a car, glucose would be the gasoline.
Glucose gives humans energy, we basically run on glucose, among other things, the same way a car would run on gas.
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Your answer is D. Since there is little to no magnetic field to wire, if it is copper which most wires are, there will be no voltage in a wire.
Answer:
Four moles of the cation
Explanation:
2Rb2CrO4(s)<--------> 4Rb^+(aq) + 2CrO4^2-(aq)
Now looking at the reaction equation, it can be seen that one mole of rubidium chromate contains two moles of rubidium ions and one mole of chromate ions.
The dissolution of two moles of rubidium chromate should then yield four moles of rubidium ions and two moles of chromate ions since the ratio of ions present is 2:1.
This explains the reaction equation written above for the dissolution of two moles of rubidium chromate as shown.
