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
Explanation:
a. Adding a catalyst
no effect
.( Catalyst can only change the activation energy but not the free energy).
b. increasing [C] and [D]
Increase the free energy
.
c. Coupling with ATP hydrolysis
decrease the free energy value
.
d.Increasing [A] and [B]
decrease the free energy.
Answer:MnCO3+2H2O----->MnO2+ HCO3-+2e-+3H+
Explanation:The equation to be balanced is
MnCO3 ------> MnO2+HCO3-
The oxidation number of Mn changes from +2 in MnCO3 to +4 in MnO2
Therefore two electrons must be added to the right as shown below:
MnCO3 -------> MnO2+ HCO3-+ 2e-Now,there is one negative charge HCO3- and 1 negative charge on the two electrons making a total of -3 charges on the right. There is zero charge on the left.
To balance the equation,add3H+on the right,to cancel out the charges.
MnCO3 --------> MnO2+HCO3-+2e-+3H+
Adding H2O to balance Hydrogen and Oxygen atoms:
MnCO3+2H2O ------->MnO2+HCO3-+2e-+3H+
Number of moles = 5 x 10^24 / 6.02 x 10^23 = 8.305 moles. Volume= moles x 22.4 = 186.032 liters. Hope this helps!
Answer:
When the operation of the voltaic cell, which is formed of an aluminum and silver strip takes place, the atom of aluminum loses three of its electrons and the Al3+ formed moves within the solution. The Al3+ ion gets dissolved within the solution and the electrons lost in the process moves through the wire and get acquired by the ions of silver, which then get reduced to solid Ag resulting in the mass gain of silver strip.
