Answer: 406 hours
Explanation:
where Q= quantity of electricity in coloumbs
I = current in amperes = 39.5 A
t= time in seconds = ?
The deposition of copper at cathode is represented by:
Coloumb of electricity deposits 1 mole of copper
i.e. 63.5 g of copper is deposited by = 193000 Coloumb
Thus 19.0 kg or 19000 g of copper is deposited by = 
Coloumb
(1hour=3600s)
Thus it will take 406 hours to plate 19.0 kg of copper onto the cathode if the current passed through the cell is held constant at 39.5 A
Answer:4
Explanation:To balance the equation you need to make the number of each element equivalent in both sides.
To start add a 2 in front of the MnO2 which balances the Mn.
Then balance the oxygen by adding a 4 in front of H20.
The H then needs a 8 as it’s coefficient.
Answer:
Use the activity formula,
T1/2 = 4.468 x 10^9 yr x 365 x 24 x 3600 = 1.409 x 10^17 sec
l = ln(2)/T1/2 = ln(2)/1.409 x 10^17 = 4.91932697 x 10^-18 s-1
DN/Dt = lN, 265 = 4.91932697 x 10^-18 x N
<u><em>N = 5.38 x 10^19 nuclei</em></u>
Answer:
heat flows from the object that has more thermal more energy to the object with less thermal energy
The correct answer is 7 I just took the test :)
