Answer: The amperage needed to complete the electrolysis in the given time is 9.08 A
Explanation:
We are given:
Moles of electron = 1 mole
According to mole concept:
1 mole of an atom contains
number of particles.
We know that:
Charge on 1 electron = ![1.6\times 10^{-19}C](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E%7B-19%7DC)
Charge on 1 mole of electrons = ![1.6\times 10^{-19}\times 6.022\times 10^{23}=96500C](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E%7B-19%7D%5Ctimes%206.022%5Ctimes%2010%5E%7B23%7D%3D96500C)
moles of calcium =![\frac{\text {given mass}}{\text {Molar mass}}=\frac{99.47g}{40g/mol}=2.49moles](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%20%7Bgiven%20mass%7D%7D%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D%5Cfrac%7B99.47g%7D%7B40g%2Fmol%7D%3D2.49moles)
![Ca^{2+}+2e^-\rightarrow Ca](https://tex.z-dn.net/?f=Ca%5E%7B2%2B%7D%2B2e%5E-%5Crightarrow%20Ca)
is required to deposit = 1 mole
Thus 1 mole is deposited by = 193000 C
2.49 moles will be deposited by =![\frac{193000}{1}\times 2.49=480570C](https://tex.z-dn.net/?f=%5Cfrac%7B193000%7D%7B1%7D%5Ctimes%202.49%3D480570C)
To calculate the time required, we use the equation:
![I=\frac{q}{t}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7Bq%7D%7Bt%7D)
where,
I = current passed = ?
q = total charge = ![480570C](https://tex.z-dn.net/?f=480570C)
t = time required = 14.7 hours =52920 s ( 1hour=3600s)
Putting values in above equation, we get:
![I=\frac{480570C}{52920s}\\\\I=9.08A](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B480570C%7D%7B52920s%7D%5C%5C%5C%5CI%3D9.08A)
Hence, the amperage needed to complete the electrolysis in the given time is 9.08 A
Answer:
g(x) has the greater slope
1 mL =
L
so $155$ mL =$155\times10^{-3}$ L
Now, $1$ gallon = $3.785$ L , or $1$ L= $\frac1{3.785}$ gallons
so $155\times10^{-3}$ L= $155\times10^{-3}\times \frac{1}{3.785}=40.95\times 10^{-3}$ gallons
and this is for each day, so just divide by 1 (day) to get the ratw gallons/day
i.e. $40.95\times 10^{-3}$ gal/day
since the unit for the heat of fusion is kJ/mol, you're going to have to convert the grams into moles in order to cancel out the unit. After that, you can solve like normal.
Answer:
Draw the curved arrows showing a proton transfer reaction, and draw the products of that proton transfer. Do not include the Li counterion, and lone pairs are not required in the products and the question is hsown below:
Explanation:
The proton from water is abstracted by butyl carbanion and hydroxide ion is formed from water.
The reaction is shown below in the attachment.