Electricity grids will produce surplus power
1) 2
2) 2
3) there are 2 atoms of hydrogen and 1 atom of oxygen, altogether is 3
4) subscript tells how many atoms of each element are present in the molecule
5) No, it is not balanced because the oxygen atom is not equal on both sides of the reactants and products. This wouldn’t be considered balanced as there are two oxygen atoms in the reactants section and only one oxygen on the products section.
It is:
2 H2 + O2 -> 2 H2O
Both have a continuous light spectra the fluorescent source makes a spectra with more intense bands of mercury
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Answer: The mass of lead deposited on the cathode of the battery is 1.523 g.
Explanation:
Given: Current = 62.0 A
Time = 23.0 sec
Formula used to calculate charge is as follows.
![Q = I \times t](https://tex.z-dn.net/?f=Q%20%3D%20I%20%5Ctimes%20t)
where,
Q = charge
I = current
t = time
Substitute the values into above formula as follows.
![Q = I \times t\\= 62.0 A \times 23.0 sec\\= 1426 C](https://tex.z-dn.net/?f=Q%20%3D%20I%20%5Ctimes%20t%5C%5C%3D%2062.0%20A%20%5Ctimes%2023.0%20sec%5C%5C%3D%201426%20C)
It is known that 1 mole of a substance tends to deposit a charge of 96500 C. Therefore, number of moles obtained by 1426 C of charge is as follows.
![Moles = \frac{1426 C}{96500 C/mol}\\= 0.0147 mol](https://tex.z-dn.net/?f=Moles%20%3D%20%5Cfrac%7B1426%20C%7D%7B96500%20C%2Fmol%7D%5C%5C%3D%200.0147%20mol)
The oxidation state of Pb in
is 2. So, moles deposited by Pb is as follows.
![Moles of Pb = \frac{0.0147}{2}\\= 0.00735 mol](https://tex.z-dn.net/?f=Moles%20of%20Pb%20%3D%20%5Cfrac%7B0.0147%7D%7B2%7D%5C%5C%3D%200.00735%20mol)
It is known that molar mass of lead (Pb) is 207.2 g/mol. Now, mass of lead is calculated as follows.
![No. of moles = \frac{mass}{molar mass}\\ 0.00735 = \frac{mass}{207.2 g/mol}\\mass = 1.523 g](https://tex.z-dn.net/?f=No.%20of%20moles%20%3D%20%5Cfrac%7Bmass%7D%7Bmolar%20mass%7D%5C%5C%200.00735%20%3D%20%5Cfrac%7Bmass%7D%7B207.2%20g%2Fmol%7D%5C%5Cmass%20%3D%201.523%20g)
Thus, we can conclude that the mass of lead deposited on the cathode of the battery is 1.523 g.
The correct answer is B.
1 mol O2 x 15.999 O2/ 1 mol O2 = 15.999 O2
16 O2 when rounded.