Answer : The cell potential for this reaction is 0.50 V
Explanation :
The given cell reactions is:

The half-cell reactions are:
Oxidation half reaction (anode): 
Reduction half reaction (cathode): 
First we have to calculate the cell potential for this reaction.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B2.303RT%7D%7BnF%7D%5Clog%20%5Cfrac%7B%5BZn%5E%7B2%2B%7D%5D%7D%7B%5BPb%5E%7B2%2B%7D%5D%7D)
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = 
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = +0.63 V
= cell potential for the reaction = ?
= 3.5 M
= 
Now put all the given values in the above equation, we get:


Therefore, the cell potential for this reaction is 0.50 V
Answer:
a) The theoretical yield is 408.45g of 
b) Percent yield = 
Explanation:
1. First determine the numer of moles of
and
.
Molarity is expressed as:
M=
- For the 
M=
Therefore there are 1.75 moles of 
- For the 
M=
}{1Lsolution}[/tex]
Therefore there are 2.0 moles of 
2. Write the balanced chemical equation for the synthesis of the barium white pigment,
:

3. Determine the limiting reagent.
To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:
- For the
:

- For the
:

As the
is the smalles quantity, this is the limiting reagent.
4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.

5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:
Percent yield = 
Percent yield = 
The real yield is the quantity of barium white pigment you obtained in the laboratory.
Answer
D
Explanation:
They take up usable forms of nitrogen found in soil
1 mole=6.02 x 10^23 atoms so how many moles are there in 3.0 x 10^23 we will cross multiply, 1 x 3.0 x 10^23 / 6.02 x 10 ^23. Which will give us 0.498 moles.
Hope this helped