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 = ?
![[Zn^{2+}]](https://tex.z-dn.net/?f=%5BZn%5E%7B2%2B%7D%5D)
= 3.5 M
![[Pb^{2+}]](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D)
= 
Now put all the given values in the above equation, we get:
Therefore, the cell potential for this reaction is 0.50 V
Number 2 is the correct answer
Answer:
50 g of S are needed
Explanation:
To star this, we begin from the reaction:
S(s) + O₂ (g) → SO₂ (g)
If we burn 1 mol of sulfur with 1 mol of oxygen, we can produce 1 mol of sulfur dioxide. In conclussion, ratio is 1:1.
According to stoichiometry, we can determine the moles of sulfur dioxide produced.
100 g. 1mol / 64.06g = 1.56 moles
This 1.56 moles were orginated by the same amount of S, according to stoichiometry.
Let's convert the moles to mass
1.56 mol . 32.06g / mol = 50 g
Size (length+width) approx.
Answer:
The mass of this 25 mL supercritical CO2 sample has a mass of 11.7g
Explanation:
Step 1: Given data
The supercritical CO2 has a density of 0.469 g/cm³ (or 0.469 g/mL)
The sample hasa volume of 25.0 mL
Step 2: Calculating mass of the sample
The density is the mass per amount of volume
0.469g/cm³ = 0.469g/ml
The mass for a sample of 25.0 mL = 0.469g/mL * 25.0 mL = 11.725g ≈ 11.7g
The mass of this 25 mL supercritical CO2 sample has a mass of 11.7g
