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
Answer:
A
Explanation:
Mass is a raw measurement, weight is effected by gravity
Answer:
Explanation:
We must do the conversions
mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of CO₂ ⟶ volume of CO₂
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
Mᵣ: 180.16
C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O
m/g: 24.5
(a) Moles of C₆H₁₂O₆
(b) Moles of CO₂
(c) Volume of CO₂
We can use the Ideal Gas Law.
pV = nRT
Data:
p = 0.960 atm
n = 0.8159 mol
T = 37 °C
(i) Convert the temperature to kelvins
T = (37 + 273.15) K= 310.15 K
(ii) Calculate the volume
Answer: There are 4.8 x 1024 hydrogen atoms in 2.0 moles of CH4
