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
Reflection, refracting, and the energy levels of molecular orbitals
Answer:
Heat required = 13,325 calories or 55.75 KJ.
Explanation:
To convert a water to steam at 100 degree celsius to vapor, we have to give latent heat of vaporization to water
Which equals ,
Q = mL,
Where, m is the mass of water present
L = specific latent heat of vaporization
Here , m= 25 gram
L equals to 533 calories (or 2230 Joules)
So, Q = 25×533 = 13,325 Calories
Or , Q = 55,750 Joules = 55.75 KJ
so, Heat required = 13,325 calories or 55.75 KJ.
Answer:
your answer gonna be 20 miles
1. A. All the elements in the column have similar chemical properties.
2. Substances on the periodic table cannot be broken down into other substances and are therefore elements.
