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:
5.3%
Explanation:
Let the volume be 1 L
volume , V = 1 L
use:
number of mol,
n = Molarity * Volume
= 0.8846*1
= 0.8846 mol
Molar mass of CH3COOH,
MM = 2*MM(C) + 4*MM(H) + 2*MM(O)
= 2*12.01 + 4*1.008 + 2*16.0
= 60.052 g/mol
use:
mass of CH3COOH,
m = number of mol * molar mass
= 0.8846 mol * 60.05 g/mol
= 53.12 g
volume of solution = 1 L = 1000 mL
density of solution = 1.00 g/mL
Use:
mass of solution = density * volume
= 1.00 g/mL * 1000 mL
= 1000 g
Now use:
mass % of acetic acid = mass of acetic acid * 100 / mass of solution
= 53.12 * 100 / 1000
= 5.312 %
≅ 5.3%
Answer:- Third choice is correct, 17.6 moles
Solution:- The given balanced equation is:
Al_2(SO_4)_3+6KOH\rightarrow 2Al(OH)_3+3K_2SO_4
We are asked to calculate the moles of potassium hydroxide needed to completely react with 2.94 moles of aluminium sulfate.
From the balanced equation, there is 1:6 mol ratio between aluminium sulfate and potassium hydroxide.
It is a simple mole to mole conversion problem. We solve it using dimensional set up as:
2.94molAl_2(SO_4)_3(\frac{6molKOH}{1molAl_2(SO_4)_3})
= 17.6 mol KOH
So, Third choice is correct, 17.6 moles of potassium hydroxide are required to react with 2.94 moles of aluminium sulfate.
The alkali metals are first column, alkali earth 2nd, halogens 2nd to last, and noble gases last. Hope it helps!
