Answer:
0.01836 M
Explanation:
Again the reaction equation is;
Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s)
E°cell= 0.77 V
Ecell= 0.78 V
[Mn2+] = 0.040 M
[Fe2+] = the unknown
n=2
From Nernst's equation;
Ecell= E°cell- 0.0592/n log Q
0.78= 0.77 - 0.0592/2 log [Fe2+] /[0.040]
0.78-0.77= - 0.0592/2 log [Fe2+] /[0.040]
0.01/ -0.0296= log [Fe2+] /[0.040]
-0.3378= log [Fe2+] /[0.040]
Antilog(-0.3378) = [Fe2+] /[0.040]
0.459= [Fe2+] /[0.040]
[Fe2+] = 0.459 × 0.040
[Fe2+] = 0.01836 M
The total volume of water that would be removed will be 75 mL
<h3>Dilution equation</h3>
Using the dilution equation:
M1V1 = M2V2
In this case, M1 = 500 mL, V1 = 10.20 M, M2 = 12 M
Substitute:
V2 = 500 x 10.20/12
= 425 mL
The final volume in order to arrive at 12 M HNO3 would be 425 mL from the initial 500 mL. Thus, the total amount of water that will be removed by evaporation can be calculated as:
500 - 425 = 75 mL
More on dilution can be found here: brainly.com/question/7208939
Equation for Half life :
A = a(0.5)^(t/h)
A is current amount, "a" is initial amount, h is halflife, t is time
5 = 40(0.5)^(t/1.3x10^9)
5/40 = (0.5)^(t/1.3x10^9)
take the log of both sides , power rule
Log(5/40) = (t/1.3x10^9) * Log(0.5)
(1.3x10^9) * Log(5/40) / Log(0.5) = t
3.9x10^9 years = t
And if you think about what a half life is, the time it take for the amount to reduce to half.
40/2 = 20
20/2 = 10
10/2 = 5
It went through 3 half-lifes
3 * 1.3x10^9 = 3.9x10^9 years
Hmm.... Oxygen support combustion. Other than that, rusting occur when oxygen is combined with metal. When iron rusts, it combines with oxygen. Hope it helps.
Let the mass of the solute be x
So, the equation would be
x/2.5+x ×100 = 23.22
x/2.5+x = 23.22/100
100x = 58.05 + 23.22x
100x - 23.22x = 58.05
76.78x = 58.05
x =0.756 ≈ 0.76 litres
