What is the molarity of a solution in which 211g sodium hydrogen carbonate is dissolved in a 10.0L solution?
1 answer:
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Answer:
Ecell = +0.25V
Explanation:
the half-cell reactions for a voltanic cell
cathode(reduction): 2H⁺(aq) + 2e⁻ ------- H₂(g)
anode(oxidation): 2AgCl(s) ------- 2Ag⁺(aq) + 2Cl⁻ + 2e⁻
we have the standard cell potential E⁺cell = 0.18V at 80C respectively
Q = [H⁺]/[Cl⁻]
sub for [H+] = 0.10M and [Cl-] = 1.5M
Q= 0.1M/1.5M
Q = 0.067
Ecell = E⁺cell - logQ
= 0.18 - log 0.067
0.18- 0.059(-1.174)
Ecell = +0.25V
The question is incomplete, the table of the question is given below
Answer:
I) xA= 0.34, yA= 0.55
ii) 76.2 mole % vapor
iii) Percentage of vapor volume = 98%
Explanation:
i) xA= 0.34, yA= 0.55
xA= 0.34, yA= 0.55
ii) 0.50 = 0.55 nv + 0.34 nL
Therefore, nV = 0.762 mol vapor and nL = 0.238 mol liquid
This shows 76.2 mole % vapor
iii) ρA= 0.791 g/cm3 and, ρE = 0.789 g/cm3
Therefore, ρ = 0.790 g/cm3
Now, we have:
MA = 58.08 g/mol and ME= 46.07 g/mol
So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol
1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor
Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3
Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3
Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %
