Write the symbol for every chemical element that has atomic number greater than 55 and atomic mass less than 144.0 u
1 answer:
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Answer:
This solution is quite lengthy
Total system = nRT
n was solved to be 0.02575
nH20 = 0.2x0.02575
= 0.00515
Nair = 0.0206
PH20 = 0.19999
Pair = 1-0.19999
= 0.80001
At 15⁰c
Pair = 0.4786atm
I used antoine's equation to get pressure
The pressure = 0.50
2. Moles of water vapor = 0.0007084
Moles of condensed water = 0.0044416
Grams of condensed water = 0.07994
Please refer to attachment. All solution is in there.
Answer:
a) [H₃O⁺] = 1.8x10⁻⁵ M
b) pH = 4.75
c) % rxn = 3.5x10⁻³ %
Explanation:
a) The dissociation reaction of HCN is:
HCN(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CN⁻(aq)
0.5 M - x x x
The dissociation constant from the above reactions is given by:
By solving the above quadratic equation we have:
x = 1.75x10⁻⁵ M = 1.8x10⁻⁵ M = [H₃O⁺] = [CN⁻]
Hence, the [H₃O⁺] is 1.8x10⁻⁵ M.
b) The pH is equal to:
Then, the pH of the HCN solution is 4.75.
c) The % reaction is the % ionization:
Therefore, the % reaction or % ionization is 3.5x10⁻³ %.
I hope it helps you!