Answer:
126.0g of water were initially present
Explanation:
The electrolysis of water occurs as follows:
2H₂O(l) ⇄ 2H₂(g) + O₂(g)
<em>Where 2 moles of water produce 2 moles of hydrogen and 1 mole of oxygen.</em>
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To find the mass of water we need to determine moles of oxygen and hydrogen, thus:
<em>Moles Hydrogen:</em>
14.0g H₂ ₓ (1mol / 2g H₂) = 7 moles H₂
<em>Moles Oxygen:</em>
112.0g O₂ ₓ (1mol / 32g) = 3.5 moles O₂
Based on the chemical equation, the moles of water initially present were 7 moles (That produce 7 moles H₂ and 3.5 moles O₂). The mass of 7 moles of H₂O is:
7 moles H₂O * (18g / mol) =
<h3>126.0g of water were initially present</h3>
<u>Given:</u>
Concentration of HNO3 = 7.50 M
% dissociation of HNO3 = 33%
<u>To determine:</u>
The Ka of HNO3
<u>Explanation:</u>
Based on the given data
[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M
The dissociation equilibrium is-
HNO3 ↔ H+ + NO3-
I 7.50 0 0
C -2.48 +2.48 +2.48
E 5.02 2.48 2.48
Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23
Ans: Ka for HNO3 = 1.23