Answer:
AWWWWWWWWWWWWWWWW SOOO NICEEE
When a 0. 1 m aqueous solution of hydrocyanic acid, HCN, reaches equilibrium, the ka for hydrocyanic acid is 3.969 x 10⁻¹⁰.
<h3>What is ka value?</h3>
It's the value of equilibrium constant for the dissociation of ions into a solution. The more the Ka value the more will be dissociation.
Ka = [H₃O⁺]² / [HCN] [H₃O⁺]
The pH is 5.20
-log [H₃O⁺] = 5.20
Putting antitlog both side.
The value will be 6.30 x 10⁻⁶
Ka = (6.30 x 10⁻⁶)² / 0.1 - 6.30 x 10⁻⁶
0.1 - 6.30 x 10⁻⁶ = 0.1
Ka = 3.969 x 10⁻¹⁰
Thus, the Ka value for hydrocyanic acid is 3.969 x 10⁻¹⁰.
To learn more about ka value, refer to the link:
brainly.com/question/2796803
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Answer:
T2 = 2843.1 oK. This is a huge temperature. Check it for errors.
Explanation:
Remark
This is the same question as the other one I've answered. Only the numbers have been altered.
Givens
v1 = 56 mL
P1 = 1 atm
T1 = 273o K
v2 = 162
P2 = 3.6 atm
T2 = ?
Formula
Vi * P1 / T1 = V2 * P2/T2
Solution
Rearrange the formula so T2 is on the left
T2 = V2 P2 * T1 / (V1 * P1) Now just put the numbers in.
T2 = 162 * 3.6* 273 / (56 *1)
T2 = 159213.6/56
T2 = 2843.1
Answer:
HPO42- (aq) + NH4+ (aq)
Explanation:
This is a conjugate acid-base pair. Please forgive if my answers are incorrect. I myself am quite unsure.
