For ksp of silver sulfate (ag2so4) at 1. 2 ✕ 10-5, the solubility of silver sulfate is mathematically given as
x = 5.26*10^{-4}
<h3>What is the
solubility of silver
sulfate in 0.15 M AgNO3?</h3>
Generally, the equation for the Chemical reaction is mathematically given as
Ag2SO4 ⇄ 2Ag+ + SO4 2-
Therefore
Ksp = [Ag+ ]2 *[SO4 2-]
1.2*10-5 = ( 0.15 +2x)2*(x)
x = 5.26*10^{-4}
In conclusion, the solubility of silver sulfate is
x = 5.26*10^{-4}
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brainly.com/question/16416932
Nitrous acid<span> dissociates as follows:
</span>
HNO₂(s) ⇄ H⁺(aq) + NO₂⁻(aq)
According to the equation, an acid constant has the following form:
Ka = [H⁺] × [NO₂⁻ ] / [HNO₂]
From pH, we can calculate the concentration of H⁺ and NO₂⁻:
[H⁺] = 10^-pH = 10^-2.63 = 0.00234 M = [NO₂⁻]
Now, the acid constant can be calculated:
Ka = 0.00234 x 0.00234 / 0.015 = 3.66 x 10⁻⁴
And finally,
pKa = -log Ka = 3.44
Explanation:
<h2>The three major branches of natural science are</h2><h2>1.Physical,the study of universe.</h2><h2>2.Chemistry,the study of matter.</h2><h2>3.Biology,the study of life and living or organisms.</h2><h2 /><h2>have a good day.</h2><h2>I hope this answer can help you</h2>
Relative formula mass C₅H₁₁ = 71
Now divide the molar mass by the RFM = 142.32 / 71 = 2
Now C₍₅ₓ₂₎H₍₁₁ₓ₂) = C₁₀H₂₂
Hope that helps
