Acidic because anything over a ph of 7 would be basic but anything below 7 would be acid and if it has a ph of 7 it would be neutral....Hope I helped
a. AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)
b. Ksp AgBr = s²
c. 5 x 10⁻¹³ mol/L
<h3>Further explanation</h3>
Given
solubility AgBr = 7.07 x 10⁻⁷ mol/L
Required
The dissolution reaction
Ksp
The solubility product constant
Solution
a. dissolution reaction of AgBr
AgBr(s)⇒ Ag⁺(aq) + Br⁻(aq)
b. Ksp
Ksp AgBr = [Ag⁺] [Br⁻]
Ksp AgBr = (s) (s)
Ksp AgBr = s²
c. Ksp AgBr = (7.07 x 10⁻⁷)² = 5 x 10⁻¹³ mol/L
Chemical Equation Balancer
Let us say that R is the major enantiomer, while
S is the minor enantiomer, therefore the formula for enantiomeric excess (ee)
is:
ee = (R – S) * 100%
Let us further say that the fraction of R is x (R
= x), and therefore fraction of S is 1 – x (S = 1 – x), therefore:
75 = (x – (1 – x)) * 100
75 = 100 x – 100 + 100 x
200 x = 175
x = 0.875
Summary of answers:
R = major enantiomer = 0.875 or 87.5%
<span>S = minor enantiomer = (1 – 0.875) = 0.125 or
12.5%</span>
