Answer:
1.76
Explanation:
There is some info missing. I think this is the original question.
<em>A chemist dissolves 660.mg of pure hydroiodic acid in enough water to make up 300.mL of solution. Calculate the pH of the solution. Be sure your answer has the correct number of significant digits.</em>
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Step 1: Calculate the molarity of HI(aq)
M = mass of solute / molar mass of solute × liters of solution
M = 0.660 g / 127.91 g/mol × 0.300 L
M = 0.0172 M
Step 2: Write the acid dissociation reaction
HI(aq) ⇄ H⁺(aq) + I⁻(aq)
HI is a strong acid, so [H⁺] = 0.0172 M
Step 3: Calculate the pH
pH = -log [H⁺]
pH = -log 0.0172
pH = 1.76
The answer for this issue is:
The chemical equation is: HBz + H2O <- - > H3O+ + Bz-
Ka = 6.4X10^-5 = [H3O+][Bz-]/[HBz]
Let x = [H3O+] = [Bz-], and [HBz] = 0.5 - x.
Accept that x is little contrasted with 0.5 M. At that point,
Ka = 6.4X10^-5 = x^2/0.5
x = [H3O+] = 5.6X10^-3 M
pH = 2.25
(x is without a doubt little contrasted with 0.5, so the presumption above was OK to make)
Answer: I think that you have to find it in your lesson
Explanation:
Answer:
95% Copper and 5% Zinc
Explanation:
Easy stuff read in the chemistry book...
