Answer:
Ka = 0.1815
Explanation:
Chromic acid
pH = ?
Concentration = 0.078 M
Ka = ?
HCl
conc. = 0.059M
pH = -log(H+)
pH = -log(0.059) = 1.23
pH of chromic acid = 1.23
Step 1 - Set up Initial, Change, Equilibrium table;
H2CrO4 ⇄ H+ + HCrO4−
Initial - 0.078M 0 0
Change : -x +x +x
Equilibrium : 0.078-x x x
Step 2- Write Ka as Ratio of Conjugate Base to Acid
The dissociation constant Ka is [H+] [HCrO4−] / [H2CrO4].
Step 3 - Plug in Values from the Table
Ka = x * x / 0.078-x
Step 4 - Note that x is Related to pH and Calculate Ka
[H+] = 10^-pH.
Since x = [H+] and you know the pH of the solution,
you can write x = 10^-1.23.
It is now possible to find a numerical value for Ka.
Ka = (10^-1.23))^2 / (0.078 - 10^-1.23) = 0.00347 / 0.0191156
Ka = 0.1815
The pair of both compounds that have the same empirical formula are C6H12O6 and HC2H3O2. The answer is letter D. <span>H2O and H2O2, BaSO4 and BaSO3 and FeO and Fe2O3 do not have the same empirical formula.</span>
Answer:
6 valence electrons
Explanation:
The atom you have given is a sulfur atom. It has 6 valence electrons.
<span>A: best time to go fishing at sea</span>
The reaction between the reactants would be:
CH₃NH₂ + HCl ↔ CH₃NH₃⁺ + Cl⁻
Let the conjugate acid undergo hydrolysis. Then, apply the ICE approach.
CH₃NH₃⁺ + H₂O → H₃O⁺ + CH₃NH₂
I 0.11 0 0
C -x +x +x
E 0.11 - x x x
Ka = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
Since the given information is Kb, let's find Ka in terms of Kb.
Ka = Kw/Kb, where Kw = 10⁻¹⁴
So,
Ka = 10⁻¹⁴/5×10⁻⁴ = 2×10⁻¹¹ = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
2×10⁻¹¹ = [x][x]/[0.11-x]
Solving for x,
x = 1.483×10⁻⁶ = [H₃O⁺]
Since pH = -log[H₃O⁺],
pH = -log(1.483×10⁻⁶)
<em>pH = 5.83</em>
