- log [H+] = 8
log [H+] = -8
10^(log [H+]) = 10^-8
[H+] = 1x10^-8 M
Answer:
The answer to your question is letter b) X
Explanation:
Data
W = 5 units
X = 6 units
Chemical reaction
2W + 3X ⇒ 3Y + Z
To determine which reactant is the limiting reactant, we must use proportions:
-Theoretical proportion W / X = 2 / 3 = 0.67
- Experimental proportion W / X = 5 / 6 = 0.83
As the experimental proportion is higher than the theoretical proportion, we conclude that the amount of W is higher in the experiment so the limiting reactant is X.
Answer:
1. wa. Molecular form.
2. wa. Molecular form.
3. sa. Ionic form.
Explanation:
1. hydrocyanic acid
HCN is a weak acid. That means that in the following equilibrium, the molecular form (HCN(aq)) is favored, and that's how it should be written in solution.
HCN(aq) ⇄ H⁺(aq) + CN⁻(aq)
2. hypochlorous acid
HClO is a weak acid. That means that in the following equilibrium, the molecular form (HClO(aq)) is favored, and that's how it should be written in solution.
HClO(aq) ⇄ H⁺(aq) + ClO⁻(aq)
3. hydrochloric acid
HCl is a strong acid. That means that it completely dissociates and it should be written in the ionic form (H⁺(aq) + Cl⁻(aq)) when it is in solution.
HCl(aq) → H⁺(aq) + Cl⁻(aq)
Answer:
The answer to your question is the letter d. S
Explanation:
Data
Change of +4 in the oxidation number
Chemical reaction
K₂Cr₂O₇ + H₂O + S ⇒ KOH + Cr₂O₃ + SO₂
Process
1.- Calculate the oxidation numbers following the rules.
Some rules
H = +1 O = -2 Alkali metals = + 1 Alkali earth metals = +2
K₂⁺¹Cr₂⁺⁶O₇⁻² + H₂⁺¹O⁻² + S⁰ ⇒ K⁺¹O⁻²H⁺¹ + Cr₂⁺³O₃⁻² + S⁺⁴O₂⁻²
Elements that changed their oxidation numbers
Cr₂⁺⁶ ---------------- Cr₂⁺³
S⁰ --------------- S⁺⁴
5600 why because you take away and then you add then yeah
