You're going to divide the mass of chlorine within the compound by the mass of the compound, and then multiply the result by 100 to get the answer
C. a change of state. It can be a physical or a chemical state of change.
The solution for this problem is:
Get into moles first. .0560 grams over 540.8 grams per mole = 1.04 x l0^-4 moles
Sr3(As04)2 = 3 Sr++(aq) plus 2 As04^-3(aq)
Ksp = (Sr++)^3(As04^-3)^2
(Sr++) = 3 X 1.04 x l0^-4= 3.11 x l0^-4
(As04^-3) = 2 x 1.04 x l0^-4= 2.07 x l0^-4
Ksp = (1.04 x l0^-4)^3 (2.07 x l0^-4)^2 which equals 4.82 x 10^-20
Answer:
I hope this helps, sorry if it doesnt. Not really good with chemistry, hope you have a great day/night :}
Explanation:
A mole (mol) is the amount of a substance that contains 6.02 × 10 23 representative particles of that substance.
12.00 g C-12 = 1 mol C-12 atoms = 6.022 × 1023 atoms • The number of particles in 1 mole is called Avogadro's Number (6.0221421 x 1023).
Answer:
<h3>1. 10 e⁻</h3>
Oxidation numbers
I₂O₅(s): I (5+); O(2-)
CO(g): C(2+); O(2-)
I₂(s): I(0)
CO₂(g): C(4+); O(2-)
<h3>2. 4 e⁻</h3>
Oxidation numbers
Hg²⁺(aq): Hg(2+)
N₂H₄(aq): N(2-); H(1+)
Hg(l): Hg(0)
N₂(g): N(0)
H⁺(aq): H(1+)
<h3>3. 6 e⁻</h3>
Oxidation numbers
H₂S(aq): H(1+); S(2-)
H⁺(aq): H(1+)
NO₃⁻(aq): N(5+); O(2-)
S(s): S(0)
NO(g): N(2+); O(2-)
H₂O(l): H(1+); O(2-)
Explanation:
In order to state the total number of electrons transferred we have to identify both half-reactions for each redox reaction.
1. I₂O₅(s) + 5 CO(g) → I₂(s) + 5 CO₂(g)
Oxidation: 10 e⁻ + 10 H⁺(aq) + I₂O₅(s) → I₂(s) + 5 H₂O(l)
Reduction: 5 H₂O(l) + 5 CO(g) → 5 CO₂(g) + 10 H⁺(aq) + 10 e⁻
2. 2 Hg²⁺(aq) + N₂H₄(aq) → 2 Hg(l) + N₂(g) + 4 H⁺(aq)
Oxidation: N₂H₄(aq) → N₂(g) + 4 H⁺(aq) + 4 e⁻
Reduction: 2 Hg²⁺(aq) + 4 e⁻ → 2 Hg(l)
3. 3 H₂S(aq) + 2H⁺(aq) + 2 NO₃⁻(aq) → 3 S(s) + 2 NO(g) + 4H₂O(l)
Oxidation: 3 H₂S(aq) → 3 S(s) + 6 H⁺(aq) + 6 e⁻
Reduction: 8 H⁺(aq) + 2 NO₃⁻(aq) + 6 e⁻ → 2 NO(g) + 4 H₂O
