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
Water can act as both an acid and a base in a solution and undergoes auto ionization where it can be both an acid and a base to itself. The ph of water is 6.5 to 8.5.
Hope this helped :)
The sample contains 1.00 mol H₂O.
Moles of water = 18.0 g H₂O × 1 mol H₂O/18.0 g H₂O = 1.00 mol H₂O
Answer:
Look at explanation
Explanation:
Size of particles increases, rate of reaction decreases, and vise versa
Catalyst causes the rate of reaction to increase
Concentration increases, rate of reaction increases, and vise versa
Temperature increases, rate of reaction increases, and vise versa
Answer:
B) ![k_{sp} =[Mg_{aq} ]*[OH_{aq}]^{2}](https://tex.z-dn.net/?f=k_%7Bsp%7D%20%3D%5BMg_%7Baq%7D%20%5D%2A%5BOH_%7Baq%7D%5D%5E%7B2%7D)
Explanation:
Ksp is the solubility product constant.
For a generic chemical equation
The Ksp for this generic chemical equation is
![k_{sp} =[A_{aq} ]*[B_{aq}]^{2}](https://tex.z-dn.net/?f=k_%7Bsp%7D%20%3D%5BA_%7Baq%7D%20%5D%2A%5BB_%7Baq%7D%5D%5E%7B2%7D)
So the Ksp fot the Mg(OH)2 is
