Using the law of dilution :
Mi x Vi = Mf x Vf
2.00 x Vi = 0.15 x 100.0
2.00 x Vi = 15
Vi = 15 / 2.00
Vi = 7.5 mL
hope this helps!
Answer: 18.0152 milliliters
Explanation:
Hi, to answer this question we have to apply the next formula:
Water volume = water mass / water density
Since 1 mol of water weights 18.0152 grams
Replacing with the values and solving:
Water volume = 18.0152 g / 1 g /ml = 18.0152 milliliters
Feel free to ask for more if needed or if you did not understand something.
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
I think sand is just glass and other stuff
