Answer:
1.09 L
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the volume in liters of a 0.360 mol/L barium acetate solution that contains 100 g of barium acetate. Be sure your answer has the correct number of significant digits.</em>
<em />
The molar mass of barium acetate is 255.43 g/mol. The moles corresponding to 100 grams are:
100 g × (1 mol/255.43 g) = 0.391 mol
0.391 moles of barium acetate are contained in an unknown volume of a 0.360 mol/L barium acetate solution. The volume is:
0.391 mol × (1 L/0.360 mol) = 1.09 L
2C3H8+ 702--->6CO2+8H20
FROM Equation above 2 moles of C3H8 reacted with 7 moles of oxygen to form 6 moles of c02 plus 8 molesof H2O
the moles of c3H8 reacted is = MASS/ R.F.M
THE R.F.M =48+8=44
Number of moles is hence 0.025/44=5.68x10^-4
since ratio of C3H8 to O2 is 2:7 Therefore moles of O2 reacted is 1.989 x10^-3
mass= r.f.m x number of moles
(1.989x10^-3) x 32 =0.064g
A molecule can<span> possess </span>polar bonds<span> and still be </span>non polar. If the polar bonds<span> are evenly (or symmetrically) distributed, the </span>bond<span> dipoles cancel and </span>do<span> not create a molecular dipole.</span>
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
The alcoholic can seek rehab
