Answer:
0.0325M S₂O₃²⁻
Explanation:
Based on the reaction:
S²⁻ + ZnCO₃ → ZnS(s) + CO₃²⁻
<em>1 mole of S²⁻ reacts per mole of ZnCO₃</em>
when the reaction occurs, the sulfide ions are precipitated keeping in solution just S₂O₃²⁻ ions.
Then, these ions are titrated with iodine, thus:
2 S₂O₃²⁻ + I₂ → S₄O₆²⁻ + 2I⁻
That means 2 moles of thiosulfate react per mole of iodine.
Moles of iodine spent are:
5.20x10⁻³L ₓ (0.01000 mol / L) = 5.20x10⁻⁵ moles of I₂
5.20x10⁻⁵ moles of I₂ ₓ (2 moles S₂O₃²⁻ / 1 mole I₂) =
1.04x10⁻⁴ moles of S₂O₃²⁻
As dilution factor of the S₂O₃²⁻ solution was 50.0mL / 20.0mL = 2.5. Moles of the initial solution X are:
1.04x10⁻⁴ moles of S₂O₃²⁻ ₓ 2.5 = 1.60x10⁻⁴ moles of S₂O₃²⁻. In 20.0mL:
1.60x10⁻⁴ moles of S₂O₃²⁻ / 0.0200L =
<em>0.0325M S₂O₃²⁻</em>
<em></em>
Salts dissociate when dissolved in water:
PbSO4 -> Pb2+ + SO4 2-
So there are 2 different particles. 1 mol
of PbSO4 produces 1 mol of Pb2+, and 1 mol of SO4 2- (one of each).
Kps = [Pb2+] [SO4 2-]
We will call “s” the molarity of PbSO4
So the molarity of Pb2+ is also “s” (same
number)
And the molarity of SO4 2- is also “s”
(same number)
Kps = s*s = s2
1.82
× 10-8 = s2
So,
in order to find s, we have to make the square root of 1.82 × 10-8, which is: 0,00013491 M
26, protons and nuetrons will always be the same
Answer:
The percentage yield is 78.2g
Explanation:
Given, mass of propane = 42.8 g , sufficient O2 percent yield = 61.0 % yield.
Reaction - C3H8(g)+5O2(g)------> 3CO2(g)+4H2O(g)
First we need to calculate the moles of propane
Moles of propane = 
g.mol-1
= 0.971 moles
So, moles of CO2 from the moles of propane
1 mole of C3H8(g) = 3 moles of CO2(g)
So, 0.971 moles of C3H8(g) = ?
= 2.913 moles of CO2
So theoretical yield = 2.913 moles 
44.0 g/mol
= 128.2 g
So, the actual mass of CO2 = percent yield theoretical yield / 100 %
= 61.0 % 128.2 g / 100 %
= 78.2 g
the mass of CO2 that can be produced if the reaction of 42.8 g of propane and sufficient oxygen has a 61.0 % yield is 78.2 g
