Answer:
Weight percent = 29.5%
Molality = 2.460 mol/kg
Mole fraction = 0.042
Explanation:
The weight percent (%w) of a substance is the mass of it divided by the total mass of the solution, and then multiplied by 100%, so:
%w = [44.5/(44.5 + 106.5)] *100%
%w = 29.5%
The molality (W) of a solution is the number of moles of the solute divided by the mass (in kg) of the solvent. In this case, water is the solvent, and AgNO₃ the solute (water is generally the solvent, and the solvent is always in a larger amount). The number of moles is the mass divided by the molar mass, and the molar mass of AgNO₃ is 169.87 g/mol:
n = 44.5/169.87 = 0.2620 mol
W = 0.2620/0.1065 = 2.460 mol/kg
The mole fraction (x) of a substance is the number of moles of it divided by the total number of moles of the solution. The molar mass of water is 18 g/mol, so the number of moles of water is:
n = 106.5/18 = 5.92 moles
Thus the mole fraction of AgNO₃ is:
x = 0.2620/(0.2620 + 5.92) = 0.042
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Answer:
The answer to your question is 9.075 g of CO₂
Explanation:
Data
mass of C₃H₈ = 39 g
mass of O₂ = 11 g
Balanced chemical reaction
C₃H₈ + 5O₂ ⇒ 3CO₂ + 4H₂
-Calculate the molar mass of the reactants
C₃H₈ = (12 x 3) + (8 x 1) = 36 + 8 = 44 g
O₂ = (16 x 2) = 32 g
-Calculate the limiting reactant
theoretical yield C₃H₈ / O₂ = 44/5(32) = 44/ 160 = 0.275
experimental yield C₃H₈/O₂ = 39/11 = 3.5
From the previous result, we conclude that the limiting reactant is O₂ because the experimental yield was higher than the theoretical yield.
-Calculate the mass of CO₂
160 g of O₂ ----------------- 3(44) g of CO₂
11 g of O₂ ------------------ x
x = (11 x 3(44)) / 160
x = 1452 / 160
x = 9.075 g of CO₂
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the declaration of rights and grievances was written as protest to the Stamp Act, which was in 1765