Data:
weight of water before heating = 0.349
weight of hydrate before heateing = 2.107
Formula:
Weight percent of water = [ (weight of water) / (weight of the hydrate) ] * 100
Solution:
Weight percent of water = [ 0.349 / 2.107] * 100 ≈ 16.6 %
Answer: 16.6%
Answer:
Explanation:
The equation for mass is:
We plug in the given values into the equation:
Answer:
The molality of the solution is 0.3716 mol/kg
The number of moles of solute is 0.0157 mol
The molecular weight of the solute is 129.30 g/mol
The molar mass of the solute is 129.32 g/mol
Explanation:
m (molality of the solution) = ∆T/Kf = (43.17 - 40.32)/7.67 = 0.3716 mol/kg
Number of moles of solute = molality × mass of solvent in kilogram = 0.3716 × 0.04219 = 0.0157 mol
Molecular weight of solute = mass/number of moles = 2.03/0.0157 = 129.3 g/mol
When Kf = 7.66 °C.kg/mol
Molar mass = 2.03 ÷ (2.85/7.66 × 0.04219) = 129.32 g/mol
Answer:
The reaction can produce 287 grams of iron(II) carbonate
Explanation:
To solve this question we must find the moles of iron(II) chloride that react. Using the chemical equation we can find the moles of iron(II) carbonate and its mass -Molar mass FeCO3: 115.854g/mol-
<em>Moles FeCl2:</em>
1.24L * (2.00mol / L) = 2.48 moles FeCl2
As 1 mol FeCl2 produce 1 mol FeCO3, the moles of FeCO3 = 2.48 moles
<em>Mass FeCO3:</em>
2.48mol * (115.854g / mol) =
<h3>The reaction can produce 287 grams of iron(II) carbonate</h3>
