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
The rate at which a radioactive<span> isotope decays is measured in </span>half-life. The termhalf-life<span> is defined as the time it takes for one-</span>half<span> of the atoms of a </span>radioactive material<span> to disintegrate. </span>Half-lives<span> for </span>various radioisotopes<span> can range from a few microseconds to billions of years.
</span>.
back at it again with that answer
.
zane
The balanced equation would be (1)BaCl2 + (1)H2SO4 --> (1)BaSO4 + (2)HCl2
Then you should know that the coefficients stand for moles.
The thing is I'm not sure if H2SO4 is 35 ml or .200 m.
Also, is this topic stoichiometry?
Answer:
% purity of limestone = 96.53%
Explanation:
Question (4).
Weight of impure CaCO₃ = 25.9 g
Molecular weight of CaCO₃ = 40 + 12 + 3(16)
= 100 g per mole
We know at S.T.P. number of moles of CO₂ = 1 and volume = 22.4 liters
From the given reaction, 1 mole of CaCO₃ reacts with 1 mole or 22.4 liters of
CO₂.
∵ 22.4 liters of CO₂ was produced from CaCO3 = 100 g
∴ 1 liter of CO₂ will be produced by CaCO₃ = 
∴ 5.6 liters of CO₂ will be produced by CaCO₃ = 
= 25 g
Therefore, % purity of CaCO₃ = 
= 
= 96.53 %
Answer:
I hope this helps 52.2 g/mol
Explanation:
1) Solve for the moles using PV = nRT:
n = PV / RT
n = [(700.0 mmHg / 760.0 mmHg atm¯1) (48.0 L)] / [(0.08206 L atm mol¯1 K¯1) (293.0 K)]
n = 1.8388 mol
2) Divide the grams given (96.0) by the moles just calculated above:
96.0 g / 1.8388 mol = 52.2 g/mol
