Combined gas law is
PV/T = K (constant)
P = Pressure
V = Volume
T = Temperature in Kelvin
For two situations, the combined gas law can be applied as,
P₁V₁ / T₁ = P₂V₂ / T₂
P₁ = 3.00 atm P₂ = standard pressure = 1 atm
V₁ = 720.0 mL T₂ = standard temperature = 273 K
T₁ = (273 + 20) K = 293 K
By substituting,
3.00 atm x 720.0 mL / 293 K = 1 atm x V₂ / 273 K
V₂ = 2012.6 mL
hence the volume of gas at stp is 2012.6 mL
Moles (mol) = mass (g) / molar mass (g/mol)
Mass of NaCl = 21.7 g
Molar mass of NaCl = <span>58.4 g/mol
Hence, moles of NaCl = </span>21.7 g / 58.4 g/mol = 0.372 mol
Hence moles of NaCl in the mixture is 0.372 mol.
Let's assume that mixture has only given compounds and free of impurities.
Then, we can present this as a mole percentage.
mole % = (moles of desired substance / Total moles of the mixture) x 100%
Hence,
mole % of NaCl = (moles of NaCl / Total moles of the mixture) x 100%
Total moles of mixture = moles of NaCl + KCl + LiCl
Mass of KCl = 3.74 g
Molar mass of NaCl = 74.6 g/mol
Hence, moles of NaCl = 3.74 g / 74.6 g/mol = 0.050 mol
Mass of NaCl = <span>9.76 g
</span>Molar mass of NaCl = 42.4 g/mol
Hence, moles of NaCl = 9.76 g / 42.4 g/mol = 0.230 mol
Total moles = 0.372 mol + 0.050 mol + 0.230 mol = 0.652 mol
mole % of NaCl = (moles of NaCl / Total moles of the mixture) x 100%
= (0.372 mol / 0.652 mol) x 100%
= 57.06%
Hence, mixture has 57.06% of NaCl as the mole percentage.
Answer - A, B, D, E
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Answer:
13.26 g has 4 significant figures
Free energy is the answer i hope this helped
