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:
C Group 11
Explanation:
Group 11, by modern IUPAC numbering, is a group of chemical elements in the periodic table, consisting of copper (Cu), silver (Ag), and gold (Au).
These elements show highest electrical conductivity.
Answer: 2 grams
Explanation:
chemical formula of C6H:
C6H
relative atomic mass of
C:12
H:1
C6:72
H1:12
relative formula mass of
C6H1=73
molar mass of glucose:
73g
the molar mass is the mass of one mole of a substance.
in
26.2g
of C6H there are
26.2 divide by 73g moles present.
there are 0.358 or 35.8*10-1
I gram of NH3 is 17.03052
2 grams of NH3 is 35
Answer:
1.35 × 10⁴ kg/m³ at 22 °C; 1.34 × 10⁴ kg/m³ at 100 °C
Explanation:
The cubic expansivity (γ) of a liquid is the fractional change in volume per unit change in temperature.
Multiply by V₀ΔT and transpose
ΔV = γV₀ΔT
and
V = V₀ + ΔV
===============
<em>At 0 °C
</em>
Assume you have 1 m³ of Hg
ρ = m/V Multiply by V and transpose
m = ρV
ρ = 1.36 × 10⁴ kg/m³
m = 1.36 × 10⁴ × 1 = 1.36 × 10⁴ kg
===============
<em>At 22 °C
</em>
Assume that you have 1 m³ of Hg
γ = 180 × 10⁻⁶ K⁻¹
ΔT = 22 °C – 0 °C = 22 °C
ΔV = 180 × 10⁻⁶ × 22
ΔV = 3.96 × 10⁻³ m³ Calculate volume
V = 1 + 0.00396
V = 1.00396 m³ Calculate density
ρ = 1.36 × 10⁴/1.00396
ρ = 1.35 × 10⁴ kg/m³
===============
<em>At 100 °C
</em>
ΔT = 100 °C – 0 °C = 100 °C
ΔV = 180 × 10⁻⁶ × 100
ΔV = 0.0180 m³ Calculate volume
V = 1 + 0.0180
V = 1.0180 m³ Calculate density
ρ = 1.36 × 10⁴/1.0180
ρ = 1.34 × 10⁴ kg/m³
