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il63 [147K]
3 years ago
14

What is the answer to...

Chemistry
1 answer:
adoni [48]3 years ago
8 0

Answer: short answer short 0.65 / 5.37

15% NaOH by mass is 15g of NaOH in (100-15)g of solution.

No. Of moles of NaOH in solution = 0.65

No of moles of Water in solution = 4.72

Mole Fraction =  

(

(Moles of Solvent)

(Moles Of Solute)+(Moles of Solvent)

)

OR

(

(Moles of Solvent)

Total Number Of Moles

)

= 0.65 / 5.37

= 0.121

Explanation:

Hope this helped

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Oxygen has 8 plus-charged protons. The number of minus-charged electrons is
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Mike mixes two chemicals in a container. The container quickly becomes
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How many moles, kmols in: 100 g of CO2, 1 litre of ethyl alcohol of density 0.789 g/cm3 and a) 1.5m3 of O2 at 25°C and 1 atm. b)
Nutka1998 [239]

Explanation:

1) Mass of carbon dioxide = 100 g

Molar mass of carbon dioxide = 44 g/mol

Moles of carbon dioxide =\frac{100 g}{44 g/mol}=2.273 moles

1 mol = 0.001 kmol

2.273 moles= 2.273 × 0.001 kmol = 2.273\times 10^{-3} kmol

2) 1 liter of ethyl alcohol of density 0.789 g/cm^3

Volume of ethyl alcohol ,V= 1 L = 1000 mL

Density of ethyl alcohol =d = 0.789 g/cm^3

1 cm^3=1 mL

Mass of ethyl alcohol = m

m=d\times V=0.789 g/cm^3\times 1000 mL=789 g

Molar mass of  ethyl alcohol = 46 g/mol

Moles of ethyl alcohol = \frac{789 g}{46 g/mol}=17.152 mol

17.152 mol=17.152\times 0.001 kmol=1.7152\times 10^{-4} kmol

3) Volume of oxygen gas,V =1.5 m^3=1500 L

1 m^3= 1000 L

Temperature of the gas = T= 25°C = 298.15 K

Pressure of the gas ,P= 1 atm

Moles of oxygen gas = n

PV=nRT

n=\frac{RT}{PV}=\frac{0.0821 atm L/mol K\times 298.15 K}{1 atm\times 1500 L}=0.01632 mol

0.01632 mol = 0.01632 × 0.001 kmol=1.632\times 10^{-5} kmol

4) Volume water in mixture = 1 L

Density of water =  1000 kg/m^3=\frac{1,000,000 g}{1000 L}=1000 g/L

Mass of water = 1000 g/L\times 1 L = 1000 g

Volume of alcohol = 2.5 L

Density of alcohol =  789 kg/m^3=\frac{789000 g}{1000 L}=789 g/L

Mass of alcohol = 789 g/L\times 2.5 L = 1972.5 g

Mass of mixture = 1000 g + 1972.5 g = 2972.5 g

Mass percentage of water :

\frac{1000 g}{2972.5 g}\times 100=33.64\%

Mass percentage of alcohol :

\frac{1972.5 g}{2972.5 g}\times 100=66.36\%

Moles of water :

n_1=\frac{1000 g}{18 g/mol}=55.55 mol

Moles of alcohol =

n_2=\frac{1972.5 g}{46 g/mol}=42.88 mol

Mole fraction of water :

\chi_1=\frac{n_1}{n_1+n_2}=\frac{55.55 mol}{55.55 mol+42.88 mol}=0.5644

Mole fraction of alcohol :

\chi_2=\frac{n_2}{n_1+n_2}=\frac{42.88 mol}{55.55 mol+42.88 mol}=0.4356

3 0
2 years ago
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