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tekilochka [14]
3 years ago
13

HELP FAST PLZ!!!!

Chemistry
1 answer:
ra1l [238]3 years ago
3 0
D  ecosystem i think good luck 
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Which of the following is the correct definition for electronegativity?
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Read 2 more answers
Bauxite is an ore that contains the element aluminum. If you obtained 108 grams of aluminum from an ore sample that initially we
Stels [109]

Answer:

53% aluminium is present.

Explanation:

Given data:

Mass of aluminium in bauxite = 108 g

Total mass of sample = 204 g

Percentage of aluminium = ?

Solution:

Formula:

Percentage = Mass of aluminium / total mass of sample× 100

Now we will put the values in formula:

Percentage = 108 g/204 × 100

Percentage = 0.53  × 100

Percentage = 53%

So in given 204 g of bauxite there is 53% aluminium is present.

8 0
3 years ago
Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21 ∘C∘C. Express the pressure
Volgvan

The question is incomplete, complete question is ;

A deep-sea diver uses a gas cylinder with a volume of 10.0 L and a content of 51.8 g of O_2 and 33.1 g of He. Calculate the partial pressure of each gas and the total pressure if the temperature of the gas is 21°C.Express the pressures in atmospheres to three significant digits separated by commas.

Answer:

Partial pressure of the oxygen gas is 3.91 atm.

Partial pressure of the helium gas is 20.0 atm

Total pressure of the gases is 24.0 atm

Explanation:

Moles of oxygen gas = n_1=\frac{51.8}{32 g/mol}=1.619 mol

Moles of helium gas = n_2=\frac{33.1 g}{4 g/mol}=8.275 mol

Total moles of gas = n_1+n_2=(1.619 +8.275 ) mole=9.894 mol

Volume of the cylinder = V = 10.0 L

Total pressure in the cylinder = P = ?

Temperature of the gas in cylinder = T = 21°C = 21 + 273 K = 294 K

PV = nRT ( ideal gas equation )

P=\frac{nRT}{V}

=\frac{9.894 mol\times 0.0821 atm L/mol K\times 294 K}{10.0 L}

P = 23.88 atm ≈ 23.9

Partial pressure of the individual gas will be determined by the help of Dalton's law:

partial pressure = Total pressure × mole fraction of gas

Partial pressure of the oxygen gas

p_{1}=P\times \chi_{1}=P\times \frac{n_1}{n_1+n_2}

p_1=23.88 atm\times \frac{1.619 mol}{9.894 mol}=3.91 atm

Partial pressure of the helium gas

p_{2}=P\times \chi_{2}=P\times \frac{n_2}{n_1+n_2}

p_2=23.88 atm\times \frac{8.275 mol}{9.894 mol}=19.97 atm\approx 20.0 atm

6 0
3 years ago
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