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laiz [17]
2 years ago
13

Oxygen gas was produced in a reaction and collected over water. A 136.1 mL mL sample of gas was collected over water at 25C and

1.06 atm. The vapor pressure of water is 23.76 mmHg at 25C. Find the mass of ocygen gas collected in a reaction and collected over water. A 136.1 mL sample of gas was collected over water at 25C and 1.06 atm. The vapor pressure of water is 23.76 mmHg at 25C. Find the mass of oxygen gas collected.
Chemistry
1 answer:
Xelga [282]2 years ago
4 0

Answer:

Explanation:

We shall find volume of gas at NTP or at 273 K , 760 mm of Hg .

Pressure of given gas = 1.06 x 760 mm of Hg less vapor pressure of water .

= 805.6 - 23.76 = 781.84 mm of Hg

For it we use gas law formula ,

P₁V₁ / T₁ = P₂V₂ / T₂

781.84 x 136.1 / ( 273 + 25 ) = 760 x V₂ / 273

= 128.26 mL .

= 128.26  x 10⁻³ L .

22.4 L of oxygen will have mass of 32 g

128.26 x 10⁻³ L of oxygen will have mass of 32 x 128.26 x 10⁻³ / 22.4 g

= 183.22 mg .

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For all three questions, we will use the fact that

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  • 0.203 = (moles of solute)/0.175
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Considering the hydrochloric acid solution, if we have 0.035523 mol, then:

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<u />

2) If there is 20.3 mL = 0.0203 L, then:

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  • moles of solute = 0.16646 mol

This means that the molarity of the diluted solution is:

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3) If we need 1.50 L of 0.700 M solution, then:

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Considering the 9.36 M acid solution, from which we need 1.05 mol of perchloric acid from,

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8 0
1 year ago
How many moles of magnesium, Mg, are there in 4.75 grams of magnesium?
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Molar mass Mg = 24.3 g/mol

1 mole mg ------------ 24.3 g
?? moles mg --------- 4.75 g

4.75 x 1 / 24.3 => 0.195 moles of Mg

hope this helps!
3 0
3 years ago
What are the terms that descripe these values for the element oxygen?
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Answer:

16 is the mass number. 8 is the atomic number.

5 0
2 years ago
Use the reaction given below to solve the problem that follows: Calculate the mass in grams of aluminum oxide produced by the re
bearhunter [10]

Answer:  28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}   

\text{Moles of} Al=\frac{15.0g}{27g/mol}=0.556moles

The balanced chemical equuation is:

4Al+3O_2\rightarrow 2Al_2O_3  

According to stoichiometry :

4 moles of Al produce == 2 moles of Al_2O_3

Thus 0.556 moles of Al will produce=\frac{2}{4}\times 0.556=0.278moles  of Al_2O_3

Mass of Al_2O_3=moles\times {\text {Molar mass}}=0.278moles\times 102g/mol=28.4g

Thus 28.4 g of aluminum oxide is produced by the reaction of 15.0 g of aluminum metal.

7 0
2 years ago
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