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zavuch27 [327]
3 years ago
15

How many moles of a gas sample are in a 20.0 L container at 373 K and 203 kPa? The gas constant is 8.31 L-kPa/mol-K. 0.33 moles

3.05 moles 1.30 moles 0.65 moles
Chemistry
2 answers:
Gennadij [26K]3 years ago
8 0
We need to use the ideal gas formula--> PV= nRT 
P, pressure= 203 kPa
V, volume= 20.0 Liters
n, moles= ?
R, constant= <span>8.31 L-kPa/mol-K
T, temperature= 373 K

now, lets plug these values into the formula

(203)(20.0)= (n)(8.31)(373)

n= 1.31 moles</span>
Degger [83]3 years ago
3 0

<u>Answer:</u> The number of moles of the gas is 1.31 mol.

<u>Explanation:</u>

To calculate the number of moles of gas, we use the equation given by ideal gas, which is:

PV=nRT

where,

P = pressure of the gas = 203 kPa

V = volume of the gas = 20 L

n = Number of moles of gas = ? mol

R = Gas constant = 8.31\text{L kPa }mol^{-1}K^{-1}

T = temperature of the gas = 373 K

Putting values in above equation, we get:

203kPa\times 20L=n\times 8.31\text{L kPa }\times 373K\\\\n=1.31mol

Hence, the number of moles of the gas is 1.31 mol.

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Acetic acid weighs 60g contains 40% carbon, 6.67% hydrogen, and 55.33% oxygen. Calculate it’s molecular formula.
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C₂H₄O₂ is the molecular formula of acetic acid

Explanation:

Acetic acid have a molecular wight of 60 g/mol.

The correct mass percentages for acetic acid are 40% carbon, 6.67% hydrogen and 53.33 % oxygen.

To find the molecular formula knowing the elements mass percentages we use the following algorithm:

we divide each percentage by the atomic weight of the element

for carbon 40 / 12 = 3.33

for hydrogen 6.67 / 1 = 6.67

for oxygen 53.33 / 16 = 3.33

now we divide each value obtained by the smallest value, which is 3.33

for carbon 3.33 / 3.33 = 1

for hydrogen 6.67 / 3.33= 2

for oxygen 3.33 / 3.33 = 1

empirical formula for acetic acid is CH₂O

molecular formula for acetic acid is (CH₂O)ₓ

30x g/mol = 60 g/mol (given by the problem)

x = 2

so the molecular formula for acetic acid is C₂H₄O₂

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Taking into account the reaction stoichiometry, 2.13 grams of magnesium was dissolved in the solution.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Mg²⁺(aq) + Na₂CO₃(aq) → MgCO₃(s) + 2 Na⁺(aq)

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Mg²⁺: 1 mole
  • Na₂CO₃: 1  mole
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  • Na⁺: 2 moles

The molar mass of the compounds is:

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  • MgCO₃: 84.3 g/mole
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Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

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The following rule of three can be applied: If by reaction stoichiometry 1 mole of Na₂CO₃ react with 24.3 grams of magnesium, 0.0877 moles of Na₂CO₃ react with how much mass of magnesium?

mass of magnesium=\frac{0.0877 moles of Na_{2}C O_{3}x24.3 grams of magnesium }{1 mole of Na_{2}C O_{3}}

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Finally, 2.13 grams of magnesium was dissolved in the solution.

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