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Sindrei [870]
3 years ago
9

A sample of dry gas weighing 2.1025 g is found to occupy 2.850 L at 22.0oC and 740.0 mm Hg. How many moles of gas are present?

Chemistry
1 answer:
soldi70 [24.7K]3 years ago
4 0

0.11 moles of the gas are present in the sample of dry gas.

Explanation:

Data given:

mass of the gas  = 2.1025 grams

volume of the gas = 2.850 litres

temperature =  22 degrees (273.15+22) = 295.15 K

Pressure = 740 mm Hg or 0.973 atm

moles of the gas =?

R = 0.08206 atmL/Mole K

From the ideal gas law the number of moles can be calculated in the sample of dry gas. Number of moles will be determined by the pressure exerted, volume and temperature of the gas.

The formula:

PV = nRT

n = \frac{PV}{RT}

putting the values in the above equation:

n = \frac{2.850 x 0.973}{0.08206 X 295.15}

  = 0.11 moles

0.11 moles of the dry gas is present in the sample given.

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The characteristic odor of pineapple is due to ethyl butyrate, a compound containing carbon, hydrogen, and oxygen. combustion of
aev [14]

Answer: <em>C₃H₆O</em>


Explanation:


1)  First, calculate the chemical composition, which you do using balance mass using the data given.


2) Data:


i) mass of ethyl butyrate: 2.22 mg

ii) mass of CO₂: 5.06 mg

iii) mass of H₂O: 2.06 mg


3) Solution


i) All the carbon in CO₂ produced was present in the ehtyl butyrate sample.


ii) The molar mass of CO₂ is 44.01 g/mol


iii) Use proportionality:


12.01 g C / 44.01 g CO₂ = x / 5.06 mg C  ⇒ x = 1.38 mg  C


iv) All the H in H₂O was present in the original sample of ethyl butyrate


v) The molar mass of H₂O is 18.015 g/mol


vi) Use proportionality


2.016 g H / 18.015 g H₂O = x / 2.06 mg H ⇒ x = 0.23 mg H


vii) Mass balance:


mass of O in the sample = mass of the sample - mass of H - mass of C


mass of O = 2.22 mg - 1.38 mg - 0.23 mg = 0.61 mg O.


viii) Composition:

C = 1.38 mg

H = 0.23 mg

O = 0.61 mg


Which in terms of composition is the same that:

C = 1.38 g

H = 0.23 g

O = 0.61 g


ix) Divide each by the molar atomic mass of the corresponding element, to obtain the number of moles:


C = 1.38 g / 12.01 g/mol = 0.115 mol

H = 0.23 g / 1.008 g/mol = 0.228 mol

O = 0.61 g / 16 g/mol = 0.0381 mol


x) Divide each by the least number of moles to obtain mole proportion


C = 0.115 / 0.0381 = 3.02 ≈ 3

H = 0.228 / 0.0381 = 1.98 ≈ 5.98 ≈ 6

O = 0.0381 / 0.0381 = 1


xi) Therefore the empirical formula searched is <em>C₃H₆O</em>.


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Answer:

Most likely Potential Energy

Explanation:

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