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Shtirlitz [24]
2 years ago
10

What is the temperature of 12.20 mol of gas in a 18.35 l tank at 16.40 atm?

SAT
1 answer:
NeX [460]2 years ago
3 0

The temperature of the given mole and pressure of gas is 300.6K.

Given the data in the question;

  • Amount of gas; n = 12.20mol
  • Volume of gas; V = 18.35L
  • Pressure; P = 16.40atm

Temperature; T = \ ?

<h3>The Ideal Gas Law</h3>

The Ideal gas law or general gas equation emphasizes on the state or behavior of a hypothetical ideal gas. It states that  pressure multiplied by volume is equal to moles multiply by the universal gas constant multiply by temperature. It is expressed;

PV = nRT

Where P is pressure, V is volume, n is the amount of substance, T is temperature and R is the ideal gas constant ( 0.08206Latm/molK 0

The determine the temperature of the gas, we substitute our given values inro the equation above.

PV = nRT\\\\T = \frac{PV}{nR}\\ \\T = \frac{16.40atm\ *\ 18.35L  }{12.20mol\ * \ 0.08206Latm/molK} \\\\T = \frac{300.94Latm}{1.001132molLatm/molK} \\\\T = 300.6K

Therefore, the temperature of the given mole and pressure of gas is 300.6K.

Learn more about Ideal Gas Law: brainly.com/question/4147359

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3 years ago
Determine the empirical formula of a compound containing 47. 37 grams of carbon, 10. 59 grams of hydrogen, and 42. 04 grams of o
Effectus [21]

Considering the definition of empirical formula, the empirical formula is C₃H₈O₂ and the molecular formula is C₉H₂₄O₆.

<h3>Definition of empirical formula</h3>

The empirical formula is the simplest expression to represent a chemical compound, which indicates the elements that are present and the minimum proportion in whole numbers that exist between its atoms, that is, the subscripts of chemical formulas are reduced to the most integers. small as possible.

<h3>Empirical formula in this case</h3>

In this case, you have:

  • Carbon (C): 47.37 grams
  • Hydrogen (H): 10.59 grams
  • Oxygen (O):  42.04 grams

Then it is possible to calculate the number of moles of each atom in the molecule, taking into account the corresponding molar mass:

  • Carbon (C): \frac{47.37 grams}{12 \frac{g}{mole} }= 3.95 moles
  • Hydrogen (H): \frac{10.59 grams}{1 \frac{g}{mole} }= 10.59 moles
  • Oxygen (O):  \frac{42.04 grams}{16 \frac{g}{mole} }= 2.63 moles

The empirical formula must be expressed using whole number relationships, for this the numbers of moles are divided by the smallest result of those obtained. In this case:

  • Carbon (C): \frac{3.95 moles}{2.63 moles }= 1.5
  • Hydrogen (H): \frac{10.59 moles}{2.63 moles }= 4
  • Oxygen (O):  \frac{2.63 moles}{2.63 moles }= 1

Finally, since in the empirical formula the numbers of moles must be expressed in whole numbers, the ratio of atoms is multiplied by some number with which everything is obtained in simple whole numbers (in this case by 2):

  • Carbon (C): 1.5× 2= 3
  • Hydrogen (H): 4× 2= 8
  • Oxygen (O):  1× 2= 2

Therefore the C: H: O mole ratio is 3: 8: 2

Finally, the empirical formula is C₃H₈O₂.

<h3>Molecular formula</h3>

To obtain the molecular formula you must relate its molecular weight (PMc) with the molecular weight of the empirical formula (PMfe).

PMfe= PM (C₃H₈O₂)= 76 g/mole

The molar mass of the compound (PMc) was determined to be 228.276 g/mol. Dividing this mass PMc by the molar mass of the empirical formula PMfe gives:

228.276 g/mol ÷ 76 g/mole= 3

This means that in the molecular formula there are 3 unit formulas (empirical formulas), so it is necessary to multiply the number of all atoms by 3.

Then, the molecular formula is C₉H₂₄O₆.

Learn more about empirical formula:

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brainly.com/question/13112542?referrer=searchResults

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