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Mama L [17]
3 years ago
11

A gas occupies a volume of 2.4 L at 14.1 kPa. What volume will the gas occupy at 84.6 kPa?

Chemistry
1 answer:
Naddik [55]3 years ago
3 0

Answer:

  • <u>0.40 L</u>

Explanation:

Boyle's law for gases states that, at constant temperature, the volume and pressure of a fixed amount of gas are inversely related.

Mathematically, that is:

  • PV = constant

  • P₁V₁ = P₂V₂

Here, you have:

  • V₁ = 2.4 L
  • P₁ = 14.1 Kpa
  • P₂ = 84.6 KPa
  • V₂ = ?

Then, you can solve for V₂:

  • V₂ = P₁V₁ / P₂

Substitute and compute:

  • V₂ = 14.1 KPa × 2.4L / 84.6 KPa = 0.40 L ← answer
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Naturally occurring element X exists in three isotopic forms: X-28 (27.979 amu, 92.21% abundance), X-29 (28.976 amu 4.70% abunda
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<u>Answer:</u> The average atomic mass of X is 28.09 amu

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i .....(1)

  • <u>For isotope 1:</u>

Mass of isotope 1 = 27.979 amu

Percentage abundance of isotope 1 = 92.21 %

Fractional abundance of isotope 1 = 0.9212

  • <u>For isotope 2:</u>

Mass of isotope 2 = 28.976 amu

Percentage abundance of isotope 2 = 4.70 %

Fractional abundance of isotope 2 = 0.0470

  • <u>For isotope 3:</u>

Mass of isotope 3 = 29.974 amu

Percentage abundance of isotope 3 = 3.09 %

Fractional abundance of isotope 3 = 0.0309

Putting values in equation 1, we get:

\text{Average atomic mass of X}=[(27.979\times 0.9212)+(28.976\times 0.0470)+(29.974\times 0.0309)]

\text{Average atomic mass of X}=28.09amu

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3 years ago
How many moles of nitrogen are present at STP if the volume is 846L
Artyom0805 [142]
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Alternatively, you can go the long route and use the ideal gas law to solve for the number of moles of nitrogen given STP conditions (273 K and 1.00 atm). From PV = nRT, we can get n = PV/RT. Plugging in our values, and using 0.08206 L•atm/K•mol as our gas constant, R, we get n = (1.00)(846)/(0.08206)(273) = 37.8 moles, which confirms our answer.
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