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3 years ago

PLEASE HELP ASAP

Chemistry

1 answer:

swat323 years ago

7 0

Answer:

12.34 amu

Explanation:

Let the 1st isotope be A

Let the 2nd isotope be B

Let the 3rd isotope be C

From the question given above, the following data were obtained:

1st Isotope (A):

Mass of A = 12.32 amu

Abundance (A%) = 19.5%

2nd isotope (B):

Mass of B = 13.08 amu

Abundance (B%) = 26.23%

3rd isotope (C):

Mass of C = 11.99 amu

Abundance (C%) = 54.27%

Atomic mass of X =?

The atomic mass of the element X can be obtained as follow:

Atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]

= [(12.32 × 19.5)/100] + [(13.08 × 26.23)/100] + [(11.99 × 54.27)/100]

= 2.402 + 3.431 + 6.507

= 12.34 amu

Thus, the atomic mass of the element X is 12.34 amu

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A 1.93-mol sample of xenon gas is maintained in a 0.805-L container at 306 K. Calculate the pressure of the gas using both the i

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Explanation :

First we have to calculate the pressure of gas by using ideal gas equation.

$PV=nRT$

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V = Volume of $Xe$ gas = 0.805 L

n = number of moles $Xe$ = 1.93 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $Xe$ gas = 306 K

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$P=60.2atm$

Now we have to calculate the pressure of gas by using van der Waals equation.

$(P+\frac{an^2}{V^2})(V-nb)=nRT$

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V = Volume of $Xe$ gas = 0.805 L

n = number of moles $Xe$ = 1.93 mole

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $Xe$ gas = 306 K

a = pressure constant = $4.19L^2atm/mol^2$

b = volume constant = $5.11\times 10^{-2}L/mol$

Now put all the given values in above equation, we get:

$(P+\frac{(4.19L^2atm/mol^2)\times (1.93mole)^2}{(0.805L)^2})[0.805L-(1.93mole)\times (5.11\times 10^{-2}L/mol)]=1.93mole\times (0.0821L.atm/mol.K)\times 306K$

$P=44.6atm$

Therefore, the pressure of the gas using both the ideal gas law and the van der Waals equation is, 60.2 atm and 44.6 atm respectively.

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