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BlackZzzverrR [31]
4 years ago
7

A gas mixture of Ne and Ar has a total pressure of 4.00 atm and contains 16.0 mol of gas. If the partial pressure of Ne is 2.75

atm, how many moles of Ar are in the mixture?
A) 11.0
B) 5.00
C) 6.75
D) 9.25
E) 12.0
Chemistry
1 answer:
Stolb23 [73]4 years ago
4 0

Answer:

5 moles of Argon is present in the mixture.

Explanation:

Total pressure of the gaseous mixture = 4 atm

Total number of moles = 16

Partial pressure of Ne = 2.75 atm

By Dalton's law of partial pressure, the total pressure of gaseous mixture is the sum of partial pressures of individual gases which are non-reactive.

Hence:

P_{total}=P_{Ar}+P_{Ne}\\4=P_{Ar}+2.75\\P_{Ar}=1.25\ atm

Also :

Partial pressure = mole fraction*total pressure

P_{Ar}=X_{Ar}P_{total}

X_{Ar}=\frac{1.25}{4}=0.3125

\frac{n_{Ar}}{n_{total}}=0.3125\\n_{Ar}=5

∴Number of moles of Argon = 5

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<u>Answer:</u>

<u>For a:</u> The edge length of the unit cell is 314 pm

<u>For b:</u> The radius of the molybdenum atom is 135.9 pm

<u>Explanation:</u>

  • <u>For a:</u>

To calculate the edge length for given density of metal, we use the equation:

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where,

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Z = number of atom in unit cell = 2  (BCC)

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N_{A} = Avogadro's number = 6.022\times 10^{23}

a = edge length of unit cell =?

Putting values in above equation, we get:

10.28=\frac{2\times 95.94}{6.022\times 10^{23}\times (a)^3}\\\\a^3=\frac{2\times 95.94}{6.022\times 10^{23}\times 10.28}=3.099\times 10^{-23}\\\\a=\sqrt[3]{3.099\times 10^{-23}}=3.14\times 10^{-8}cm=314pm

Conversion factor used:  1cm=10^{10}pm  

Hence, the edge length of the unit cell is 314 pm

  • <u>For b:</u>

To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

R=\frac{\sqrt{3}a}{4}

where,

R = radius of the lattice = ?

a = edge length = 314 pm

Putting values in above equation, we get:

R=\frac{\sqrt{3}\times 314}{4}=135.9pm

Hence, the radius of the molybdenum atom is 135.9 pm

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