Chemistry, 30.11.2019 05:31 froyg1234

An unknown diatomic gas has a density of 3.164 g/l at stp. what is the identity of the gas?

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Answer from: asdf334asdf334

The identity of the diatomic gas is Chlorine (Cl)

From the question, we are to determine the identity of a gas that has a density of 3.164 g/L at stp.

To determine the identity of the diatomic gas, we will determine the mass of 1 mole of the gas

From the formula

$Density = \frac{Mass}{Volume }$

Then,

Mass = Density × Volume

Also, at standard temperature and pressure (STP), 1 mole of any gas will occupy a volume of 22.4 L

∴ Mass of the diatomic gas = 3.164 × 22.4

Mass of 1 mole of the diatomic gas = 70.8736 g

Now, we will determine the atomic mass of the gas

Since, the gas is a diatomic gas,

The atomic mass of the gas = $\frac{70.8736}{2}$

The atomic mass of the gas = 35.4368 g/mol

The atomic mass of the gas ≅ 35.44 g/mol

This is reasonably close to the atomic mass of chlorine ( At. mass of Cl = 35.453 g/mol)

Hence, the identity of the diatomic gas is Chlorine (Cl)

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Answer from: jadeossowski3590

The identity of the gas is Cl₂ (chlorine)

Explanation:

STP conditions are:

1 atm → Pressure

273.15 K → T°

So, we must use the Ideal Gas Law to get the moles.

Before that, we will think density as data where 3.164 g of diatomic gas are contained in 1 L of volume.

P.V = n . R . T

1 atm . 1 L = n. 0.082 L.atm/mol.K . 273.15K

1 L.atm / (0.082 L.atm/mol.K . 273.15K) = n

0.0446 mol = n

This quantity of diatomic gas, are 3.164 g so the molar mass will be:

Mass / mol = molar mass

3.164 g / 0.0446 mol = 70.9 g/m

The element (a diatomic molecule), which has that molar mass in the periodic table is the Cl₂.

1 Cl = 35.45 g/m

Cl₂. = 70.9 g/m

Answer from: Quest

The kinetic energy of the liquid increases. hope this !

Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

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