Gaseous compound Q contains only xenon and oxygen. When a 0.100 g sample of Q is placed in a 50.0-mL steel vessel at 0°C, the pressure is 0.229 atm. What is the likely formula of the compound?

A. XeO

B. $XeO_4$

C. $Xe_2O_2$

D. $Xe_2O_3$

E. $Xe_3O_2$

Answer: The chemical formula of the compound is $XeO_4$

Explanation:

To calculate the molecular mass of the compound, we use the equation given by ideal gas equation:

PV = nRT

Or,

$PV=\frac{w}{M}RT$

where,

P = Pressure of the gas = 0.229 atm

V = Volume of the gas = 50.0 mL = 0.050 L (Conversion factor: 1 L = 1000 mL)

w = Weight of the gas = 0.100 g

M = Molar mass of gas = ?

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = Temperature of the gas = $0^oC=273K$

Putting value in above equation, we get:

$0.229\times 0.050=\frac{0.100}{M}\times 0.0821\times 273\\\\M=\frac{0.100\times 0.0821\times 273}{0.229\times 0.050}=195.4g/mol\approx 195g/mol$

The compound having mass as 195 g/mol is $XeO_4$

Hence, the chemical formula of the compound is $XeO_4$

5 0

3 years ago

What is the number of molecules present in 1.12 dm^3 of nitrogen gas at STP

Anuta_ua [19.1K]

Answer:

$\huge\boxed{\sf No.\ of\ molecules = 3 * 10\²\² \ molecules}$

Explanation:

Given Data:

Volume = v = 1.12 dm³ = 1.12 L

Density of nitrogen at STP = D = 1.25 g / L

Molar mass = M = 14 * 2 = 28 g / mol

Avogadro's Number = $\tt{N_{A}}$ = 6.023 * 10²³ mol⁻¹

Required:

No. of molecules = ?

Formula:

$\tt{No. \ of \ molecules = \frac{Density * Volume}{Molar\ Mass} * N_{A}}$

Solution:

No. of molecules = (1.25*1.12) / 28 * (6.023 * 10²³)

No. of molecules = ( 1.4 / 28 ) * 6.023 * 10²³

No. of molecules = 0.05 * 6.023 * 10²³

No. of molecules = 0.3 * 10²³

No. of molecules = 3 * 10²² molecules

$\rule[225]{225}{2}$

Hope this helped!

4 0

3 years ago

a sample of argon gas is cooled and its volume went from 380. mL to 250. mL. If its final temperature was -55.0C, what was its o

inn [45]

Answer:

The initial temperature was 58.4°C

Explanation:

Given the following data:

initial volume = V₁ = 380 mL = 0.38 L

final volume = V₂ = 250 mL = 0.25 L

final temperature = T₂ = -55°C = 218 K

According to Charles's law, the volume of a gas is directly proportional to the temperature (in Kelvin). The mathematical expression is:

V₁/T₁= V₂/T₂

So, we calculate the initial temperature (V₁) as follows:

T₁ = T₂/V₂ x V₁ = 218 K/(0.25 L) x 0.38 L = 331.36 K ≅ 331.4 K

Finally, we convert the initial temperature from K to °C:

T₁= 331.4 K - 273 = 58.4°C

6 0

3 years ago