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Mrrafil [7]
3 years ago
14

How does the density of a gas depend on the molar mass of the gas?

Chemistry
1 answer:
AlladinOne [14]3 years ago
6 0

Answer:

The density of the ideal gas is directly proportional to its molar mass.

Explanation:

Density is a scalar quantity that is denoted by the symbol ρ (rho). It is defined as the ratio of the mass (m) of the given sample and the total volume (V) of the sample.

\rho = \frac{m}{V}                          ......equation (1)

According to the ideal gas law for ideal gas:

PV = nRT                                       ......equation (2)

Here, V is the volume of gas, P is the pressure of gas, T is the absolute temperature, R is Gas constant and n is the number of moles of gas

As we know,

The number of moles: n = \frac{m}{M}

where m is the given mass of gas and M is the molar mass of the gas

So equation (2) can be written as:

PV = \frac{m}{M}RT

⇒ PM= \frac{m}{V} RT

⇒ \frac{PM}{RT}= \frac{m}{V}             ......equation (3)

Now from equation (1) and (3), we get

\frac{PM}{RT}= \frac{m}{V} = \rho  

⇒ Density of an ideal gas: \rho = \frac{PM}{RT}  

⇒ <em>Density of an ideal gas: ρ ∝ molar mass of gas: M</em>

<u>Therefore, the density of the ideal gas is directly proportional to its molar mass. </u>

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a flask was filled with SO2 at a partial pressure of 0.409 atm and O2 at a partial pressure of 0.601 atm. The following gas-phas
jarptica [38.1K]

Answer:

The equilibrium partial pressure of O2 is 0.545 atm

Explanation:

Step 1: Data given

Partial pressure of SO2 = 0.409 atm

Partial pressure of O2 = 0.601 atm

At equilibrium, the partial pressure of SO2 was 0.297 atm.

Step 2: The balanced equation

2SO2 + O2 ⇆ 2SO3

Step 3: The initial pressure

pSO2 = 0.409 atm

pO2 = 0.601 atm

pSO3 = 0 atm

Step 4: Calculate the pressure at the equilibrium

pSO2 = 0.409 - 2X atm

pO2 = 0.601 - X atm

pSO3 = 2X

pSO2 = 0.409 - 2X atm = 0.297

 X = 0.056 atm

pO2 = 0.601 - 0.056 = 0.545 atm

pSO3 = 2*0.056 = 0.112 atm

Step 5: Calculate Kp

Kp = (pSO3)²/((pO2)*(pSO2)²)

Kp = (0.112²) / (0.545 * 0.297²)

Kp = 0.261

The equilibrium partial pressure of O2 is 0.545 atm

3 0
3 years ago
A beaker is filled to the 500 mL mark with alcohol. What increase in volume (in mL) does the beaker contain when the temperature
Aliun [14]

Answer:

"1.4 mL" is the appropriate solution.

Explanation:

According to the question,

  • v_0=500
  • \alpha =1.12\times 10^{-4}
  • \Delta \epsilon = 25

Now,

Increase in volume will be:

⇒ \Delta V = \alpha\times v_0\times \Delta \epsilon

By putting the given values, we get

           =1.12\times 10^{-4}\times 500\times 25

           =1.12\times 10^{-4}\times 12500

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8 0
3 years ago
A 48.3 mL sample of gas in a cylinder is warmed from 22 C, 87 C to 0.0 C what is the final temperature
snow_lady [41]

The answer is:

the volume stays the same. it is the pressure that increases

7 0
3 years ago
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enyata [817]

Answer:

b)5l x 10kg  c)10kg + 9l   (Not sure for the last 1)

6 0
3 years ago
Write the equilibrium expression of each chemical equation.<br> 2H2S(g) 2H2(g) + S2(g)
hodyreva [135]

Answer:

<u>[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2</u>

Explanation:

2H2S(g)⇋2H2(g)+S2(g)2H2S(g)⇋2H2(g)+S2(g)

The equilibrium constant expression in terms of concentrations is:

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6 0
3 years ago
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