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kumpel [21]
1 year ago
13

A 10-liter container has 2 moles of oxygen at a pressure of 92 kpa. The effective speed (rms) of the oxygen molecules in the gas

is?
Chemistry
1 answer:
bezimeni [28]1 year ago
4 0

The effective speed (rms) of the oxygen gas is 293.68 m/s.

<h3></h3><h3>What is Root-mean-square velocity?</h3>

Root mean square velocity is the square root of the mean of squares of the velocity of individual gas molecules

v_{rms}=\sqrt[]{\frac{3RT}{M} }

<em>where </em>R = universal gas constant

M = molar mass of the gas in kg/mol

T = temperature in Kelvin

According to the ideal gas law,

PV = nRT

RT = \frac{PV}{n}

Substitute in the rms velocity formula,

v_{rms} = \sqrt[]{\frac{3PV}{nM} }

P = 92 kPa, V = 10 L, n = 2 moles and M = 32 x 10⁻³ kg/mol

v_{rms} = \sqrt[]{\frac{3\times92\times10}{2\times32\times10^-^3} }

=293.68 m/s

Thus, the effective speed (rms) of O₂ gas is 293.68 m/s.

Learn more about Root-mean-square velocity:

brainly.com/question/15995507

#SPJ4

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8 0
3 years ago
If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is Group of answer choices
BaLLatris [955]

Answer:

The water is completely vaporized at this stage.

Explanation:

The complete question is

If 5.0 kJ of energy is added to a 15.5-g sample of water at 10.°C, the water is

-boiling

-completely vaporized

-frozen solid

-decomposed

-still a liquid

Energy added = 50 kJ = 50000 J

mass of water = 15.5 g = 0.0155 kg

temperature of water = 10 °C

We know that the energy posses by a mass of water at a given temperature is given as

H = mcT

where H is the energy possessed by the mass of water

m is the mass of the water

c is the specific heat capacity of water = 4200 J/ kg- °C

T is the temperature of the water

substituting values, the energy of this amount of water is

H = 0.0155 x 4200 x 10 = 651 J

If 50 kJ is added to the water, the energy increases to

50000 J + 651 J = 50651 J

Temperature of this water at this stage will be gotten from

H = mcT

we solve for the new temperature

50651 = 0.0155 x 4200 x T

50651 = 65.1 x T

T = 50651/65.1 = 778.05 °C

This temperature is well over 100 °C, which is the vaporization temperature of water, but less than 3000 °C for its molecules to decompose.

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Complete the statement<br>qxy- bxy+cxy= xy( )​
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4 years ago
A mixture of methane and carbon dioxide gases contains methane at a partial pressure of 431 mm Hg and carbon dioxide at a
KatRina [158]

Answer:

XCH₄ = 0.461

XCO₂ = 0.539

Explanation:

Step 1: Given data

  • Partial pressure of methane (pCH₄): 431 mmHg
  • Partial pressure of carbon dioxide (pCO₂): 504 mmHg

Step 2: Calculate the total pressure in the container

We will sum both partial pressures.

P = pCH₄ + pCO₂

P = 431 mmHg + 504 mmHg = 935 mmHg

Step 3: Calculate the mole fraction of each gas

We will use the following expression.

Xi = pi / P

XCH₄ = pCH₄/P = 431 mmHg/935 mmHg = 0.461

XCO₂ = pCO₂/P = 504 mmHg/935 mmHg = 0.539

3 0
3 years ago
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