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Anton [14]
2 years ago
10

For this exercise, you can simulate the described conditions by changing the values in the run experiment tool of the simulation

. to be able to measure the effects on the gas volume, slide the rspd bar in the properties box to v (l) so that the volume bar turns yellow. this way, volume becomes the only dependent variable. note that when volume is selected with rspd, its value can no longer be directly controlled. suppose a piston automatically adjusts to maintain a gas at a constant pressure of 14.20 atm . for the initial conditions, there are 0.04 mol of helium at a temperature of 240.00 k . this gas occupies a volume of 0.06 l under those conditions. what volume will the gas occupy if the number of moles is increased to 0.07 mol (n2) from the initial conditions? what volume will the gas occupy if the temperature is increased to 340.00 k (t2) from the initial conditions?
Chemistry
2 answers:
Serggg [28]2 years ago
4 0

Answer: The volume of the gas when number of moles are increased is 0.105L.

And the volume of the gas when temperature is increased is 0.085L

Explanation: The pressure of the gas is held constant which is 14.20 atm

We are given Initial conditions, which are:

V = 0.06L

T = 240 K

n = 0.04 mol

  • When the number of moles of gas are increased, we will use Avogadro's Law, which says that the volume is directly proportional to the number of moles of gas at constant temperature and pressure.

Mathematically,

\frac{V_1}{n_1}=\frac{V_2}{n_2}

Final conditions are:

n_2=0.07mol

V_2=?L

Putting values in above equation, we get

\frac{0.06}{0.04}=\frac{V_2}{0.07}

V_2=0.105L

  • When the temperature of the gas is increased, we will use Charles's Law, which says that the volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

Mathematically,

\frac{V_1}{T_1}=\frac{V_2}{T_2}

Final conditions are:

T_2=340K

V_2=?L

Putting values in above equation, we get

\frac{0.06}{240}=\frac{V_2}{340}

V_2=0.085L

Furkat [3]2 years ago
3 0
1) ideal gas law: p·V = n·R·T.
p - pressure of gas.
V -volume of gas.
n - amount of substance.
R - universal gas constant.
T - temperature of gas.
n₁ = 0,04 mol, V₁ = 0,06 l.
n₂ = 0,07 mol, V₂ = 0,06 · 0,07 ÷ 0,04 = 0,105 l.
2) V₁ = 0,06 l, T₁ = 240,00 K.
T₂ = 340,00 K, V₂ = 340 · 0,06 ÷ 240 = 0,05 l.
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If 43.1 g of O2 and 6.8 g of CO2 are placed in a 13.7 L container at 34 degrees C, what is the mixture of gasses?
jonny [76]

Answer:

The total pressure of the gas mixture = 2.76 atm

Note: The question is not complete. The complete question is as follow:

If 43.1 g of O2 and 6.8 g of CO2 are placed in a 13.7 L container at 34 degree Celsius , what is the pressure of the mixture of gases?

Explanation:

Mass of O₂ gas = 43.1 g, molar mass of O₂ gas = 32.0 g/mol

Number of moles of O₂ gas = 43.1/32.0 = 1.347 moles

Mass of CO₂ gas = 6.8 g, molar mass of CO₂ gas = 44.0 g

Number of moles of CO₂ gas = 6.8/44 = 0.155 moles

Total number of moles of gas mixture, n = (1.347 + 0.155) = 1.502 moles

Volume of gas mixture, V = 13.7 L

Temperature of gas mixture, T = 34 °C = (273.15 + 34) K = 307.15 K

Pressure of gas mixture = ?

Molar gas constant, R = 0.0821 liter·atm/mol·K.

Using the ideal gas equation: PV =nRT

P = nRT/V

P = (1.502 × 0.0821 × 307.15) / 13.7

P = 2.76 atm

Therefore, the total pressure of the gas mixture = 2.76 atm

5 0
2 years ago
The electronegativity values of carbon, hydrogen, and nitrogen are compared in the table.
Fofino [41]

Answer:

C. CH₄ is less than NH₃ because the NH bond is more polar than the CH bond

Explanation:

The intermolecular forces between ammonia is far stronger than for methane. Between the molecules of ammonia we have the presence of hydrogen bonds. This bond is absent in methane.

Hydrogen bonds are one of the strongest intermolecular forces. It is as a result of the electrostatic attraction between the hydrogen atom of one molecule and the electronegative atom N, O and F of another molecule.

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The strength of the hydrogen bond depends on the electronegativity of the combining atoms.

5 0
2 years ago
Consider this equation: 2.524 g (5.1 × 106 g) ÷ (6.85 × 103 g) = ? How many significant figures should the result have?
weqwewe [10]

Answer:

7

Explanation:

2.524g(5.1)(106)g

(6.85)(103)g

=

1364.4744g2

705.55g

=

1364.4744g

705.55

=

1364.4744g

705.55

=1.933916g

6 0
3 years ago
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dsp73
I think the answer is b
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