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geniusboy [140]
3 years ago
10

An aerosol can contains 800.0 mL of compressed gas at 10.2 atm pressure. When the gas is sprayed into a large plastic bag, the b

ag inflates to a volume of 4.14 L. What is the pressure of gas (atm) inside the plastic bag if the temperature did not change and all of the gas was transferred completely?
Chemistry
2 answers:
Roman55 [17]3 years ago
5 0

Answer:

The pressure of the gas inside the plastic bag is 1.971 atm

Explanation:

Since temperature did not change, it means it is constant. This will then be Boyle's law. Thus

P1V1 =P2V2

Where P1 is the initial pressure = 10.2 ATM, P2 is the final pressure which is to be calculated.

V1 is the initial volume = 800.0 mL and V2 is the final volume of the gas = 4.14 L

From the formula,

P2 = P1V1/V2

= 10.2×800÷4.14

= 1.971 atm

Klio2033 [76]3 years ago
5 0

Answer:

Pressure of gas in the plastic bag is 1.97 atm

Explanation:

It is possible to answer this question by using Boyle's law: Boyle's law describes that, for ideal gases, the pressure of the gas is inversely proportional to the volume it occupies. Is expressed as:

P1V1 = P2V2

Where:

P1 is the initial pressure of the gas (In the problem, 10.2 atm).

V1 is initial volume (800.0mL)

V2 is final volume (4.14L ≡ 4140mL)

And P2 is the final pressure.

Replacing:

10.2 atm×800.0mL = 4140mL×P2

<em>P2 = 1.97 atm</em>

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What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)
lbvjy [14]

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

\Delta T = i \times Kf \times m =   1 \times 6.90 \°C/m  \times 1.324m = 9.14  \°C

where,

  • i: van 't Hoff factor (1 for non-electrolytes)
  • Kf: cryoscopic constant
  • m: molality

The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:

T = 80.26 \° C - 9.14 \° C = 71.12 \° C

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

Learn more: brainly.com/question/2292439

3 0
2 years ago
Calculate the standard entropy of vaporization of ethanol, C2H5OH, at 285.0 K, given that the molar heat capacity at constant pr
MArishka [77]

Answer: The standard entropy of vaporization of ethanol is 0.275 J/K

Explanation:

C_2H_5OH(l)\rightleftharpoons C_2H_5OH(g)

Using Gibbs Helmholtz equation:

\Delta G=\Delta H-T\Delta S

For a phase change, the reaction remains in equilibrium, thus \Delta G=0

\Delta H=T\Delta S

Given: Temperature = 285.0 K

\Delta H=78.3J/mol

Putting the values in the equation:

78.3J=285.0K\times \Delta S

\Delta S=0.275J/K

Thus  the standard entropy of vaporization of ethanol is 0.275 J/K

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Answer:

Below.

Explanation:

Coal gas is a mixture of a variety of gases: inflammable gases including, hydrogen, methane, ethylene, carbon monoxide and volatile hydrocarbons and small amounts of non flammable gases like nitrogen and carbon dioxide.

Water gas consists mainly of carbon monoxide and hydrogen.

Producer gas is similar to water gas and consists mainly of carbon monoxide and hydrogen together with nitrogen and carbon dioxide.

Natural gas occurs naturally and consists mainly of methane with small amounts of other hydrocarbon gases.

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The average distance between nitrogen and oxygen atoms is 115 pm in a compound called nitric oxide. what is this distance in cen
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