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klasskru [66]
3 years ago
12

Which of the following is a variable property for a gas ? Select all that apply.

Chemistry
2 answers:
disa [49]3 years ago
8 0
C) volume
.........
The volume of gas depends on e.g. temperature and pressure.
Molodets [167]3 years ago
4 0

The correct answer is volume and shape.

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In which state of matter do particles have the least amount of energy
kogti [31]

Answer:

solid

Explanation:

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The side chains of amino acids may contain..
salantis [7]

The correct answer is d)all of above

examples for a are tyrosine for b lysine and for c isoleucine

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A container of gas has a volume of 280 mL at a temperature of 22 Celsius if the pressure remains constant what is the volume 44
d1i1m1o1n [39]

Answer:

300.9mL

Explanation:

Given parameters:

V₁  = 280mL

T₁ = 22°C

T₂  = 44°C

Unknown:

V₂ = ?

Solution:

To solve this problem, we apply Charles's law;

  it is mathematically expressed as;

            \frac{V_{1} }{T_{1} }  = \frac{V_{2} }{T_{2} }

We need to convert the temperature to kelvin;

       T₁ = 22°C = 22 + 273 = 295K

       T₂  = 44°C = 44 + 273  = 317K

Input the parameters and solve;

        \frac{280}{295}   = \frac{V_{2} }{317}

  V₂ x 295 = 280 x 317

  V₂ = 300.9mL

5 0
3 years ago
Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an
garri49 [273]

Explanation:

The given data is as follows.

      T = 120^{o}C = (120 + 273.15)K = 393.15 K,  

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So,            x_{1} = 0.5   and   x_{2} = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

               p^{sat}_{1} (393.15 K) = 9.2 bar

               p^{sat}_{1} (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

                 p_{1} = x_{1} \times p^{sat}_{1}

                            = 0.5 \times 9.2 bar

                             = 4.6 bar

and,           p_{2} = x_{2} \times p^{sat}_{2}

                         = 0.5 \times 10.5 bar

                         = 5.25 bar

Therefore, the bubble pressure will be as follows.

                           P = p_{1} + p_{2}            

                              = 4.6 bar + 5.25 bar

                              = 9.85 bar

Now, we will calculate the vapor composition as follows.

                      y_{1} = \frac{p_{1}}{p}

                                = \frac{4.6}{9.85}

                                = 0.467

and,                y_{2} = \frac{p_{2}}{p}

                                = \frac{5.25}{9.85}

                                = 0.527  

Calculate the dew point as follows.

                     y_{1} = 0.5,      y_{2} = 0.5  

          \frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}

           \frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}

             \frac{1}{P} = 0.101966 bar^{-1}              

                             P = 9.807

Composition of the liquid phase is x_{i} and its formula is as follows.

                   x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{9.2}

                               = 0.5329

                    x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}

                               = \frac{0.5 \times 9.807}{10.5}

                               = 0.467

4 0
3 years ago
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Ulleksa [173]
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3 years ago
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