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grandymaker [24]
3 years ago
6

Type the correct answer in each box.

Chemistry
1 answer:
Musya8 [376]3 years ago
6 0

Answer:

1 . 26

2. 55.854

Explanation:

because atomic number is the number of proton in other words smaller number and the mass number is the number of neutrons and electrons in other words bigger number .

hope this make sense :)

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If a certain gas occupies a volume of 19 L when the applied pressure is 9.5 atm, find the pressure when the gas occupies a volum
kakasveta [241]

Answer: 37.6 atm

Explanation:

Given that,

Initial volume of gas (V1) = 19L

Initial pressure of gas (P1) = 9.5 atm

Final volume of gas (V2) = 4.8L

Final pressure of gas (P2) = ?

Since pressure and volume are given while temperature remains the same, apply the formula for Boyle's law

P1V1 = P2V2

9.5 atm x 19L = P2 x 4.8L

180.5 atm•L = 4.8L•P2

Divide both sides by 4.8L

180.5 atm•L/4.8L = 4.8L•P2/4.8L

37.6 atm = P2

Thus, the final pressure is 37.6 atmospheres.

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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

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