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Softa [21]
3 years ago
9

1. If 100 mL of a gas, originally at 760 torr, is compressed to a pressure of 120 kPa

Chemistry
1 answer:
marshall27 [118]3 years ago
3 0

Answer : The final volume will be, 84.4 mL

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure = 760 torr

P_2 = final pressure = 120 kPa = 900 torr   (1 kPa = 7.5 torr)

V_1 = initial volume = 100 mL

V_2 = final volume = ?

Now put all the given values in the above equation, we get:

760torr\times 100mL=900torr\times V_2

V_2=84.4mL

Therefore, the final volume will be, 84.4 mL

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Which of the following is an example of a decomposition reaction
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Arsine, AsH3, is a highly toxic compound used in the electronics industry for the production of semiconductors. Its vapor pressu
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Answer : The value of \Delta H of vaporization for arsine and its normal boiling point is, 17694.3  J/mol and 209.9 K respectively.

Explanation :

First we have to calculate the value of \Delta H of vaporization for arsine.

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of arsine at -111.95^oC = 35 torr

P_2 = vapor pressure of arsine at -83.6^oC = 253 torr

T_1 = temperature of arsine = -111.95^oC=273+(-111.95)=161.05K

T_2 = temperature of arsine = -83.6^oC=273+(-83.6)=189.4K

\Delta H_{vap} = heat of vaporization = ?

R = universal constant = 8.314 J/mol.K

Now put all the given values in the above formula, we get:

\ln (\frac{253}{35})=\frac{\Delta H_{vap}}{8.314}\times (\frac{1}{161.05}-\frac{1}{189.4})

\Delta H_{vap}=17694.3J/mol

Now we have to calculate the normal boiling point of arsine.

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of arsine at -111.95^oC = 35 torr

P_2 = vapor pressure of arsine at normal boiling point = 760 torr

T_1 = temperature of arsine = -111.95^oC=273+(-111.95)=161.05K

T_2 = normal boiling point of arsine = ?

\Delta H_{vap} = heat of vaporization = 17694.3  J/mol

R = universal constant = 8.314 J/mol.K

Now put all the given values in the above formula, we get:

\ln (\frac{760}{35})=\frac{17694.3J/mole}{8.314J/K.mole}\times (\frac{1}{161.05}-\frac{1}{T_2})

T_2=209.9K

Hence, the value of \Delta H of vaporization for arsine and its normal boiling point is, 17694.3  J/mol and 209.9 K respectively.

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It is true because protons are positively charged and electrons are negatively charged. if there is the same number of protons and electrons, both charges cancel out.
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