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Viktor [21]
3 years ago
11

A gas with a volume of 4.0L at a pressure of 205 kPa is allowed to expand to a volume of 12000 mL. What is the pressure in atmos

pheres in the container if the temperature remains constant?
Chemistry
1 answer:
Leni [432]3 years ago
8 0

Answer: The pressure in atmospheres is 0.674 in the container if the temperature remains constant.

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}     (At constant temperature and number of moles)

P_1V_1=P_2V_2  

where,

P_1 = initial pressure of gas  = 205 kPa

P_2 = final pressure of gas = ?

V_1 = initial volume of gas  = 4.0 L

V_2 = final volume of gas = 12000 ml = 12 L    (1L=1000ml)  

205\times 4.0=P_2\times 12  

P_2=68.3kPa=0.674atm        (1kPa=0.0098atm)

Therefore, the pressure in atmospheres is 0.674 in the container if the temperature remains constant.

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Answer:

0.384\ \text{L}

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P_1 = Initial pressure = 931 torr = 931\times \dfrac{101.325}{760}=124.12\ \text{kPa}

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V_1 = Initial volume = 350 mL

V_2 = Final volume

From the Boyle's law we have

P_1V_1=P_2V_2\\\Rightarrow V_2=\dfrac{P_1V_1}{P_2}\\\Rightarrow V_2=\dfrac{124.12\times 350}{113}\\\Rightarrow V_2=384.44\ \text{mL}=0.384\ \text{L}

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Chlorine has two stable isotopes, Cl-35 and Cl-37. If their exact masses are 34.9689 amu and 36.9695 amu, respectively, what is
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X(Cl-35) = 75.95% => Answer 'A'

Explanation:

34.9689·X(Cl-35) + 36.9695·X(Cl-37) = 35.45; X = fractional abundance

X(Cl-35) + X(Cl-37) = 1  ⇒  X(Cl-37) = 1 - X(Cl-25)

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34.9689·X(Cl-35) + 36.9695 - 36.9695·X(Cl-35) = 35.45

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