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lina2011 [118]
3 years ago
12

A balloon of hydrogen is subjected to vacuum. The initial pressure and volume of hydrogen is 0.95 atm and 0.55 L. Calculate the

final pressure if the final volume is 1.22 L? ​
Chemistry
1 answer:
Scilla [17]3 years ago
7 0

Answer:

0.43 atm

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 0.95 atm

Initial volume (V₁) = 0.55 L

Final volume (V₂) = 1.22 L

Final pressure (P₂) =?

The final pressure of the gas can be obtained by using the Boyle's law equation as follow:

P₁V₁ = P₂V₂

0.95 × 0.55 = P₂ × 1.22

0.5225 = P₂ × 1.22

Divide both side by 1.22

P₂ = 0.5225 / 1.22

P₂ = 0.43 atm

Therefore, the final pressure of the gas is 0.43 atm

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Answer:

V_2 = 5.07L

Explanation:

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\\V_2 = \frac{P_1 V_1 T_2}{T_1 P_2} = \frac{3.00atm \cdot 2.00L \cdot 273K}{323K \cdot 1.00atm} = 5.07L

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Ice is actually frozen ______
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Explanation:

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A sample of carbon dioxide at RTP is 0.50 dm3. How many grams of carbon dioxide do we have?
prohojiy [21]

Answer:

0.924 g

Explanation:

The following data were obtained from the question:

Volume of CO2 at RTP = 0.50 dm³

Mass of CO2 =?

Next, we shall determine the number of mole of CO2 that occupied 0.50 dm³ at RTP (room temperature and pressure). This can be obtained as follow:

1 mole of gas = 24 dm³ at RTP

Thus,

1 mole of CO2 occupies 24 dm³ at RTP.

Therefore, Xmol of CO2 will occupy 0.50 dm³ at RTP i.e

Xmol of CO2 = 0.5 /24

Xmol of CO2 = 0.021 mole

Thus, 0.021 mole of CO2 occupied 0.5 dm³ at RTP.

Finally, we shall determine the mass of CO2 as follow:

Mole of CO2 = 0.021 mole

Molar mass of CO2 = 12 + (2×16) = 13 + 32 = 44 g/mol

Mass of CO2 =?

Mole = mass /Molar mass

0.021 = mass of CO2 /44

Cross multiply

Mass of CO2 = 0.021 × 44

Mass of CO2 = 0.924 g.

3 0
3 years ago
What statement would beast describe a digital signal
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4 years ago
34. An electron in the 3d state of the hydrogen atom has a radial part of the wave function of the form A r2 exp(-r/3ao). What i
nikdorinn [45]

Answer: Option (e) is the correct answer.

Explanation:

Formula to calculate radius is as follows.

          p(r) = Ar(r^{2}e^{\frac{-r}{3a_{o}}}

                = Ar^{3}e^{\frac{-r}{3a_{o}}}

     \frac{dp(r)}{dr} = 0

            Ar^{3}e^{\frac{-r}{3a_{o}}}(\frac{-1}{3a_{o}} + Ae^{\frac{-r}{3a_{o}}}(3r^{2}) = 0

                     \frac{Ar^{3}}{3a_{o}}e^{\frac{-r}{3a_{o}}}

                      = 3Ar^{2}e^{\frac{-r}{3a_{o}}}

                       r = 9a_{o}

Thus, we can conclude that most likely radius at which the electron would be found is 9a_{o}.

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3 years ago
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