Chemistry College

balloon has a volume of 2 L at sea level. if final pressure is halved. what is the final volume of the balloon?​

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Answer 1
4. p1v1=p2v2. p1=x v1=2L so p2=1/2x v2=4.

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Venus's atmosphere, while primarily CO2, is also 3.5% nitrogen gas (i.e. mole fraction of 0.035). What is the partial pressure of nitrogen on Venus in kPa given that the total atmospheric pressure is 1334 psi?

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The partial pressure of nitrogen on Venus is approximately 321.914 kPa.

To find the partial pressure of nitrogen on Venus, we need to calculate the partial pressure using the mole fraction of nitrogen and the total atmospheric pressure. First, we convert the total atmospheric pressure from psi to kilopascals (kPa) since the mole fraction is given in terms of kPa.

1 psi = 6.89476 kPa

Therefore, the total atmospheric pressure on Venus is:

1334 psi × 6.89476 kPa/psi = 9197.53 kPa

Next, we can calculate the partial pressure of nitrogen using the mole fraction. The mole fraction of nitrogen is given as 0.035, which means that nitrogen makes up 3.5% of the total moles of gas in the atmosphere.

The partial pressure of nitrogen is given by:

Partial pressure of nitrogen = Mole fraction of nitrogen × Total atmospheric pressure

Partial pressure of nitrogen = 0.035 × 9197.53 kPa

Partial pressure of nitrogen = 321.914 kPa

Therefore, the partial pressure of nitrogen on Venus is approximately 321.914 kPa.

It's important to note that the given atmospheric composition of Venus's atmosphere and the total atmospheric pressure are approximate values and can vary depending on specific conditions and measurements.

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what is the photoelctric effect?

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Explanation:

It is the emission of electron from a metal under the effect of light is known as photo electric effect

I hope this imformation help full for you

5.86 ■ Liquid oxygen for use as a rocket fuel can be produced by cooling dry air to −183°C, where the O2 condenses. How many liters of dry air at 25°C and 750 torr would need to be processed to produce 150 L of liquid O2 at −183°C? (The mole fraction of oxygen in dry air is 0.21, and the density of liquid oxygen is 1.14 g/mL.)

Answers

Approximately 631.5 liters of dry air at 25°C and 750 torr would need to be processed to produce 150 liters of liquid [tex]O_2[/tex] -183°C.

To solve this problem, we need to consider the ideal gas law and the molar volume of gases.

First, we can calculate the number of moles of oxygen in 150 L of liquid [tex]O_2[/tex] at -183°C. To do this, we divide the mass of liquid oxygen by its molar mass:

Mass of liquid oxygen = volume of liquid oxygen * density of liquid oxygen = 150 L * 1.14 g/mL = 171 g

Molar mass of oxygen (O2) = 32 g/mol

Number of moles of oxygen = mass of oxygen / molar mass of oxygen = 171 g / 32 g/mol ≈ 5.34 mol

Since the mole fraction of oxygen in dry air is given as 0.21, we can calculate the total moles of dry air needed to produce 5.34 mol of oxygen:

Moles of dry air = moles of oxygen / mole fraction of oxygen = 5.34 mol / 0.21 ≈ 25.43 mol

Now, we can use the ideal gas law to calculate the volume of dry air at 25°C and 750 torr (convert to atm) that corresponds to 25.43 mol:

PV = nRT

P = 750 torr * (1 atm / 760 torr) ≈ 0.987 atm

V = volume of dry air (unknown)

n = 25.43 mol

R = 0.0821 L·atm/(mol·K)

T = 25°C + 273.15 = 298.15 K

Solving for V:

V = nRT / P = (25.43 mol)(0.0821 L·atm/(mol·K))(298.15 K) / 0.987 atm ≈ 631.5 L

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