Butane (C4H10) has a heat of vaporization of 22.44 kJ/mole and a normal boiling point of -0.4 C. A 250 mL sealed flask contains 0.65 g of butane at -22 C.

How much butane is present as a liquid at -22C?

If the butane is warmed to 25 C, how much liquid butane is present?

Check my thinking on this carefully.

I looked up the density of liquid butane. It is 0.573 g/mL and since -22C is well below the boiling point the volume occupied by the liquid is v = m/d = 0.65/0.573 = 1.13 mL for the first part.

For the second part, I am assuming the jar was opened, the butane inserted (at -22 C) then the jar was closed with the pressure of 1 atm.
So the volume of the container is 250-1.13 = 248.9 mL. How much of the butane can evaporate when the T is raised to 25 C? That is
n = PV/RT = 1 atm x 0.2489/0.08206*298 = 0.01. You had 0.011 mols initially. The difference is 0.011-0.01 = 0.001 left as a liquid. That translates to
0.001 x 58 = approx 0.06 g left as a liquid or about 0.06/0.573 or about 0.1 mL. Check my thinking.

Ah, the magical world of butane! Let's dive in and find some answers to your questions.

To determine how much butane is present as a liquid at -22°C, we need to compare the actual temperature with the normal boiling point of butane (-0.4°C). Since -22°C is way colder than the boiling point, we can safely assume that all the butane is in its liquid state. So, the amount of liquid butane at -22°C is equal to the total amount of butane initially present in the flask, which was given as 0.65 g.

Now, let's bring a little warmth into the picture, shall we? If the butane is warmed to 25°C, we need to check the boiling point again to determine if it's still in its liquid form. The boiling point of butane is still -0.4°C, so at 25°C, we can say that all the butane has now vaporized into a gaseous state. Therefore, there would be no liquid butane left at 25°C.

Remember, butane can be a bit moody when it comes to temperature, but we can always rely on its boiling point to guide us through its transitions.

To determine how much butane is present as a liquid at -22°C, we need to compare the given temperature to the normal boiling point of butane (-0.4°C).

Since -22°C is lower than the boiling point of butane, all of the butane will be in the liquid phase at -22°C.

To find out how much liquid butane is present, we'll use the formula:

n = m / M

Where:
n is the number of moles of butane
m is the mass of butane
M is the molar mass of butane

Given:
Mass of butane (m) = 0.65 g
Molar mass of butane (M) = 58.12 g/mol

Using the formula:
n = 0.65 g / 58.12 g/mol = 0.0112 mol

Therefore, at -22°C, there are 0.0112 moles of liquid butane present.

Next, let's calculate the amount of liquid butane at 25°C.

To do this, we need to determine if the temperature is above the boiling point of butane. Since the boiling point of butane is -0.4°C, 25°C is clearly above the boiling point.

At temperatures above the boiling point, all of the butane will be in the vapor phase.

Hence, there will be no liquid butane present at 25°C.

To determine the amount of butane present as a liquid at a certain temperature, we can use the formula:

n = mass / molar mass

Where:
n is the number of moles of butane
mass is the mass of butane
molar mass is the molar mass of butane

First, let's solve for the number of moles of butane present as a liquid at -22°C.

Step 1: Convert the mass of butane to grams:
0.65 g

Step 2: Calculate the number of moles using the molar mass of butane:
The molar mass of butane (C4H10) = (12.01 g/mol x 4) + (1.01 g/mol x 10) = 58.12 g/mol

n = 0.65 g / 58.12 g/mol
n ≈ 0.0112 moles

Therefore, at -22°C, approximately 0.0112 moles of butane are present as a liquid.

Now, let's determine the amount of liquid butane at 25°C.

The normal boiling point of butane is -0.4°C, and its heat of vaporization is 22.44 kJ/mole.

Step 1: Calculate the heat required to convert all the liquid butane to gas:
ΔH = n x heat of vaporization

ΔH = 0.0112 moles x 22.44 kJ/mole (Note: Convert kJ to J for consistent units)
ΔH = 251.328 J

Step 2: Convert the heat required to grams using the specific heat capacity of butane:
If we assume that butane has a specific heat capacity of 2.5 J/g°C (a reasonable estimate for organic compounds), we can use the formula:

q = m x c x ΔT

where:
q is the heat in joules
m is the mass in grams
c is the specific heat capacity in J/g°C
ΔT is the change in temperature

Since we want to find the mass, we rearrange the formula: m = q / (c x ΔT)

m = 251.328 J / (2.5 J/g°C x (25°C - (-22°C)))
m = 251.328 J / 2.5 J/g°C x 47°C
m ≈ 2.66 g

Therefore, at 25°C, approximately 2.66 grams of butane are present as a liquid.