Gases - AP Chemistry

The equilibrium expression indicates that one mole of COCl2(g) will decompose into CO (g) and Cl2 (g). One mole of gas is present on the left side of the equation, and two moles of gas are present on the right side of the equation. The increases in pressure is due to the increase in the number of gas molecules in the reaction chamber.

The van't Hoff factor for CaCl2 is 3 since the molecule dissociates in solution to 3 ions. T is in absolute temperature (25+273=298). This makes the osmotic pressure:

3*2M*0.082*298.

Since we have the density of the mystery gas, we can rearrange the ideal gas law so that the remaining factors are equal to the density of the gas.

We can redefine moles of gas as mass over molar mass.

We can now rearrange the equation to solve for density (mass per volume), which is given in the question.

Using this set up and the values for standard temperature and pressure, we can solve for the molar mass.

First, identify the necessary parameters.

We must use the ideal gas equation:

and since the molar gas constant given uses Kelvin in it's definition, we must convert to Kelvin as follows:

Since the pressure given is gauge pressure, we must convert to the actual pressure as follow:

So therefore the pressure we want to use is

Now that we have listed the given quantities we can see that in order to get the mass of needed, we must first find the number of moles of needed to fill the volume of the airbag, then use the molar ratios from the above chemical equation to convert moles of to moles of , after which we can use the molar mass of to calculate the grams of needed to inflate the airbag.

In order to calculate the number of moles of we can employ the Ideal gas law: to find n (the number of moles of gas; which in this case the only gas involved is

Plug in the given quantities that have been converted into the correct units and quantities:

Solve.

Now that we know the moles of needed we can use stoichiometry, and the information in the following equation:

To determine the number of moles of required and then convert the moles of to the grams of using the molar mass of . The molar mass of can be found by using a periodic table and adding the molar masses of the constituent elements as shown below:

Molar mass of

Given that we calculated the moles of to fill the volume of the airbag to be: , and the molar ratios from the chemical equation we can set up the final steps of the calculation as shown below:

Therefore in order to inflate this driver's side airbag it would need to contain at least 68.04g of

Most gases exist as diatomic molecules in nature. , , and are all common gases found in the atmosphere and the air we breathe.

Ozone, , is a rare exception to the diatomic bonding principle, and is also found in significant quantities in the atmosphere.

Monatomic oxygen, , has only six valence electrons and is relatively unstable; this compound would not be found naturally in the atmosphere.

This question does not acually deal with the ideal gas law, but only with the standard volume of a gas at STP. At standard temperature and pressure, one mole of gas has a volume of 22.5L.

We know that initially we have four moles of octane and excess oxygen. This means that octane will be the limiting reagent for the combustion reaction, since oxygen is in excess.

The total moles of reactant gas when utilizing all four moles of octane is 54, while the total moles of product gas is 68.

There is a total change in moles of:

Since we end the reaction with 14 more moles of gas than we started, the volume must increase:

The change in volume is .

We use the ideal gases equation to know the mol of water we need in the gas phase to generate a pressure of in a flask at .

And the mass is:

If we have less than that amount of water the liquid will evaporate completely but there will not be enough molecules to reach the vapor pressure. More than of water will allow to reach the vapor pressure and there will be also a liquid leftover. Remember that the vapor pressure of a liquid at a fixed temperature is a constant.

Using the molecular mass we calculate the number of mol of :

At this moment we could apply the percentage yield, however let's live it for the end. Each mole of decomposed will generate of gas. Then the total pressure assuming yield will be:

Since only of the decomposes, the real total pressure will be:

In this question, we're given the mass and volume of an unknown hydrocarbon filling a container. We're asked to determine the identity of this compound.

To solve this problem, it's important to realize that for any ideal gas at STP (standard temperature and pressure), of the gas will equate to of that gas. For this question, we're told that there is present.

Now that we know how many moles of gas are in the container, we can use this information, together with the mass provided to us in the question stem, to determine the molecular mass of the unknown compound.

Looking at the answer choices, the only hydrocarbon that matches this molecular mass is ethane, .

Rate of effusion:

and must be used because they exist as bimolecular molecules. The correct answer is that Oxygen gas will effuse 4 times slower than hydrogen gas.

According to Graham's law, the rate of diffusion of a gas molecule is inversely proportional to the root square of that molecule's mass. Because molecule B has a smaller mass than molecule A, it will have a higher rate of diffusion.

Note: this also applies to finding the rate of effusion.

This question is notably difficult, as it may not be immediately apparent what concept is being tested. As the vapors of the compounds mix and react, we are able to establish the distance the each vapor has traveled from the cotton ball into the tube in the given amount of time. The tube is 20cm, and the reaction takes place 6cm from the mystery compound cotton ball. From this, we can establish that in an equal amount of time the HCl vapor traveled 14cm and the mystery compound traveled 6cm.

In order to solve this problem, we use Graham's law to compare molar masses to the rates of diffusion of the two gases.

Since HCl moved 14cm to the right before interacting with the mystery compound, we know that the mystery compound moved only 6cm to the left. As a result, the diffusion ratio is 2.33.

Now, we need to find the square root of the inversed molar masses, which equals this diffusion ratio.

So, the molar mass of the mystery compound is 198 grams per mole. This makes sense, because larger gases will move more slowly compared to lighter gases.

The rate of effusion for two gases can be compared to one another using the following equation:

Here, the effusion rates are inversely proportional to the square root of the molecular masses of the gases in question. Because the relationship is to the square roots of the molecular masses, we will not observe a 2:1 ratio of effusion for neon compared to argon.

We will, however, see that more neon effuses out of container A compared to the amount of argon because neon is the lighter gas and will thus have a faster effusion rate. As a result, there will be more argon than neon in container A after the pinhole is plugged. This results in argon having a larger partial pressure than neon in container A.

We can compare the effusion rates of these gases using the following equation.

By calling neon "gas 1" and argon "gas 2," we can compare the effusion rates of the two gases by plugging their molecular masses into the equation.

This proportion is equal to the rate of neon effusion over the rate of argon effusion, giving the ratio of neon atoms to argon atoms in container B.

As a result, 141 atoms of neon gas will effuse out of the pinhole for every 100 argon gas atoms. Keep in mind that the heavier gas will effuse at a slower rate than the lighter gas; thus, we would expect there to be more neon than argon in container B.

At a particular temperature, the average kinetic energy of all gaseous molecules is equal. Since hydrogen gas has the lowest mass out of these gases, it will have the highest average velocity. This means that it will exit out of the tiny hole at a rate faster than the other gases. Conversely, bromine, which has the most mass compared to the other gases, will exit the hole the slowest.

This relationship is mathematically represented in Graham's law:

As the mass increases, the rate of effusion decreases.

The rate of effusion for a gas is inversely proportional to the square-root of its molecular mass (Graham's Law).

The gas with the lowest molecular weight will effuse the fastest.

Oxygen:

Nitrogen:

Carbon dioxide:

Sulfur dioxide:

Helium:

The lightest, and therefore fastest, gas is helium.

The rate of effusion of a gas is inversely proportional to the square root of the molecular weight of the gas.

The lighter a gas is, the faster it will effuse; the heavier a gas is, the slower it will effuse.

Of all the choices, helium (He) has the lowest molecular weight (atomic weight in this case), so it will have the highest rate of effusion.

In order to answer this question, let's start by considering what effusion is and what things affect it. Effusion is the movement of a gas through a tiny hole that separates two different spaces. Because the gas particles move around in random directions with an average speed that is dependent on the temperature of the sample, lighter gas particle will move faster than heavier gas particles. This is because at a given temperature, all gas particles in a sample will have the same average kinetic energy. Consequently, we would expect gas particles with a higher molar mass to effuse more slowly than gases with a lower molar mass. This means that Gas B should effuse faster than Gas A.

The next step is to actually calculate how much greater Gas B effuses compared to Gas A. To do this, we'll need to use the following equation:

Since we know that Gas A is times heavier than Gas B, we can plug this into the equation to solve for the ratio of Gas B's rate of effusion to that of Gas A.

Therefore, Gas B effuses times faster than Gas A.

Standard temperature and pressure is 00C and 1 atm