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AP Chemistry DBQ/Free Response. Perform the following calculations.

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Transcript

00:04

Here's your shmoop du jour, brought to you by balloons. Aside from being kicked in a

00:08

sensitive area, they're still the number one way to talk in a high-pitched voice.

00:15

A party is thrown with balloons. The balloon salesman tells the party planners that he

00:19

is out of helium cylinders, but he has a stock of other gases: hydrogen, argon, and chlorine.

00:26

The party planners see an unlabeled gas cylinder in the back of the store. They are curious

00:30

as to which gas it is. They fill a balloon to 1 liter and weigh it.

00:35

NOTE: We do NOT recommend filling balloons with random, unlabeled gas. These guys are

00:40

professional idiots...do not try this at home... Anyway, after subtracting the weight of the

00:45

balloon, they find that the weight of the gas in the balloon is 0.819 grams

00:50

It is 27 degrees Celsius outside with standard atmospheric pressure. Assume ideal behavior

00:57

for all gases. Air has a density of 1.161 grams per Liter under these conditions.

01:03

So for Part A, we need to perform the following calculations:

01:07

1 - Calculate the density in grams per Liter of hydrogen.

01:11

2 - Calculate the density in grams per Liter of argon.

01:14

3 - Calculate the density in grams per Liter of chlorine.

01:21

First things first: Calculate the density in grams per liter of hydrogen.

01:25

Okay, so this problem mentions that we're working with gases, and that we can assume

01:29

ideal behavior for all gases. So...we can assume that none of them is going

01:33

to be resting their elbows on the dinner table. We should automatically be thinking about

01:38

the ideal gas law... PV equals nRT. This law allows us to find certain properties

01:44

about an ideal gas...where p equals pressure, v equals volume in liters, n equals number

01:50

of moles, R is the ideal gas constant, and T is temperature in Kelvin.

01:58

What's an ideal gas, you might ask? It's a hypothetical gas in which atoms are perfectly

02:03

elastic and there are no intermolecular attractive forces between molecules. They're all...VERY

02:09

attracted to each other. But back to solving the problem.

02:14

Ok, so we're given standard atmospheric pressure, which means that P equals 1 atmosphere, the

02:20

gas constant R equals .0821, and the temperature equals 27 degrees Celsius.

02:27

But we have to convert temperature to Kelvin...

02:29

and to do so, we can add 273 to 27...to get 300 Kelvin.

02:35

Plugging all these numbers into the equation,

02:37

we find that V equals n times .0821 times 300. But that doesn't look anything like density,

02:44

which is in units: grams per liter. So what else do we know about hydrogen gas

02:49

that can help us? It's molecular weight -- bingo.

02:53

We can calculate the molecular weight of hydrogen gas, which is 2 times the weight of two hydrogen

02:58

atoms, using the periodic table. This equals 2.106 grams per mole.

03:04

Using dimensional analysis, to convert grams per mole into grams per liter, we know that

03:08

we have to have moles on the top, and liters on the bottom, which is equivalent to n over V.

03:16

Now we know what to do with our ideal gas

03:18

law equation. N over V equals 1 over .0821 times 300, or .0406 moles per Liter.

03:30

Multiplying this by the molecular weight of hydrogen gas, we can convert our value into density.

03:36

So we have .0406 moles per liter times 2.106 grams per

03:40

mole equals .0855 grams per liter. But we can't forget about significant figures

03:47

either! All the numbers we're given in the problem have three sig figs, so our answer

03:52

should have three as well. So great.

03:56

Now we just need to do the same thing

03:57

for the other 2 calculations... Let's find ourselves the density of argon.

04:02

Because we are assuming ideal gas behavior, this calculation will be the same as the calculation

04:06

in part 1 to obtain the value for n over v... ... .0406 moles per liter.

04:12

For ideal gases, this value is independent of the identity of the gas.

04:16

We use the molecular weight of argon -- 39.948 grams per mole...which we can find from the

04:22

periodic table again....to convert this value to density.

04:26

0.0406 moles per liter times 39.948 grams per mole equal 1.62 grams per liter.

04:35

Okay, and now for the density of chlorine. Same deal, but this time we use the molecular

04:39

weight of chlorine - 70.905 grams per mole. We can multiply .0406 times 70.905 to get

04:47

2.88 grams per liter.

04:52

OK...Now for part B of this question...

04:55

Assuming the party planners want their balloons to float, which of the three listed gases

04:59

should they purchase? Justify your answer.

05:04

Well, if we ever want something to float in

05:06

another, we need the thing we want to float to have a LOWER density than the other.

05:14

Since air has a density of 1.161 grams per Liter, as given in the problem...we want the

05:19

gas that has a density less than that....

05:21

Hydrogen has a density of 0.0855 grams per Liter, so that's our answer!

05:31

Time for Part C...

05:33

Let's take question 1 first... what is the molecular weight and the identity of the gas

05:37

in the unlabeled cylinder, assuming it is monatomic and acts as an ideal gas?

05:44

Ok...so this question is asking us to figure out the molecular weight and identity of the

05:48

unknown gas. We're given that the weight of the gas in the balloon is 0.819 grams.

05:53

But molecular weight is in the unit: grams per mole...so all we have to do is figure

05:57

out the number of moles, using the ideal gas law again, PV equals nRT.

06:03

Rearranging the equation to find moles, n equals PV divided by RT.

06:10

We have all the information we need to solve: standard pressure -- P equals 1 atmosphere...

06:16

volume -- V equals 1 Liter, gas constant -- (R = 0.0821 liters atmospheres over moles times

06:24

Kelvin), and temperature -- T equals 300 Kelvin.

06:28

We plug in the numbers and solve for n:

06:30

1 times 1 over .0821 times 300 equals .0406 moles.

06:38

So the molecular weight is grams per mole.... .819 grams divided by .0406 moles equals 20.2

06:44

grams per mole. Looking at the periodic table, the element

06:47

with that molecular weight is Neon.

06:51

Now the question wants to know...will it float?

06:54

Well, like part B of this question, the key to figuring out if a gas will float is if

06:59

it is less than the density of air. The density of neon is in grams per liter...so

07:06

.819 grams divided by 1 liter is .819 grams per liter.

07:11

0.819 grams per liter is less than 1.161 grams per liter...so yeah, it totally floats!

07:20

Finally... Part D... Question 1 of part D... Write the balanced,

07:25

net ionic equation for the reaction that occurs when chlorine gas is bubbled into a solution

07:30

of sodium bromide.  Alright, to be honest, we know this question

07:33

is kinda unrelated to the rest of the problem. But sometimes the test makers are like that,

07:37

and we just have to... deal with it. So to find the net ionic equation, we first have to start

07:41

with the balanced molecular equation. Chlorine gas into sodium bromide...chlorine

07:46

gas is a diatomic molecule, so we have Cl2. The formula for sodium bromide is NaBr.

07:55

This is an example of a single displacement chemical equation...so the chlorine replaces

07:59

the bromide to make sodium chloride gas and bromine gas...which is a diatomic molecule.

08:06

So we have two bromine atoms, not just one. To balance this equation, we can add a 2 in

08:11

front of the sodium bromide and 2 in front of the NaCl.

08:16

To find the net ionic equation, we have to break all the soluble electrolytes into their

08:21

ions... ...which means the sodium ions on both sides

08:24

need to be taken out of their salts.

08:28

Just like a normal math equation, we can cancel

08:30

out the two sodium ions on the left and right...and we're left with chlorine gas plus 2 bromide

08:35

ions, which makes bromine gas and 2 chlorine ions.

08:40

Number 2 in part D. Last question...asks us which species is oxidized in the equation.

08:48

Well, oxidation occurs when an element becomes positive because they are LOSING electrons.

08:53

Good for them. We don't become positive when we lose ANYTHING.

08:58

If we look at our net ionic equation, the bromide ions are being oxidized because, on

09:02

the left, they have a negative charge, but then after the ions react with the chlorine

09:06

gas, they have no charge. They are becoming more positive, or being oxidized!

09:11

Phew. So thanks for hanging in there with us for that one. It was a beast.

09:15

Hello? Oh...don't tell us we have to say all that all over again...

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