Molar Volume of a Gas

HW due today: Gas problem set #1 & 2.
HW: Molar Volume of a Gas Lab is due Wednesday! Also, there is a WA reading 5.4.
 *If you have questions from the two worksheets, we will go over it in class tomorrow.


Hello, this is Eunice! Today in class we did the Molar Volume of a Gas Lab the entire class period. Our lab goal is to measure the molar volume of hydrogen gas at STP. If you don't know what molar volume is, it is the volume occupied by one mole of a gas is called the molar volume. During this experiment, we measured the molar volume of hydrogen gas as standard temperature and pressure. The reaction is Mg(s) + 2HCl(aq) --> MgCl_2(aq) + H<sub>2(g).


First, we got a piece of magnesium and tied it around a copper wire like a ball.  
Then, my partner got 10 mL of hydrochloric acid and poured it at the bottom of a eudiometer. Next, we filled a beaker with 600mL of tap water.  
Later, we used a smaller beaker to put in water into the eudiometer to the top. We put in the copper wire with magnesium inside, plugged it with a rubber stopper, and flipped the eudiometer over into the tap water. There was a lot of bubbling and sizzling inside, so we waited until there were no bubbles left. Then we measured the volume inside, and we got 26.3 mL. Share your trial data with your table and make sure you have both trial 1 and trial 2. 
After you record the length of Mg ribbon, mass of Mg, evidence of chemical reaction, volume of H2 gas, barometric pressure, room temperature, and water vapor pressure, go on to the calculations part! 

BTW, remember:
Total Pressure= pressure of H2O + pressure of H2.
10cm of Mg=.1407grams
Barometric pressure=30.04 in Hg.</sub> 

                                     Get 600mL of water and be ready to flip over the eudiometer!

                                                 This is not a great picture. But you should tie the magnesium
                                                with copper wire like a ball. And put in rubber stopper.

                                           Flip over the eudiometer quickly and carefully.
                                            You will see lots of bubbles!! (H2)

                                                  This is an awesome picture of the bubbles!
                                                         The magnesium disappeared.
                        
                     

                       

Next Scribe is: BENYA C

 

  

Gas Stoichiometry

Due Today: Ideal Gas Law worksheet
Due Tomorrow: Ideal vs. Combined worksheet
                                       Gas Stoich #1 worksheet

Today we learned how to do gas stoichiometry. Basically, we use stoichiometry to solve the equations of the different gas laws.

Today's demo was all about pressure of gas in a cannon.

                                                                                                                      

 After the demo, we had to answer a couple questions about it that would lead to gas stoichiometry:

 CaC₂ + 2H₂O --> C₂H_2(g) + Ca(OH) ₂

1.) How many moles of gas are formed in the cannon?

     To solve this problem, we use gram-->mol stoich:

 

     0.41g x 1mol CaC₂ x 1mol C₂H_{2  = 0.0064mol}

                        64 g         1 mol CaC₂

 

2.) What is the pressure of the gas that was formed in the cannon?

      To solve this problem, we first found all the variables of the ideal gas equation: PV=nRT. We used this equation because there is only one set of conditions.

 

       T = room temperature = 23°C = 296K

       n = 0.0064

       V = pr² h = p(2.68)² (41) = 870 cm³ = 870mL = 0.87L

       R = 0.00821 (atm/mol K)

The equation to solve for P is:

       (P)(0.87) = (0.0064)(0.00821)(296)

                    P = 0.18atm

 

This example should help with the Gas Stoich homework sheet. The other sheet for homework is review for the quiz tomorrow :)

 

 

 

 

 

 

 

 

 

 

Next Scribe: Eunice C.

 

 

 

 

 

 

 

 

 

 

Due Today: Pressure Worksheet

Due Monday: Ideal Gas Law Worksheet

Today, we learned many new things about gases like ideal gas law and saw the properties of gases in action by seeing many cool demos. 

First, we talked about the many laws that gases and pressure follow.

The first law is Boyle's Law: 

Boyle's law states that pressure is inversely proportional to volume. In other words, as Pressure goes up, volume goes down, and as pressure goes down, volume goes up. 

This can be written as:  P1V1=P2V2

 Mr. Lieberman Showed us a real life example of this by putting a balloon in a pressure chamber. When Mr. Lieberman decreased the pressure in the chamber, the volume of the balloon went up and the balloon got bigger.

                                                       

 When Mr. Lieberman increased the pressure in the chamber, the balloon shrunk and the balloon went back to normal size because the pressure went up so the volume went down.

The next law that Mr. Lieberman talked about was Charles's Law, which states that: The volume of a gas is directly proportional to the temperature, and extrapolates to zero at zero Kelvin. In other words, this law says that as temperature goes up, volume goes up and as temperature goes down, the volume goes down. 

To show us this law, Mr. Lieberman poured extremely cold liquid Nitrogen into a container and put 2 very big balloons into the container. The baloons volume decreased dramatically because of the extremely low temperature. 

 

Here is the Equation: V1 =  V2

                                  T1      T2
For the law to mathematically work, the temperature must be in Kelvins!
Converting from Celsius to Kelvins is easy: Just add 273.15 to degrees Celsius.
20 degrees Celsius would become 293.15 Kelvins.

The next law Mr. Lieberman showed us was Gay Lussac's Law.
This law states: The pressure and temperature of a gas are directly related. In other words, as temperature goes up, pressure goes up, and as temperature goes down, pressure goes down.

To show us this law, Mr Lieberman put a pressure measurement instrument in the liquid nitrogen container. The Pressure started at around 15 psi, the normal pressure measurement on ground level.

But then after being in the extremely cold liquid nitrogen, the pressure decreased dramatically.

 The equation for this law is:  P1 =  P2

                               T1     T2

Like Charles's Law, the temperature in Gay Lussac's Law needs to be in Kelvins for the law to work mathematically. 

The next law is Avagadro's Law. This law states that: For a gas a constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures). In other words, gases with the sam amount of volume will have the same number of moles.

The Formula can be written as:  V1 =  V2

                                  n1      n2

N= the number of moles and V= the volume

The last law that Mr. Lieberman taught us was the Ideal Gas Law. This is just a combination of all the laws listed above so they don't have to be individually used. 

The equation for this law is:  PV=nRT

P= pressure in either atm, or kPa

V= volume in liters

n= the amount of moles present

R= The universal gas constant. 

In other words, if atm is used for pressure, then R would be .0821 atm/ mol K(change)

If kPa is used for the pressure, then R would be 8.314 kPa/ mol K(change)

T= the temperature, in terms of Kelvin. 

In the homework that is due Monday, the Ideal Gas Law worksheet, we use the Ideal Gas Law to solve the problems. Since Mr. Lieberman already explained how to do number 1 in class, here is how to do number 2.

2. What temperature must be maintained to insure that a 1.00 L flask containing .04 moles of oxygen will show a continuous pressure of 745 mm Hg?

The first thing that should be done is to convert 745 mm Hg to atm or kPa, because the pressure has to be in one of these two units. 

Here's how to convert it: 745mm Hg    1atm             = .98 atm

                                         760mm Hg 

 You should use the conversion factor of 1 atm per 760mm Hg so you can cancel mm Hg out and get the pressure in terms of atmospheric pressure, atm.

So, P= .98atm

The volume is 1.00 Liters because in the problem it says that the flask is 1.00 Liters.
So, V= 1L

The amount of moles is .04 moles because the problems says that that is how many moles of oxygen are in the flask.
So n= .04 moles

R is .0821 because that is the gas constant of atmospheric pressure, which is what the pressure unit is in this problem.
R= .0821

Finally T is not known because it is not given to us in the problem.
So T will remain T for now because it is the variable.

So knowing all of this: (.98atm)(1.00L)= (.04moles)(.0821)(T)
After multiplying it out, you should get .98= .003218T.
Divide both sides by .003218 and you should get 305.3.
Remember, temperature has to be in terms of Kelvin for this formula, so the answer is:
305.3 Kelvin

The Next Scribe is... Renee H.

Pressure: Gases

Due today:  -The Behavior of gases worksheet  (answer key has been posted on moodle)
Due tomorrow:  -Pressure worksheet

Today, we continued our introduction to pressure and gases with some more demos.  Below are the summaries of the demos.


1. Flask and paper

In this demo, a note card was placed on top a flask
 filled with water, then the flask was turned over.
Interestingly, the water did not spill from the flask.
This demo proves that pressure is everywhere
 pushing in all directions.  

      2.   Can Crushing

In this demo we heated an aluminum can filled with a little water.

Then, we placed this can top down in to a large beaker of cold water. The can

immediately crushed when it contacted the water.  Inside the can is low pressure

because of the water vapor.  And the outside or the atmosphere is high pressure.  That is why when placed in the water the high pressure crushes the low pressure.

3.  The Vacuum

For the third demo, we put someone into a large garbage bag an tried to get an air tight seal.  Then we turned on the vacuum and removed all the air from the bag.  This removed all the pressure surrounding the person in the bag which made it feel like there was more pressure pushing down on them.  Our lucky demonstrators were Madi, Georgia, and Mr. Lieberman.

4.  Stuck to the ground




This was an interesting demo that involved a device with two levers that when flipped down, the device would stick to whatever smooth surface it was on.  Xavier came up to try to pull it off the ground with a promise of breakfast from Mr. Lieberman if he did.  Xavier removed it without a sweat!  Breakfast for him!










Later in class we went over some notes on pressure.  In theses notes we discussed the different measures of pressure along with some formulas and conversions:

Formula for pressure:

 In case you are not sure what pressure is, it is: 

--the force created by the collisions of

 molecules with the walls of a container.

The molecules in the container move in a random
direction as shown to the right.

We also talked about how a mercury barometer works
as shown below:
.

Here are the standard pressure values and units that can be used for conversions:

• 1 standard atmosphere
• 101.3 kPa (kilopascals)
• 14.7 lbs/in2
• 760 mm Hg (millimeters of mercury)
• 760 torr

These standard values can be used to convert from unit to unit  as shown below:

99.6 kPa x 1 atm/101.3 kPa = 0.983 atm

0.983 atm x 760 mm Hg/1 atm = 747 mm Hg

Next scribe: Jeremy E.

Properties of Gases Intro and Demos

Due: Nothing (b/c we had a test yesterday)

Homework: Behavior of Gases worksheet, chemthink-gases

Most of our class time today was spent reviewing the test, but we did begin our unit on Gases. 

Mr. Lieberman conducted a series of demos exhibiting different behaviors of gases. He proved that gases can go directly from the the gas state to solid state. This is called sublimation. He demoed this by heating up a closed flask of iodine and we saw that a purple/pink gas was created. 

Gases can also behave like a fluid. Fluids have the characteristics of taking the shape of their container, moving throughout their environments (ex. pipe), they flow, and you can pour them. Mr. Leiberman, exemplified this by setting up two flasks. One with baking soda and vinegar that react to form CO2 and the other, left empty. When a flame was placed in the empty beaker there was no reaction, but when it was placed in the beaker with the baking soda the flame was extinguished. Then Mr. Lieberman poured the CO2 gas in the first beaker into the second one, without pouring any of the baking soda mixture into the second beaker. Then when the flame was placed in the second beaker, the flame was extinguished. This happened because gases can behave like fluids, and the CO2 was poured into the second beaker where it remained and extinguished the flame. 

Next scribe: Daniel H. 

Solution Stoichiometry

Solution Stoich

By Megan Noone

Date: November 8, 2012

Homework: Solution Stoichiometry Sheet, (It's double-sided!)

Next Test: Tuesday November 13, 2012.

Today we learned how to apply Stoichiometry to Molarity!

For example...

What volume of .1 M Na3PO4 is required to precipitate all of the lead (II) ions from 150 ml of .250 M Pb(NO3)2?

We can solve this problem by using what we learned in Stoichiometry and combining it with what we learned from Molarity!

1) Balance the equation! 

    2 Na3PO4 + 3 Pb(NO3)2 ------> 6 NaNO3 (aq) + Pb3(PO4)2 (s) 

2) Convert the volume to moles!

          

   150ml x       1L        x 0.250 mol Pb(NO3)2

                    1000ml                    1L

3) Next convert the moles of Pb(NO3)2 to moles of Na3PO4!

   150ml x       1L        x 0.250 mol Pb(NO3)2 x 2 mol Na3PO4  

                    1000ml                    1L                   3 mol Pb(NO3)2

4) Finally, convert moles of Na3PO4 to liters! 

  150ml x       1L        x 0.250 mol Pb(NO3)2 x 2 mol Na3PO4  x              1L            

                    1000ml                    1L                   3 mol Pb(NO3)2    0.1 mol Na3PO4

When all is calculated out, it looks something like this! 

 150ml x       1L        x 0.250 mol Pb(NO3)2 x 2 mol Na3PO4  x              1L            = 0.25 L Na3PO4 

                    1000ml                   1L                   3 mol Pb(NO3)2    0.1 mol Na3PO4

See? That wasn't so hard! 

As a reward, please enjoy this picture of 2 adorable penguins! 

Good luck with the homework! 

Concentration of Solute

Date:  November 5th, 2012
Homework:  Molarity Worksheet

On Monday, we learned how to determine the amount of solute in a solution by using it's concentration. One of the ways to measure concentration is molarity, which is modeled by the following equation:

Molarity (M) = moles of solute
                       liters of solution

You can find any one of these variables with simple algebra if you know the two others:

6 = mol solute
           .25

.25 x 6  = .25 x mol solute

mol solute = 1.5



The two other ways to measure concentration are molality and % mass :

Molality (m) = mol solute
                  kilograms solvent

% mass = grams solute
               grams solution


Molarity can be used to calculate the dilutions of substances by the equation:

M₁V₁=M₂V₂

M₁ and V₁ are the initial molarity and volume of the solution. M₂ and V₂ are the molarity and volume after the dilution.

Colligative Propertie

Colligative Properties

Date: November 6th, 2012
Homework: Solution Concentration Worksheet

On Tuesday we learned about Colligative properties. Colligative properties depend only on the number of solute particles present, not on the type of particle it is. We see Colligative properties in things such as boiling points and melting points.

Our first demo showed how water reacts when other particles are added to the solution. In this demo, the students represented water and the objects in hand represented the other particles.

First the water moves freely. When a particle is added, some of the water is attracted to that particle.

Then, more particles are added to the water and less of the water moves freely.

Finally when more particles are added, the water is the water is attracted to the particle and less water can escape.

When more water is "linked", it doesn't escape and not attracted to pure solvent. The vapor pressure goes down.

The other demo we had today was a bottle of soda flash freezing. Salt was sprinkled on ice with soda in the ice. The ice had a lower freezing point and was able to cool the drink to a lower temperature without freezing it. When the soda was opened, the pressure decreased because the CO2 escaped. This resulted in the flash freezing.

THE NEXT SCRIBE WILL BE..............Megan N.

Net Ionic Equations

Date: November 2nd, 2012
Homework: Net Ionic Equations Worksheet

On Friday, we learned how double replacement reactions occur in water. Also, we learned how to write molecular equations in the form of ionic and net ionic equations.
The following reaction takes place in water:

AgNO_{3 (aq)} + NaCl _(aq) AgCl _(s) + NaNO_{3 (aq)}

_{AgCl is the precipitate formed in the solution.}

_{When the reaction occurs the soluble compound (NaNO3) is broken down into its ions (Na+ and Cl-).}

_{The ionic formula of this reaction is:}

Ag⁺ + NO₃^- + Na⁺ +Cl^-  AgCl _(s) + Na⁺ + NO₃^-

AgCl would remain in its compound form since it was formed as a precipitate.The rest of the products, if are soluble, would remain in their ionic form. In this reaction, not all the ions reacted. NO₃^- & Na⁺ did not combine to form a compound.

Therefore, the net ionic equation, an equation with only the reacting ions, is:

Ag⁺_(aq) + Cl^-_(aq) AgCl _(s)

                                                   Silver nitrate and sodium chloride are put into water.

                                         When inside the water, both compounds break into their ionic forms.

The silver and chloride ions react to form a precipitate. While the nitrate and sodium ions remain in the ionic form and don't react.

THE NET SCRIBE WILL BE.............. JOSH M.