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.

 

Solubility Rules

Hey everybody!  Not a lot went on today in class, but we did learn a new topic, Solubility rules!  People who were present on Wednesday received class time to work on the Reaction Solubility 1 worksheet or the Solubility Rules Lab.  People who were absent on Wednesday completed the Solubility Rules Lab.

Homework:
  -Due tomorrow: Reaction Solubility 1
  -Due tomorrow: Solubility Rules Lab

The worksheet we received today has the common rules for the solubility of compounds.  These charts help identify if two compounds will form a precipitate or not.

Here is a chart that helps identify different ions that are usually soluble in water, which means that they don't form a precipitate.  I use the word usually because there are exceptions to the rule.  The right side of the chart lists the exceptions that make the compound insoluble.

Example problem:

KNO₃ (aq) + NaCl (aq) →  KCl (?) + NaNO₃ (?)

1. Using our solubility rules we know that all compounds with the K+ ion are soluble  which means it dissolves in water and is therefore aqueous (aq). 

2. Then, we can determine that all nitrates (NO₃-), are also soluble and aqueous (aq).

Final Answer:  KNO₃ (aq) + NaCl (aq) → KCl (aq) + NaNO₃ (aq)

This chart identifies ions that are usually insoluble in water, which means they do form a precipitate which is a sold.  Again, I use the word usually because there are significant exceptions listed on the right side of the chart.

Example problem:

AgNO₃ (aq)+ NaOH (aq) → AgOH (?) + NaNO₃ (?)

1.  Using the chart above, you can conclude that the (OH-) ion is insoluble, and in the reaction, it will not be combined with any ion that makes it soluble.  Therefore, we can conclude that (AgOH) will form a precipitate called a solid (s).

2.  Using the first chart, you can determine that all (NO₃-)compounds are soluble in water and aqueous (aq).

Final Answer:  AgNO₃ (aq)+ NaOH (aq) → AgOH (s) + NaNO₃ (aq)

Now you should have a clear understanding of the solubility rules and how to apply them!

THE NEXT SCRIBE WILL BE:.......Fabian J.