MaVa=MbVb

Today we went over pages 9 and 10. Finish page 5 for homework along with the bottom of page 4.

Remember, Molarity and Concentration are the same thing. The formula is Moles of Solute/Liters of Solution.

We also took notes about a pH Scale.

It is a scale that determines how acidic or basic a solution is.

It ranges from 0 to 14.

Below 7- Acid

7- Neutral

Above 7- Base

We can use mathematical expressions that have been determined experimentally to solve for concentrations, plt, and pOH.

1. pH= -log [H+]

2. pOH= -log[OH-]

3. pH+pOH=14

4. [H+]=10^(-pH) <- this is how you find the molarity of the acid
5. [OH-]=10^(-pOH) <- this is how you find the molarity of the base

just use these formulas to help with the homework... that was basically all we talked about.

I would personally like to go over the reason why we keep circling OH and H in the problems. I get that it has a certain meaning, but I wasn't there that day so I am out of the loop. Thanks!

Ionic Solubility

Dear period 8,

Today in class we did the Ionic Solubility lab. You were to observe the reactions of two compounds that either formed a reaction or not. After recording the data on the first page of the lab, we were to do the following pages according to the first page.

Homework: Finish lab, and finish problem #'s 3,4,5 from the Reaction Solubility 1 notes packet.

Thank you for reading my post, have a nice day.

a.v.

Solutions

Yesterday in class we finished our Mixtures Lab. We took each of the four mixtures and completed the evaporation test and went on to the filtrate lab. To fill out the back sheet of the paper, we can use the notes we took in class on Wednesday. If you want to check if you got the correct results, they are on moodle.
We also learned about solubility and dissolving. We learned that there are two categories of solutions: the solutes and the solvents.

Solutes:
-What is being dissolved
-Usually smaller amount

Solvents:
-What deos the dissolving
-Usually the greater amount

Ionic Compounds: When they are dissolved in water they split into ions (dissociation)
Covalent Compounds: When they are dissolved in water they do not split

*When the compounds are dissolved in water you change the subscrips to (aq), which means in water*

The homework is to finish the Mixtures Lab.
[The notes Mrs. Mandarino gave us in class are on moodle, in the notes section.
If you missed ChemDay, the make up assignment can be found in the homework log section on moodle.]

Mixtures

Homogeneous


The particles are evenly distributed

Heterogeneous

The particles remain thoroughly mixed throughout the mixture while the liquid is being mixed, but later settle to the bottom.



examples:

anything with "shake before use"

salad dressing

orange juice with pulp

children's medicine

* these are all known as suspensions



suspensions

after the particles settle out, the tyndall effect may be seen in the top liquid portion





Homogenous Mixtures

Solutions

are too small to reflect light or settle out
No tyndall effect
Examples:
Coke
Kool aid
Lemonade

Colloid
Are big enough to reflect light but not big enough to settle out.
Tyndall effect is seen

Examples:
milk solution


Today Mrs. Mandarino gave us our test back and our revised grade print outs. We began the new unit taking notes on mixtures. We learned there are two types of mixtures:


Kendall K
Mixtures

Limiting Reactants

 Limiting Reactant Strategy

If a problem mentions amounts of more than one reactant, its probably a limiting reactant problem.

1. Go from each reactant to amount of product (grams of each reactant to grams of product)
2. The reactant that produces the lower amount of predicted product is the limiting reactant.  This amount of product is also the amount that can be produced.
3. The other reactant is your excess reactant.

For Example:

Fe + S  à  FeS

If I have 28g of Iron (56 g/mol) and 24g of Sulfur (32 g/mol), how much FeS can be made and what is the limiting reactant?

28 g Fe  x  1 mol Fe  x  1 mol FeS  x  87.92g FeS  =  44 g FeS

                 55.85 g Fe     1 mol Fe       1 mol FeS

24 g S  x  1 mol S  x  1 mol FeS  x  87.92g FeS  =  66 g FeS

                32.07 g S     1 mol S         1 mol FeS

Fe is the Limiting reactant because it runs out first and produces less FeS.

Into to stoich.

Today, pages 3,4,5 and 6 were stamped. We went over the answers, and they should be posted on moodle.
We were supposed to take the first stoich quiz today, but it has been moved to Monday.

We worked on pg 9, 10, 11, and 12 which were intros to stoich. Most of the pages were review from the particulate nature unit. We reviewed how to balance equations.

Remember- the same amount of each atoms had to be on both sides of the equation.

We also learned how to figure out how much of each atom you need to balance an equation.


Example: 4Fe + 3O2 -> 2Fe2O3

How many molecules of oxygen would you need to react with 380 atoms of iron?

To figure out the answer to this question, you start with what you know. You know you have 380 atoms of iron. According to the equation above, you know that for ever 3 molecules of O2 you have, their are 4Fe atoms. Therefore the equation should be like this:

380Fe x 3 molecules of O2 = 285 molecules of O2
4 molecules of Fe

This is simple stoichiometry. To use stoich, you much always start with a balanced chemical equation.

We also filled out pg 14, which is more of this simple stoich.

Pages 15 and 16 are for homework. They will be stamped on Monday.
Remember the quiz had been moved to Monday!

-Rima S.

Mole Problems

Today in chemistry we went over homework pages which was pages 1 & 2.

Then we worked on page 3 and completed the first 2 problems by converting atomic masses to calculate how many moles were present. Problems c,d and e were to be completed for homework along with page 4.Then we started the lab, 'Mole Stations Lab' where we went from station to station to convert masses to moles and moles to masses. Some converstions to remember are:
1 mole= 6.02 x 10²³ atoms or molecules.

To find the molecular mass you add the masses of all the compounds. That can be used later to find the number of moles of an element.
Danielle F.

Introduction to Moles

January 25, 2011

We began class by stamping our homework, the Mole Lab, and reviewing pages 7 and 8.
(Review)
- a compound is ionic if it contains a metal such as NaNO3, and is simply named sodium nitrate
- if the metal is a transition metal, the compound includes roman numerals, such as Fe3N2, which is named iron (II) nitride
- a compound is covalent if it contains two nonmetals such as PCl5, which includes a prefix, in this case, phosphorus pentachloride

- to calculate the molar mass of a compound, first write the compound's formula. Then take the number of atoms of the first element and multiply it by the element's atomic mass, found on the periodic table. Continue to do this with each element, and add up the masses of each element at the end. For example,
carbon dioxide --- CO2
(one carbon atom, 1 x 12.01 (the atomic mass of carbon)) + (2 x 16.00) = 44.01 g
44.01 g is the molar mass of carbon dioxide (for each mole of CO2 there is 44.01 g)

To calculate the number of moles of 50g of carbon dioxide, use the multiplication method that I can't remember the name of. It involves canceling out units of measurement by putting them diagonally across from eachother ------ x ------- (< like that)

SO...
begin the equation with 50 g of CO2.
50g x 1 mole/44.01 g --- grams cancel out, so you are left to divide 50 by 44.01, to get the answer 1.14 mol CO2

Pages 1 and 2 just went into more depth with the multiplication method with the name I can't remember.
Here's an example problem: What is the mass of 3.57 x 10^-4 mol of N2O5?
start off with what the problem gives you-
3.57 x 10 ^-4 mol N2O5 x 108.02* g/ 1 mol -- mol cancels out, leaving you to multiply 3.57 x 10^-4 by 108.0 to get .0386 g N2O5 (you might want to round that)

*you get 108.02 (the molar mass of N2O5) by (2 x 14.01) + (5 x 16.00)

Thats basically it. The homework was to finish pages 1 and 2.

Chain Reactions and Review

January 14, 2011

Dear Period 8,

On Friday we went over Chain Reactions and Nuclear Reactors.We also received the 1st semester review packet. Webassigns are extra credit and The Test is on Monday January 10, 2011.

Notes: The atomic # determines the type of element. Isotopes is the term used to describe an element written in this format. Uraniium has an unstable nucleus, so it is always found decaying. The atoms that it is found with are atoms it has transmuted into.

Link to Alpha, Beta, and Gamma review: http://gbs-moodle.glenbrook225.org/moodle/mod/resource/view.php?id=652

Chain Reactions:

Examples: Uranium-239

Mass: 239

Atomic Numbef: 92

Event 1-2-3-4-5-6-7
# of Reactions 1-2-4-8-16-32-64

Nuclear Reactors:

Control rods- absorb neutron to slow down a chain reaction

Fuel rods- contain 3% of U-235

Moderator- Slows down neutrons, so control rods can absorb them.

Fission- Splitting of atoms for energy

Fusion- Combining atoms to release energy ex. sunstars

The answers to the review guide: http://http://gbs-moodle.glenbrook225.org/moodle/mod/resource/view.php?id=666

Homework: Finish review guide, Webassigns extra credit, and study for Quest

Anson. V

Jan. 4 2011

Today Mandarino stamped pg. 3 for homework. the homework for today was pg. 9 and #3 on pg. 10. we learned about radioactive decay as well as aplpha and beta particles(also waves). we learned that isotopes are atoms with the same protons but different neutrons. we learned half-lives were the time period it takes for a substance undergoing decay to decrease by half.