Types of Reactions

Today we learned about the different kinds of chemical reactions that can occur. They are easily identifiable because each follows a specific pattern.

Synthesis:

A synthesis reaction is one in which two substances react to form a single substance.

Thus the equation A + B -------> AB

An example of a synthesis reaction is

__Ca + __CO3 -----> __CaCO3
As you can see, two substances combine to form one.
The balanced equation is

1 Ca + 1 CO3 ------> 1 CaCO3


Single Replacement

A single replacement reaction involves a compound and and element where atoms from the compound exchange places with the single element.

Thus the equation, AB + C = AC + B

An example of a single replacement reaction is
__NaCl + __I2 ------> __NaI + __Cl2

As you can see ,the iodine(non metal) traded places with the non metal in the NaCl equation, chlorine.

Balance : 2 NaCl  + 1 I2 -----> 2 NaI + 1 Cl2


Decomposition:

As the name implies, it is the breaking down of a substance into two different substances.

Therefore, the equation looks like

AB-----> A + B

An example of a decomposition reaction is

__KNO3 ---> ___KNO2 + __O2

The potassium nitrate breaks down into potassium nitrite and oxygen.

Balance! 2 KNO3 ------> 2 KNO2 + 1 O2

Here are some helpful links!
http://chemistry.about.com/library/formulabalance.pdf For extra practice!


A very visual demonstration http://www.youtube.com/watch?v=tE4668aarck

Balancing Chemical Equations

Today we learned about balancing chemical equations.

Remember: the mass of the reactants is always equal to the mass of the products. Mass cannot be created or destroyed.
Eg.

BALANCE:
__ Fe(s) + __ O2(g) -----> __Fe2O3(g)

We must always make sure that the number of atoms of each element on one side equals the number on the other side of the arrow.

Before:               After:

1 Fe                 2 Fe
2 O                  3 O

First balance the Oxygen.

The common multiple is 6 so you put 3 O2 and 2 Fe2O3
Now you have 6 oxygen atoms.

All that's left is the Fe(s). Because you put a 2 in front of Fe2O3, that means that Fe2 is multiplied by 2 which means you have 4 Fe. So that is what you put in front of Fe(s)

The balanced equation is... 4 Fe(s) + 3 O2-----> 2 Fe2O3


 Its very simple!!!!! All it requires is logic and basic multiplication skills.
http://www.youtube.com/watch?v=RnGu3xO2h74


That is a great video that we found useful
Good luck!                                                                       

Translating Word Equations and Naming Compounds.

Just a quick review..

Translating word equations:

Lets say you have the equation

Magnesium + Chlorine ----> magnesium chloride

Now since magnesium and chlorine are not compounds it is very simple to write their symbols.
Remember that chlorine is diatomic.

so Mg + Cl2 -----> magnesium chloride
This is where the charges come in!

Mg = 2 +
Cl = 1-

All you have to do is combine them in proportional ways according to their charge .

You will need 2 Cl to make one Mg. So, that means that there will be one Mg and 2 Cl

therefore the equation is MgCl2.

Naming Compounds

Simple Ionic compounds are very easy. All you do is say the first element and then add the prefix -ide to the end of the non metal and drop the original ending.

for example
NaCl is sodium chlor-ide.

CuS2 is copper sulph-ide.

For the polyatomic ions such as sulphate or phosphate you simply look up the formula unless you know it by heart,(SO4, PO4) then you just combine whatever elements you are using according to charge. Everything stays the same.

Covalent Compounds are two non metals. Here you need different prefixs depending on the number subscripts beside each element.

1-mono
2-di
3-tri
4-tetra
5-penta
6-hexa
7-hepta
8-octa
9-nona
10-deca

The second non metal in the formula always ends with -ide.

If the first non metal does not have a subscript beside it, it means that it is just one. You do not need to say mono. at the beginning.

2 examples:

NF2 is nitrogen di-fluoride

C3S4 is tricarbon tetrafluoride.

http://www.youtube.com/watch?v=ww_hvRH0Luk Naming compounds.

These are word equations ...

Review Class

Last class was a review class in preparation for the test coming up on:

Mon. Jan. 17, 2011

On the test:

Molecular Formula
Calculating Empirical Formula of an Organic Compound
Percent Composition
Molarity
Diluting Solutions to Prepare Workable Solutions
Standard Temperature and Pressure

I think thats it

GOOD LUCK

STP

STP stands for Standard Temperature and Pressure and is used to compare volumes of gases.

It is = 1 atm(atmosphere) of pressure at 0 degrees celsius or 273.15 K.

@ STP= 22.4 L IS EQUAL TO 1 MOLE OF GAS

The two conversion factors are:

1 mol of gas/ 22.4 L  & 22.4 L / 1 mol of gas

Here are some examples:

A cylinder contains 14.05 moles of gas at STP. What is the volume of the gas?

14.05 moles x 22.4 L/1 mol = 315 L

Welcome Back Fans!!!

I truly couldn't wait for this moment. The time to write another blog. May I add in 2011? Wow. Speechless. I do not think I can do this. The thrill, the hype, to do something I have not done in a month? Heck, this is going to be one greasy, sloppy, embarrassing blog. All I can say, Preet did it.
Enough about that, I hope everyone else had a fantastic break, and now, it is time to show what exactly when down last class.
Last class we learned material about "Molar Concentration", aka "Molarity of Solutions". It is the concentration measured by the number of moles of solution found in the solution, for every liter. 
The Formula for it is..........M = mol/L
To find the moles, it is........mol = M X L
and for the Volume............L= mol/M

Now of the thousands of you guys who read this blog, I am sure that plenty of you top followers are well aware that I am not a fan of showing examples, but today, why not. Here are 2 fresh from the crew! Actually one, because one is more efficient for both you the reader and me the writer. 


First Example: What is the molarity of 5.30 g of Na2CO3 dissolved in 400 mL of solution?

First you must find the number of moles contained in 5.30 g of Na2CO3
5.30 x 1mol/ 106.0 u =  0.05
Then, since M= mol/L the equation is 0.05/0.4(400 ml in L)= 0.125M

Last Example!!! Only one more to go!!! Wahoooo!!!! 


Ex. 2. How many grams of  Ca(OH)2 are needed to make 100.0 mL of 0.250 M solution?

Find out moles in by using equation mol = ML

= (0.250 M)(0.1 L) = 0.204 mol

Then convert!
0.204 x 74.1/1mol = 1.85 grams of Ca(OH)2

If you're stuck this will help :)
http://www.youtube.com/watch?v=jxSoR8RJDW0&playnext=1&list=PLCDA5A8493EB5AE7E&index=10

Dilution of Solutions

We learned about how to dilute solutions

Diluting is when water is added to a solution, which increases volume but decreases concentration of the solution(molarity)

Here is the formula needed to solve these dilution problems:

(Molarity 1) (Volume(in L)1) = (Molarity 2) ( Volume(in L) 2)

or M1L1= M2L2

Here are some examples:

1. 20.0 ml of 0.200 M NaOH solution is diluted to a final volume of 100.0 mL. Calculate the new concentration.

All you do is plug in the numbers.

20.0 mL is initial volume
0.200 M is initial molarity
100.0 mL is final volume

(0.200 M) (0.02 L) = (M2) ( 0.1 L)

= 0.200x 0.02/0.1= 0.04 M

2.  150.0 ml of 0.025 M NaOH solution is added to 150.0 mL of water. Calculate new molarity.

150.0 mL = initial volume
0.025M = initial molarity
Now, final volume is 150 mL + 150 mL = 300 mL

M1L1=M2L2

(0.025 M) (0.15 L) = (M2) ( 0.3 L)

0.025 x 0.15/ 0.3= 0.012 M


http://www.youtube.com/watch?v=0dRwHZCDBp4

Extra help!