### Feb. 27 Types of chemical reactions

During class we recalled the chemical reactions that we learned last year in science 10: Syntheis, decomposition and single replacement.

Synthesis: This is when one compound combines with another compound to create a more complicated one.  The formula would be: Compound A + Compound B = Compound C
8 Fe + S8 ---> 8 FeS

Decomposition: Opposite of a synthesis reaction.  A complicated compound breaks down into simpler ones. The formula would be: Complicated C = Compound A + Compound B
2 H2O ---> 2 H2 + O2

Single Replacement: This is when one element trades places with element in a compound: The formula would be A + BC = AC + B OR A +BC = BC + A
Mg + 2 H2O ---> Mg(OH)2 + H2

Here are some practice problems:

### Feb. 25 Translating World Equations and Balancing equations

TRANSLATING WORD EQUATIONS
When you encounter a word equations, you must specify what state each chemical is in. These are what you use to specify what state the chemical is in:
s = Solid , l = Liquid , g = Gas , aq = Aqueous
In the equation "and" or "reacts with" means +
"react to," or "produce" means ------>

Now we can use this knowledge to write our equation from words. An example: calcium hydroxide and hydrochloric acid produce calcium chloride and water.
Ca(OH)2 + HCl --> CaCl2 + H2O

The next step is to balance the equation *DONT FOREGET TO ADD THE STATES!!*:
Ca(OH)2(s) + 2 HCl(l) --> CaCl2(aq) + 2 H2O(l)

Here is a video:

BALANCING EQUATIONS

Remember back in science 9 the balancing equations worksheets? Weren't they fun?! Well now in Chemistry 11 we get to do them AGAIN!! That's right AGAIN!! In case you forgot how to do them or need some basic tips to do them, these are some steps/tips to do it:

1. Write the equation. This is just what will react; forget the coefficients for now. However, you do need to balance the charges. For example, if you were reacting copper(II) nitrate with Sodium, you would write:
Cu(NO3)2 + Na --> NaNO3 + Cu

2.Determine how any atoms are on both sides of the equation. In the example, there is 1 copper atom, 2 nitrate ions, and 1 sodium atom being reacted. Meanwhile, there is 1 sodium ion, 1 nitrate ion, and 1 copper atom being produced.

3.Add coefficients to the element or ion that needs to be increased.  Since there is (NO3)2 on the left side, you will have to make the NO3 the right side equal the left side's. You'll now have the following:
Cu(NO3)2 + Na --> 2NaNO3 + Cu

4.Continue adding coefficients until you have the same amount of atoms on each side of the equation.  Since you added 2 to the NaNO3 on the right side, there is 2 Na.  The left side has only 1 Na. You must make the Na on the right equal the Na on the left.  You would add 2 to the Na on the left side.  The final equation would be this:
Cu(NO3)2 + 2Na --> 2NaNO3 + Cu

*Hint*
When balancing an partially unknown equation, start with the elements that are rarest, and finish with hydrogen and oxygen (which form water).

Here is a website with a bunch of documents on balancing chemical equations in pdf format:

Here is a video about balancing chemical equations:

### Molar Volume of a Gas at STP

You are probably wondering what STP stands for and how it affects gas, well these are some notes:
-Gases expand and contract with changes in temperature and pressure
-STP = Standard Temperature and Pressure
-STP = 1 atmosphere of pressure and a temperature of 0°C or 273.15 K
-At STP 1 mole of gas occupies 22.4L
-Thus we can create the conversion factor :   22.4L
1 mole

Here is an example:  What is the volume of BrO3 at STP?
First step: BrO3's molar mass is 127.9 g and BrO's molar mass is 95.9 g
Then convert it to moles: 127.9 g x 1 mole = 1.33 moles
95.9 g
Next we use the answer and use it with the formula: 1.33 moles x 22.4 L = 29.8 L of BrO3.
moles

Here is a video demostration STP and its equation:

### Diluting Solutions to Prepare Workable Solutions

-Diluted solution = more water
-key idea is that the moles of solute is constant (ie the only difference is that there is more water in the less concentrated formula)

Formula: M1L1= M2L2

Eg. I have: 2.0 L of 16.0 M HCI and I need: 0.800 L of 2.00 M NCI

Do I need to use all of the 16.0 M HCI? No!!
remember: M1L1 = M2L2
16.0 M x L1 = 2.00 M x 0.800 L
16.0 M x L1 = 16.0 moles NCI
L1 = 0.100 L or 1.00 x 10^2 mL

Concentrated HCI is 11.6 mole/L.  How would you make up 250 mL of 0.500 mole/L HCI?

moles               = 0.125 mol HCI  =  1.08 x 10^-2 L
molar concentration     11.6 mol/L HCI

Here is a video of making a 6.2 mL of a 6.0M solution of KOH diluted to 50.0 mL.

### Molar Concentration or Molarity

Molar concentration:
-A homogeneous mixture where one substance is dissolved is called a solution
-In salt water salt is the solute and water is the solvent

We need to be able to calculate the concentration of solutions.  The concentration is the amount of the solute dissolved in a certain bolume of a solution.

Molar concentration or molarity is the number of moles in one lire of a solution.  We use "M" to denote concentration and it has the units of "moles/L".

Formuale:
Moles of solute (mol)
Molarity = volume of solution (L)   or simply M= mol
L

Ex. Calculate the number of grams of calcium hydroxide in 1.30 L of a 0.75 M Ca(OH)2 solution.
Moles of Ca(OH)2 = 0.75 M x 1.30 L = 0.975 mol/L of Ca(OH)2

Ex. How many moles of KCI are contained in 0.25 L of 0.80 mol/L KCI?
moles of KCI = molarity x volume = 0.80 mol/L KCI x 0.25 L
=0.20 mol/L

Ex. How many grams of MgCl2 are contained in 30.0 mL of 1.5 x 10^2 mol/L MgCl2?
moles of MgCl2 = 1.5 x 10^2 mol/L MgCl2 x .0300 L = 4.5 x 10^4 mole of MgCl2

Ex. Calculate the volume in mL of a 0.450 M NaCl solution that contains 96.45 g NaCl.

MMass= 23.0 + 35.5 = 58.5 g/mol
L = mol
L

moles of NaCl= 96.45 g = 1.64871 moles of NaCl2
58.5 g

1.64871 = 3.6638 L and since we need it in mL, the answer is 3663.8 mL
0.450

Here is a video explaining how to calculate molarity: