Sunday, January 30, 2011

Types of Reactions

there are six types of reactions
  1. Synthesis 
  2. Decomposition
  3. Simple Displacement
  4. Double Displacement
  5. Combustion
  6. Neutralization
Synthesis
  • A + B --> C
  • Eg. Ca+ O2 --> CaO
Decomposition
  • A --> B + C
  • Eg. H2 + O2--> H2O
Single Displacement
  • A = Metal,  A+ BC --> AC+ B
  • A= Non metal,  A+ BC --> BA + A
  • some elements are more relative, so you might need to check the Actitivty Series Sheet to see whether the reaction is possible
  • An element higher up on the series is able to replace the ion below it on the table

A good website to look at:
http://www.schools.utah.gov/curr/science/sciber00/8th/matter/sciber/chemtype.htm
A cool game to play
http://www.quia.com/rr/192785.html
A easier way to memorize the reactions
http://www.youtube.com/watch?v=5eWmMC7f9to&feature=related

Thursday, January 27, 2011

Balancing Equations

Last class we learnt how to balance equations. You probably remember doing this last year and if you don't, then no worries. That's why we have this blog :). Plus it's one of the easiest things to learn in chemistry. Ok time for review/notes.

The reason why we balance is not just to add extra work but it's to make sure the number of atoms on the reactant side is the same as the number of atoms as the product. Here are a few tips to help you out.

- First balance the atoms that only appear once on each side of the equation.
- Make sure to balance polyatomic ions as one group.
- Balance atoms and groups as you go, don't skip around.
- Balance atoms in elemental form last because those are super easy to balance.

Time to practice balancing equations.
Here's a simple example to get the idea of it.

H2 + O2 -> H2O
-Remember to put a subscript 2 next to diatomic elements. (e.g H2 + O2)


Ok so in the equation you see 2 hydrogen and 2 oxygen as the reactants.
On the product side you see 2 hydrogen and only 1 oxygen.
How would you balance it out to make the amount of atoms equal?
Multiply the product to make the oxygen atoms the same as the reactants.

H2 + O2 -> 2H2O

Now we have 4 hydrogen and 2 oxygen on the product side.
Since there's already 2 oxygen on the reactants side, you don't need to balance it but you must put a 1 in front to indicate that it's already balanced.
To make the hydrogen equal, just multiply it by 2 as well.

2H2 + 1O2 -> 2H2O

Now the equation is balance, both sides have an equal amount of atoms. You can check by counting.


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That's about all there is to it. If you need further explanation here's a youtube video getting into more detail about balancing equations. http://www.youtube.com/watch?v=RnGu3xO2h74

Here are some practice questions and answers. http://misterguch.brinkster.net/PRA008.pdf

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Thanks for reading :) Hope you understand how to balance equations now.

Saturday, January 22, 2011

Writing and Naming Ionic and Covalent Compounds

Review from last year EVERYONE!

  • Covalent compounds are named in different ways than  ionic compounds
  • simple covalent compounds are generally named by using prefixes to indicate how many atoms of each element are shown in the formula. 
  • the ending of the last (most negative) element is changed to -ide.
  • The prefixes used are mono-, di-, tri-, tetra-, penta-, hexa-, and so forth.
  • The mono- prefix is usually not used for the first element in the formula. 
  • The "o" and "a" endings of these prefixes are dropped when they are attached to "oxide."
 Examples:
Name the following covalent/ionic compound.
1) N2O3
2) CO
3) PH3
4) HI


Answers: 
1) dinitrogen trioxide
2) carbon monoxide
3) phosphorous trihydride
4) hydrogen monoiodide (this compound also is often called by its simple name, hydrogen iodide)


jOkE Of ThE dAY!! (: 

What do you do with a dead chemist?
Barium
get it ?
lol

Sunday, January 16, 2011

Molar Concentration / Molarity of Solutions

Molar Concentration 

is able to compare the amount of solute dissolved in a certain volume of solution.

**Need to know formula**

Molar concentration = Moles of solute (n)   /   Volume of solution (V in liter)
moles of solute= molarity X volume of solution 
Volume of solution= moles of solute/molarity

 

is the number of moles of solute in one litre of a solution. We use "M" to denote molar concentration and it has the units of "moles/L".

solute: smaller quantity being dissolved
solvent: larger quantity being dissolved 


**Example of Molarity ConcentratioN **
  • EXAMPLE 1
    EXAMPLE 2
    If 20g of NaOH is dissolved in sufficient water to produce 500 mL of solution, calculate the molar concentration in molarity.

     Molecular mass of NaOH                =  40g /mole
    20 g of NaOH                                   =  20g / 40g /mole   = 0.5 mole
    (g and g units cancel each other. mole is transposed to the numerator)
    500 mL of solution contains            = 0.5 mole
    1000 mL   contains                           = 0.5 mol  x  1000 mL / 500 mL = 1 mol
    By definition, the molar concentration is 1 mol /1000 L = 1M 

    EXAMPLE 3
    Given that conc. HCL has a density of 1.18 and has about 35% HCL dissolved what is its molar concentration (molarity) ?

    Take 100 mL of HCL. Use the density information to convert volume into mass. Its weight is given by 100mL x 1.18 g/mL = 118 g
    Amount of HCL = 118 g x 35% =  41.3g 
    Molecular mass of HCL is 36.5 g/ mole
    41.3 g = 41.3g/36.5 g/mole = 1.13 moles
    100 mL of solution contains 1.13 moles
    1000 mL contains 1.13 mole x 1000 mL /100 mL = 11.3 moles / liter = 11.3 M

    VIDEOS! (:


     


Wednesday, January 12, 2011

Molar Volume of a Gas at STP

Gas changes volume with changes in temperature and pressure
STP = Standard Temperature and Pressure
        = 1 atmosphere of Pressure and a temperature of 0 C or 272.15k

At STP, 1 mole of gas occupies 22.4L

During the calculation, factors can be written as 1 mol/ 22.4L or 22.4L/ 1mol (which kind of factor depending on which type of unit you want to reduce)

Eg. calculate the volume occupied by 29.41g of CO2

First, you have to find the number of mol
29.41g x 1 mole/ 44.0g = 0.6684 mol
then you can calculate the volumeby using the factor
0.6884 mol x 22.4L/ 1 mol = 14.97
finally check the sig. fig.!!!
there are 4 sig. fig in the question so you will need 4 sig. fig in your final answer

Review: mole conversion map
 

Monday, January 10, 2011

Diluting Solutions

Last class we learnt about diluting chemicals. It is used often when shipping because it is a lot more cost efficient. If you're wondering how chemicals have anything to do with money, it's because when you dilute something, it's in its simplest form and all the water is removed. Making it a lot lighter and easier to ship. However, even though it is more concentrated, there is still the same amount of moles of the starting solute as there is in the resulting solute. Here's the formula.

M1L1 = M2L2


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


Step 1: Find the number of moles HCl requires.
Moles HCl = 0.500 mole/L HCl x 0.250L
                  = 0.125 mol HCl


Step 2: Find the volume of 11.6mole/L HCl which contains 0.0625 mol HCl.
Volume required = moles/ molar concentration
                            = 0.125 mol HCl/ 11.6 mol/L HCl
                            = 1.08x10^-2 L


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