Writing Net Equations

Before we start with out net equations, we have to be able to identify its states.

Now, how do we do that??

WELL . . . 

u can use the table of Solubities, which was given by Ms Chen.

The anions can be found in the left hand column, the cation in the second column and then you can continue buy following its presense of absense to the word soluble or non-soluble.

if the result is soluble: then its aqueous

if the result is non soluble: then its solid

( Now were on the exciting part)

WRITING NET EQUATIONS!

FINALLY!

(woooh)

So, for starters there is only a net equation when you have a precipitation.

( keep that in mind)

Also, MAKE SURE ( when ur balancing) that aq ions that are the same on BOTH SIDES gets cancelled out.

Here are some practices.

( Like what Ms Chen once said  " Practice makes u better, but not PERFECT" )

so PRACTICE EVERYONE! (:

Fe(s) + 2AgNO₃(aq) ® 2Ag(s) + Fe(NO₃)₂(aq)

Net Ionic Equation: Fe(s) + 2Ag⁺(aq) ® 2Ag(s) + Fe²⁺(aq)

CaCO₃(s) + 2HCl(aq) ® CaCl₂(aq) + H₂O(l) + CO₂(g)

Net Ionic Equation: CaCO₃(s) + 2H⁺(aq) ® Ca²⁺(aq) + H₂O(l) + CO₂(g)

2Na3PO4 (aq) + Ca9NO3)2 (aq) -> 6NaNO3 (aq) + Ca3(PO4)2 (S)

Net equation: 3Ca 2+ (aq) + 2 PO4 3- (aq) -> Ca3(PO4)2 (s)

TFV = (TIME FOR VIDEOS!!!!!!!)

VIDEO #1

VIDEO # 2



VIDEO # 3

Enthalpy Calculation

Calculate how much energy you need or you will produce in a reaction or how much reactants you need to produce a certain amount of energy
  
Mole conversion is required in this lesson

Eg. Calculate the amount of energy released when 1.50 moles of Carbon is reacted to form the product in this reaction: 
           4 C + 6 H2 + O2 --> 2 C2H5OH + 470 KJ

Step 1.
Convert delta H in the form you need
remember a minus sign for exothermic reaction

-470 KJ     
4 mol of C

Step 2.
Put in the equation and reduce to the unit you want
calculate the answer correctly

1.50 mol of C x -   470 KJ         = - 176.25 KJ
                          4 mol of C

Step 3.
make sure you have the correct number of sig, fig.
don't forget the unit!

-176.25 KJ --> -176 KJ

A: 176 KJ of energy is released

Lab 5B

Types of Chemical Reactions
There are examples of Synthesis, decomposition, single replacement, and double replacement in this lab and from the results, we can figure out the chemical reactions.

Purpose of this lab:
To observe a variety of chemical reactions, to interpret and explan observations with balanced chemical equation, and to classify each reactions as one of the four main types.

There are 7 reactions

1. copper reacts with oxygen gas

• shiny, hard copper turns into multicoloured
• 2Cu + O2 -> 2CuO    
• it's Synthesis

2. shiny, silver nail is put in a  blue CuSo4 solution

• nail under the colution becomes copper and it's white 
• Fe + CuSo4 -> FeSo4 + Cu
• Single Replacement

3. buring blue crystal - CuSO4 ˙ 5H2O

• whitish grey colour appears on the outside
• CuSO4 ˙ 5H2O -> CuSo4 + H2O
• Decomposition

4. add water into the product of reaction3

• crystal turns blue again
• CuSo4 + H2O -> CuSO4 ˙ 5H2O
• Synthesis

5. putting CaCl2 solution and Na2CO3 together

• white precipitate settled under a layer of clear liquid as they are added together
• CaCl2 + Na2Co3 -> 2NaCl + CaCO3
• Double replacement

6. put solid zinc into HCl solution

• bubbles forming when zinc in the solution
• Zn + HCl -> ZnCl + H2
• Single replacement

7. observe H2O2

• tiny bubbles traveling toward surface
• H2O2 -> O2 +H2
• Decomposition

What did i learn??

In this lab, I learnt the differences between each types of reactions and able to identify each.  I can write the chemical equation for each reaction and balance them correctly

Endothermic and Exothermic Reactions

Last class we learnt about endothermic and exothermic reactions and how all chemistry reactions involve changes in energy. Here are some of the notes we took last class.

An exothermic reaction is when energy is released to its surroundings.
An endothermic reaction is when energy is absorbed.

Molecules are held together by chemistry bonds

• When energy is added, bonds are broken
• When energy is given off, bonds are created

- If it takes more energy to break a bond than it does to form one, then it's an endothermic reaction.
- If it takes less energy to break a bond than it does to form one, then it's an exothermic reaction.

Energy Diagrams
- We can chart potential energy of the chemicals as they change from reactants to products.
- Reactants start with a certain amount of energy, energy is added to start the reaction and then energy is released as the reaction proceeds.
- Relative amounts of energy determines if a reaction is endothermic or exothermic.

*Enthalpy, H, is the heat contained in the system.

Here are some vocabulary you should know before you can understand an energy diagram.
Energy of Reactants: Total potential energy of all reactants in the reaction
Energy of Products: Total potential energy of all products in the reaction
Activation Energy: The energy that must be added to get the reaction to progress
Delta H: The change in potential energy during the reaction. (Energy of products minus the energy of reactants)

*In this diagram substrate = energy of reactants

Energy in the Equation

- Energy absorption/release can be placed directly in the equation

- Exothermic reactions have the energy term on the right hand side of the equation and a negative delta H

- Endothermic reactions have the energy term on the left hand side of the equation and have a positive delta H

- Energy is represented as KJ in equations (kilojoules)

- Example: CH4 + 2O2 -> CO2 + 2H2O + 812KJ

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

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.

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

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

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)

VIDEOS! 

  MORE VIDEOS!

jOkE Of ThE dAY!! (: 

What do you do with a dead chemist?

Barium

get it ?

lol

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! (:
  
  
  
  
   
  
  
  
  

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

 

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