Types of Reactions

Today we learned about three different types of reactions.
These we Synthesis, Decomposition and Single Replacement.


Synthesis: A + B---> AB
Two or more elements combine to form one compound

Decomposition: AB---> A + B
One compound breaks up into two or more elements

Single Replacement: A + BC ---> AC + B   OR  A + BC---> BA + C
A metal always replaces a metal and a non-metal always replaces a non-metal.
A reaction only actually occurs if the element that is replacing the other element is higher on the activity series chart.
This means you will need an activity series chart to do these problems!

Here is a link that has examples for each type of reaction.
http://misterguch.brinkster.net/6typesofchemicalrxn.html

Here is a link to a youtube clip for those of you who are audio-visual learners!
http://www.youtube.com/watch?v=tE4668aarck

Here is a photo just for kicks

NOSTALGIA: balancing those good ol' fashion equations

ACKNOWLEDGEMENTS OF YOUR PRESENCE ARE IN ORDER,

as today we are doing something simple.

It is so simple that I could spend this entire post not even speaking about the main topic of this post.

...

when balancing equations we want to make the number of atoms equal on both sides of the reaction.

can you guess the main topic of this post is? because i can not.

Hellooo there earthlings, your's trully, the tri atomic aliens, are back with a brand new rap! NOT! Okay, so todays topic is about...anyone want to take a guess? Well, if you havent figured it out yet, this blog is about balancing equations! I promise I wont make you read alot, Ill wrap this up un a few simple steps!

To balance equations and to succeed, there are just four basic steps! After these steps are followed I assure you, you will be fine :)

Balance....

1. atoms that occue in molecules on each side

2. whole groups, such as polyatomics

3. balance as you encounter the equations

4. save the single atoms for last

......by following those four simple steps you all shall succeed! 

HERE'S A SHORT BUT HELPFUL VIDEO :) ENJOY EARTHLINGS!

http://www.youtube.com/watch?v=3N_jK-S_kHA

So for more practise use these websites! The stuff you can find on the internet is trully awesomee!

http://staff.fcps.net/jswango/unit4/reactions/balancing%20chem%20equations%20WS1.pdf

http://www.doctortang.com/Honour%20Chemistry/Worksheet%20-%20Balancing%20Chemical%20Equations.pdf

I WONDER HOW THIS WAS BALANCEDD!


ANYWHO, WE MUST RETURN TO THE MOTHER BOARD! BYE :) AND WE'RE OUT OF HERE!

Translating Word Equations

This is going to be short and simple, since I'm hoping you guys all passed science 10!
Stated below is the list that you must know in order to translate word equations :)

s = Solid
l = Liquid
g = Gas
aq = Aqueous

"and" or "reacts with" means +
"react to," "yield" or "produce" means ------>
videos on namingg covelent compounds!
http://www.youtube.com/watch?v=VokWJy_jpAc
naming ionic compounds!
http://www.youtube.com/watch?v=7Lfc6jjp1WQ&feature=relmfu

Ex. Solid sodium hydroxide reacts with chlorine gas to give aqueous nitrate and water.

NaOH(s) + Cl(g) -----> Nacl(s) + NaClO3(s) + H2O(l)

NOW CHALLENGE YOURSELF!
Example. Translate these into both sentence and word equations.
a. 2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)
b. CaH2(s) + 2H2O(l) → Ca(OH)2(aq) + 2H2(g)
c. AgNO3(aq) + KI(aq) → AgI(s) + KNO3(aq)
Answers:

a)..Two molecules of solid Zinc Sulphide reacted with three molecules of Oxygen gas will produce two molecules of solid Zinc Oxide and two molecules of Sulphur dioxide gas.

b)..One molecule of solid Calcium hydride reacted with two molecules of liquid water will produce one molecule of Calcium hydroxide solution and two molecules of Hydrogen gas.

c)..One molecule of Silver nitrate solution reacted with Potassium Iodide solution will produce one molecule of solid Silver iodide and one molecule of Potassium nitrate solution.

 

a) Zinc sulphide + oxygen gas = Zinc oxide + sulphur dioxide.

b) Calcium hydride + dihydrogen oxide = calcium hydroxide+hydrogen gas

c) Silver nitrate + potassium iodide = silver iodide + potassium nitrate

this was just a short yet simple review! hope your young minds have been refreshed :)

Molar Volume of A Gas at STP

GOOD EVENING READERS-WHO-ARE-EXTREMELY-TIRED-AND-WANT-SLEEP i curtsy in your general direction!

mother nature is an ironic vicious woman. it's a shame, she was such a nice lady before we came along

Before we begin our adventure into the land of  calculating molar volume, we must first question the blogger:

What is STP?

• standard temperature (0°C) or (273.15 k) and pressure (1 atmosphere)
• the standard condition to compare volume of gases
• where one mole of gas occupies 22.4L 

What is Molar Volume?

• the volume occupied by one mole of gas 
• 22.4L (=22.4dm³) 

IN A NUTSHELL; THIS POST IS ON CONVERSION

-note: you can only convert from mole to molar volume and vice versa-

From Moles to Molar Volume

the volume occupied by 15.85 moles of N2 is 335.0L.

From Molar Volume to Moles

There are 0.379 moles contained in the sample of CO2 gas.

TA DA!

simple post wasn't it?

OH NO!

I MUST LEAVE YOU LOVELY READERS!

i must save the aussies!

*bonus: mole/molarity review game

JY

Diluting Solutions to Prepare Workable Solutions

To ship chemicals around the world they are concentrated to their lowest form. This means they contain hardly any water. It would be pointless if they contained water because you would just be spending money on shipping water about. Because of this we need to be able to prepare solutions by adding the correct amount of water to receive the desired molarity.

Eg, I have 1L of 18.0M HCl           I need 0.600L of 3.00M HCl

**The moles you start with are the moles you end with***

Formula M1L1=M2L2
 this is because the moles stay the same and each side is  formula for calculating # Moles.

Therefore. 1x 18= 18 moles               0.600x 3= 1.8 moles

1.8/18 = 0.1

That means we need 0.1L of the original solution plus water to to make the total volume equal 0.800L

Here is a process of what you do once you've calculated how much water you need!

http://www.youtube.com/watch?v=bnQJ2q36d2E&feature=related

And if that went to fast for you here is another one

http://www.youtube.com/watch?v=A2YyIo8vSCA&feature=related

Actually the send one is far more detailed than the first one, but i like the music in the first one.

Here is a picture depicting this process

Molar Concentration of Molarity of Solutions

FIRST OFF, WELCOMEE BACK! THIS IS THE VERY FIRST BLOG OF 2011! I HOPE ALL OF YOU HAVE AN AWESOME BREAK! I KNOW YOU GUYS ALL MISSED THE ATOMIC THREE! WELL TIME TO GET TO WORK!


So last class we discussed material about molar concentration or known as molarity of solutions. Here are some notes that are a summary of last days class.


Molar concentration is concentration measured by the number of moles of solute per liter of solution.


Formula:
Molar concentration (M) =  Moles of solute (n)   /   Volume of solution (V in liter)


This formula can be flipped to find the volume and or the number of moles. 


moles of solute= molarity X volume of solution 
Volume of solution= moles of solute/molarity

Examples:

#1 Calculate the molarity when NaOH has 0.510 moles and the volume is 1.400. Be sure to use the right number of significant figures. 

M= 0.510moles of NaOH/ 1.400L

=0.364 moles/ L

#2
Q: Find out the molar concentration of acetic acid (CH₃COOH) solution when  1.24 grams of acetic acid (CH₃COOH) dissolved in 155.0 ml solution?
Solution:  Given,
mass of acetic acid (w)  =  1.24 g
volume of acetic acid (V) = 155.0 ml
                                      = (155.0 / 1000) L         {Because  1 Liter  = 1000 ml}
                                     = 0.155 L
Molar mass of acetic acid (W)  = atomic mass of C atom  + atomic mass of 2 O atoms  + atomic mass of 4 H atoms
                                                        = (12.0  + 2 * 16.00  + 4 * 1.0) g / mol                                                  
                                                        = 48.0 g / mol
And now by the help of molar concentration (molarity) formula we can determine molar concentration of  acetic acid,


Molar concentration of acetic acid (M)  = mass of acetic acid (w)   /   molar mass of acetic acid (W) *  volume of acetic acid (V)
                                                                      =  1.24 g   /  (48.0 g / mol  *  0.155 L)
                                                                      = 0.167 M
So when  1.24 grams of acetic acid (CH₃COOH) dissolved in 155.0 ml solution, molar concentration of this solution is 0.167 M.

 See you guys next class :)!