Types of Reactions

Now that you know how to balance chemical equations, let's look at classifying those reactions by type!

There are 6 general types: Synthesis, decomposition, single replacement, double replacement, combustion and neutralization.



However, in this blog, we will only cover the first 3 types. Read our next blog to learn about the last 3. :)

Ok. so..........synthesis. This is a reaction which combines 2 or more reactants (left side of arrow) to form a single product (right side of arrow)
If our 2 reactants were X and Y, and our product was Z, our synthesis equation would be

X + Y --> Z



That was easy. Now, the next one is similar. Decomposition is just synthesis, but backwards. Instead of having two reactants make one product, you have one reactant making two products. The equation would be

X --> Y + Z



For example, using real compounds, we would get...

NaCl --> Na + Cl

And that's all there is to it.

Now on to the most difficult of the three types so far: single replacement.
In this reaction, one element replaces an ion in an ionic compound.
Metals can only replace positive ions (cations) and non-metals can only replace negative ions (anions)



In a metal:
X + YZ --> Y + XZ
You can see that X replaces the Y, so Y is now the lone element, whereas X is now compounded with the Z.

In a non-metal:
X + YZ --> Z + XY
Here, X replaces the Z, because since X is a non-metal, it must replace a non-metal. And remember, if u have any of the HOFBrINCl elements that are going to be the "lonely" Z element in the product side, then you have to add a 2 to the formula, since it is diatomic.

To predict whether a specific single replacement will work, we use an activity series. This activity series show elements in order of reactivity. An element higher up the series can replace an ion below it on the table.



If the element is lower than the ion is replacing, then we say there will be no reaction (NR).

Here's a video to help you understand that.

Note that this video covers the other types, which we will cover in our next blog, so don't worry!

These are links to practice balancing and classifying reactions:
http://misterguch.brinkster.net/PRA020.doc
http://www.bishops.k12.nf.ca/science/1206/chem/reactiontypes.htm
http://chem.taysi.us/assignmentshandouts/reactions/ReactionTypesWS.pdf

Balancing Equations!!

SUP!!

K so... today's class we learned how to balance equations. Before you learn how to balance equation, you should learn how to name equations, andd... If you don't know how to do that, then i guess you should go back and REVIEW!

K... now the real deal. Well, the main goal of balancing equations is so that the number of atoms on the reactant side of the equation, can equal to the number of atoms on the product side of the equations. You may be thinking, why do they have to be equal? What is the big deal? Well.. it actually IS a big deal, cuz...as you should know, if both sides aren't equal then it would be going against the law of conservation of Mass!!!! So not following that law would definitely be a no no!

Let's try learn as we try some practice problems!

Al + O₂ ---->  Al₂O₃

*Oxygen is diatomic.

Well at first you may think that there is nothing wrong with this equation, however when you count each atoms on each side...the left side only has one aluminum atom, and two oxygen atom. Whereas the right side has two aluminum atoms and three oxygen atoms. From here, you can see that the both sides do not have equal atoms.

Now.. what we can do is to balance the equation by multiplying different atoms on each side by different amounts, in order to satisfy the law of conservation of mass!

1) Multiply oxygen on the left side by 3

  Al + 3O₂  ---->  Al₂O₃

2) Then.. we multiply the WHOLE right side by 2

  Al + 3O₂  ---->  2Al₂O₃

3) And lastly... aluminum on the left side by 4, since there is 4 aluminums on the right side of the equation! So.. the answer would like like this:

  4Al + 3O₂  ---->  2Al₂O₃

_{Well... you must think this is pretty easy.. but there may be harder ones. In the future.... you can follow these rules to make balancing easier!}

_{1) When atoms are combined together, like SO4, think of that as a group. and balance it as a whole group}

<sub>2) Balance Metal ions first, then non-metals

3) Balance single atoms last ;( poor atoms...adoms..adams HAHA ... k im not funny.

*Side note.. when P is a lonely atom by itself, you write it as P4 .</sub> 

_{*When S is by itself, that you write it as S8}

   Well... now that you have all that information stuck inside your brain... you should do some practice problems:

http://www.files.chem.vt.edu/RVGS/ACT/notes/scripts/bal_eq1.html
http://richardbowles.tripod.com/chemistry/balance.htm#part4

Here is a video to help you further understand how to balance equations...

Translating Word Equations

In our class today, we said goodbye to the mole section. And began learning about chemicals. First thing we learned was Chemical Reactions. Todays class was all about how to Translate Word Equations.

In order to specify what state each chemical is in. You need to know how to represent each state. Here is the list

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

The word "and" or "reacts with" means +
The phrase "react to," or "produce" means ------>

Now we can use all of these terms and symbols to write a chemical reaction.

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

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


And thats how you translate word equations for chemical reactions. Please feel free to check out the link on the bottom for a quick review.

Molar Volume of a Gas at STP

So...you guys all must be wondering what STP stands for. Well.... I'll tell you what it stands for because I'm such a genius! STP stands for...Standard Temperature and Pressure.

We use STP as a standard condition  to compare different sets of data, ( in our case, comparing the volume of gases). STP is most commonly used to measure gas density and volume.

The STP also corresponds to 1 atmosphere of pressure and a temperature of 0° or 273.15K. gases change in size as they expand and contract when the volume changes in temperature and pressure. We use the STP to study or test a chemical.

At STP there is 22.4 L per one mole.

With this information, we can create the conversion factor:

22.4 L of gas                                                1 mole of gas
1 mole of gas          OR switched around...    22.4 L of gas

Add caption

*This is for you to help visualize the formula.

*This is just another picture to help visualize what is going on. Because the cube has a volume of 22.4 L , so there is only 1 mole of Gas at STP











So... now for some examples...

example 1/ Calculate the volume occupied by Br2O  at STP!

Step 1 : Br20 has a molar mass of  175.8 grams
Step 2 : Convert it to moles, in order to apply it to the formula



175.8 grams  x 1 mole Br20         = 1.83 moles ( remember your sig figs!)
                        95.9 grams

Step 3:  multiply the moles with the equation we just learned!!!

1.83 moles x  22.4 L of gas    = 41.0 L of gas
                      1 mole of gas


AND BAM that is your answer! Easy eh? Well if you still don't get it... Well watch this video! Because I stink at explaining anyways :)





Now you know all about STP? well.. you should try these practice problems before you think you're all that. Cuz....who knows, STP might be harder than you think it is!

http://jc-schools.net/dynamic/science/worksheets/MolarVolumeSTPPractice.pdf

HAVE FUN!

Diluting Solutions to Prepare Workable Solutions

What is diluting?
It is when you make a liquid less concentrated by adding water or another solvent to it.



So why do we do this? 
Chemicals shipped around the world are usually in their most concentrated forms. If they were not, we would be shipping lots of water with the chemicals, making the shipping very cost ineffective.
So, once we get the chemicals, we need to make solutions of the concentration from a more concentrated source.


For example, if there were 2.00 L of KCl of 16.0 M (molarity), we don't need all of the substance in order to create a solution of 0.800 L of 2.00 M HCl. 


Here's how we calculate it.
We know that the moles of the solute is always constant, according to the Law of Conservation of Mass. We are only really adding more water to make a less concentrated solution.


A solution is prepared by dissolving a solute in a solvent.



The formula: M₁L₁ = M₂L₂ 
1st step: Figure out how much volume is needed to start (L1)
From the above example:
16.0 L1 = 2.00M X 0.800 L
16.0 L1 = 1.60
   16.0       16.0
L1 = 0.100 L (Don't forget sig figs.)


Next: We can determine how much water is necessary by subtracting L1 from the number of litres in the solution (L2).
Therefore, 0.800L - 0.100L = 0.700L
This means we need to add 0.700 L of water to the 0.100L substance in order to get our solution of 0.800L with a molarity of 2.00M.





If there are still things you don't quite understand, you can always check out these websites for more help.


http://www.ausetute.com.au/dilucalc.html
http://www.sci.sdsu.edu/classes/chemistry/chem467l/mardahl/dil.html
http://dl.clackamas.edu/ch105-04/dilution.htm


Here's a video from chemistryprofessorpc:

Molar Concentration or "Molarity" of Solutions

Welcome back to The Biggest Noob Chemists blog. We wish you all a Happy New Year and hope you're ready for a new school term. Without further ado...let's begin our lesson.

So in today's class, we wasted no time in reviewing and went straight to learning something new. In this class Ms. Chen introduced us to Molar Concentration.

Molar Concentration/Molarity is basically the number of moles of solute in one Litre of a solution. Just a heads up to all of you, we use M to denote molar concentration.



We calculate Molar Concentration by using this formula.
Molar Concentration (M) = moles of solute(mol)/Volume of solution (L)

This formula can be flipped to find Moles and Litres also.

moles of solute (mol) =  Molarity(M) x  Volume of solution(L)
                                                and
Volume of solution (L) = moles of solute (mol)/Molarity(M)



Example. What is the molarity of 5.30g of Na2 CO3 dissolved in 400.0mL solution?

1. Step is we have to convert 400.0 mL in to litres. We do this by multiplying 400 by 10^3.

2. We now have 0.4 L

3. Next we use the formula to find out the Molarity.

5.30g x 1mole/106Na2CO3 = 0.05Na2CO3/0.4L = 0.125 mol/L

Here's a youtube video:

This concludes my post. Thanks for reading.


PH