Monday April, 26th

Mixtures and Solutions

In today´s class we have begun a new unit called "Mixtures and Solutions".
We began learning such concepts and their relationship with matter.

----------------Matter------------------

l l Substances Mixtures

(can be seperated by chemical means) (Physical)

l l
----------l------------ ------------l-----------
l l l l
Element Compound Homogenous Heterogenous
(not seperated) It is made by two Can not identify parts Its parts can be identified
It is the pure form of something or more atoms
made from one specific atom

It is key to remeber that an element is singular, or just one atom and a compound is two or more atoms.

After viewing the relationships of mixtures and solutions with matter, we discussed common, every day examples of Homogenous and heterogenous.


A great example for Heterogenous is a Salad. You can see the component/ingredients (lettuce,tomato, carrot, etc), which are the parts that can be identified

A common example of Homogenous is coffe. Your not able to identify its igredients or components to make the coffe, which means you are not able to identify its parts.


Next we encountered a new term,

Phase- "the part you identify as a whole (profe Ariel)"

E.g, When you add oil to water you will be able to identify the phases as they are distinctly seperated from each substance to another, this is due to polarity (the charge of particles)

Solutions

There are two parts to a solution
1. Solute-which is the particle to be dissolved
2.Solvent- matrix where the dissolving takes place.

A solute disperses into the solvent, giving you a solution.
!!!!!!!!!!!!!!!!!!!!SOLUTIONS ARE ALWAYS LIQUIDS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!!!! IN ORDER FOR A SOLUTE TO DISSOLVE IN WATER IT HAS TO BE POLAR!!!!!



A good example of a solute and a solvent is chocolate milk

the milk is the solvent and the chocolate powder is the solute.


At the end of class we acknowledged methods for mixture seperation.

The first was filtration- which would seperate the mixture bye the size of its ingredients/components.




The second was Distillaton-is a method of seperating mixtures based on differences in their volatilities in a boiling liquid mixture.


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Stefano Pinzon

Monday March 8th, 2010

To start the class we learnt a new term,

Percent Yield

With this term, there are two types of such meaning,

1.Actual Yield: The amount of product that actually forms is carried out in the lab.
2.Theoretical Yield: the maximum amount of product that could formed from given amounts of reagents.

An example: Xgrams + Ygrams= 200g in a theoretical yield,
Xgrams+ Ygrams = 195g as a result of the actual lab, this is the actual yield.

The Percent Yield is used to compare the theoretical yield to the actual yield. And to demonstrate this we use ratios.

Example 1: If your T.Y(theoretical yield) is 200g, and your A.C(actual yield) is 195, your ratio will look like this:

195x 100%=97.5%
200

so your 97.5% is your ratio of T.Y to A.Y

Concentrations of Solutions

Before beginning the class we reviewed the meaning of Concentration

Molarity:measurement of the expressed amount of moles you have disolved into the solution.

To solve questions related to concentration, you will need to use a formula.

Formula: mass = moles=concentration=molarity
volume litres

Example: you have a 1 M solution o NaCl. How many grams of salt are in 500ml of solution?

Step1: calculate how much of molarity you have with 500ml.

1 ml= x molarity cross multiplication
1 lt=.500ml

x=.5ml

Step 2: now calculate how many grams will you get with .5ml.

1ml=58g (NaCl)
.5ml=x grams

x=29grams of salt

Answer is 29 grams of salt

*in some questions it may ask for the molarity were you do the whole same process but backwards.

Note: it is important to capitalize the M for Molarity.

Stefano Pinzon

Monday March 1st, 2010

In today's class, we came upon two new important terms. The first one:

LIMITED REAGENT: it is the reagent that determines the amount of product that can be formed by a reaction. The reaction occurs only until the limited reagent is used up.

The second term is EXCESS REAGENT: the unlimited amount of reactant. But the term is self explanatory.

In our exercises we learnt how to identify the reagents.

For example: using the analogy of recipes there can only be a limited amount of sandwiches produce with 4 slices of bread it is known to be the limited reagent.

Practice #1

Information:Copper reacts with sulfur to form copper (1) sulfide according to the following equation: Atomic Mass

2Cu+S-->Cu2S Cu=63 S=32

What is the limiting reagent when 80g of Cu reacts with 25g of S?

Step #1: If 1 mole of Cu (copper) weighs 63, how many moles are there if you have 80g

your equation should look something among the lines of this

1 MOLE of Cu = 63g using cross multiplication

X MOLES of Cu = 80g

X = 1.269 or 1.27 moles

Step #2: If one mole of S (Sulfur) weighs 32g, how many moles are there for 25g

your equation should look something like this

1 MOLE of S=32g using cross multiplication

X MOLES of S=25g

X=0.781 moles

Step#3: In this step you will be able to compare the amounts and quantities

If there is one mole of Sulfur and gives you 2 moles of Cu as a product, you will be able to calculate the end result with different amounts of moles like this:

1S--->2Cu Cross multiplication 1S--->2Cu

0.78moles of S --->X X moles--->1.27

X= 1.56 moles X=0.635 moles

Step #4: answer the question by comparing the amount of moles from the 80g of Cu (1.27) to the result you got from step #4, and depending on the contrast, in this case the answer from step #1 is less than 1.56 than your are able to say that

Answer: 1.27 moles of Cu is your limited reagent

0.78 moles of S is your excess reagent

Your process you do is not obligatory to do the same, it can be recommended, but the best method is the one you understand.

Stefano Pinzon

Miercoles 24 de febrero del 2010

Topic: Converting Moles and Grams

Today we had this VERY short topic. We learned to convert moles to grams and grams to moles. It's very easy, first we need a question: How many moles of water will you get if you react 1kg of H with excess O?

As we know, the chemical equation of water is this : 2H2 + O2 → 2H2O

We have 1kg of H but we need to convert it to grams that is equal to 1000 g.

Before resolving a rule of three, we convert the 1000g to moles:

1H2 --- 2g

x --- 1000g

(if 1 mole of H weights 2g, how mant moles weight 1000g?)

1 x 1000 = 1000 / 2 = 500 moles of H weight 1000 g

Now we resolve the rule of three:

2H2 --- 2H2O

500 moles --- x

500 x 2 = 1000 / 2 = 500 moles of H2O

And that's how you convert grams to moles.

Michelle Torrero

Viernes 19 de febrero del 2010

Miercoles 17 de febrero del 2010

oday was our first real class, and we saw a new subject: STOICHIOMETRY. Technically speaking, stoichiometry is the part of chemistry that deals with amounts and quantities, and we can use it to calculate masses, moles, and percents within a quemical equiation (In short words, stoichiometry is the math behind chemistry).

We also saw the Chemical Yield (or Reaction Yield) is the ammount of product obtained in a chemical reaction; in a chemical reaction it would be like this:

2H2 + O2 (Reactants) ------> 2H2O (Products)


The reason of the recipes (short and easy) was to explain with easier words the stoichiometry; the example we used was the Chocolate Milkshake recipe, and in that part i learned that the quantity of ingredients depended on the amount that was going to be used for all (at least that's what i understood).

Homar Hurtado

Monday, November 9th, 2009

Today we saw the topic “Chemical Balancing". Chemical Balancing is a way to balance an equation, and there are 3 methods but today we saw the Trial and error method also known as Inspection method.

We can ask first of all why we need to do it, well we need to follow the rule “matter can’t be created nor destroyed just transformed”.

To make a chemical balance first you need an equation:

Na + Cl → NaCl (It’s not balanced)

Then you need to write the amount of molecules of each element:

Na + Cl2 → NaCl

Na 1 Na 1

Cl 2 Cl 1

If you notice, the amount of Chlorine is not the same and we need to apply the rule of “matter can’t be created nor destroyed just transformed”. So we need to aggregate numbers to balance the equation:

Na + Cl2 → 2NaCl

Na 1 Na 2

Cl 2 Cl 2

Now we balanced Chlorine but now Sodium is not balanced, so we need to aggregate another number:

2Na + Cl2 → 2NaCl

Na 2 Na 2

Cl 2 Cl 2

Now the equation is balanced with the same amount of molecules before and after the arrow.

Michelle Torrero