Monday, December 13, 2010

Lab 4C December 9, 2010

On Thursday December 9, 2010, we did a lab in Chemistry Class. It was 4C: Formula of a Hydrate.

First, we heated the crucible for 5 minutes to make sure it was dry and there was no moisture contained in it. Then we took some hydrate and put it in the crucible and weighed it, and recorded it.

We heated the hydrate for 5 minutes and recorded our observations. The hydrate became bluish white in colour. After 5 minutes we weighed it again and found that it had gotten lighter because all of the water it contained was gone. It was now a anhydrous salt.

A hydrate is a water containing compound.
 Copper (II) Sulfate Pentahydrate
                                           CuSO4 * 5H2O





An anhydrous salt is a salt that has been heated up until all of the water it contains has evaporated.

We repeated the experiment twice and found that both times the mass was the same

After we added a couple of drops of water and a reaction took place.
The salt turned blue and gas was released
We learned that this water caused the salt to turn back into a hydrate.

Monday, December 6, 2010

Calculating Empirical Formula of Organic Compounds December 3rd, 2010

The empirical formula of an organic compound can be found by:

-burning the compound(reacts with O2)
- collecting and weighing products
-from mass of products, moles of each element in the original organic compound can be calculated.
What is the difference between empirical and molecular?

What is the empirical formula of a compound whose composition is 45.28% Carbon, 6.289% Hydrogen, 35.22% O and 13.21% Nitrogen?

mol C = 45.28 g x 1 mol/12.0 g C = 3.777 mol

mol H = 6.289 g H x 1 mol/ 1 g = 6.289 mol H

mole O = 35.22 g O x 1 mole/16.0 g = 2.2 mol

mole N = 13.21 g N x 1 mole/ 14.0 g = 0.944 mol



Divide by smallest molar amount

3.777/0.944= 4 x 3 = 12

6.289/ 0.944 = 6.6 x 3 = 20

2.2/ 0.944= 2.33 x3 = 7

0.944/0.944 = 1 x3 =3

= C12H20O7N3






A compound contains 92.26% Carbon and 7.74 % H

What is the empirical formula?

*Assume there are 100 g

92.26 g C x 1 mole/ 12.0 g = 7.688 mol

7.74 g H x 1 mole/1.0 g = 7.74 mol


Divide by smallest amount

7.74/7.688 = 1

7.688/7.688 = 1

Ratio of C to H = 1:1

= C1H1

= CH

http://www.youtube.com/watch?v=MpkGRCFJ_pQ

Cheers!

Empirical Formula

The Empirical formula gives the lowest term ratio of atoms (or moles) in the formula.
All Ionic compounds are empirical formula.




A compound contains 80% C and 20 % H?
What is its empirical formula?

*Assume you have 100 g

80.0 g of C x  1  mole/ 12.0 g = 6.6666 mol

20.0 g of H x 1 mole/1 = 20.0 mol

Divide by smallest mole amount


20.0/6.6666= 3

6.6666/6.6666=1

= ratio of C to H is 1 to 3 = CH3

Here are some examples to help you!!!1
http://www.chem.tamu.edu/class/majors/tutorialnotefiles/empirical.htm
A great video!!!
http://www.youtube.com/watch?v=FWozjZ20JyA


Enjoy!!!

Chapter test is December 7, 2010

Monday, November 29, 2010

Percent Composition %

WELL! Today we learned something new. Something Fresh. Something that sounds harder than it really is!
Percent Composition is a relative measure of the mass of each different element present in the compound.

A formula that may help is : % Composition = Mass of element / mass of compound x 100

Here is an example:
1) Calculate the % composition of NaCl
    1. Find its molar mass: 22.99 + 35.45 = 58.44
    2. Calculate the amount of Na: 1 Na = 22.99
    3. Calculate the weight compared to NaCl: 22.99/58.44 = 39.34%
    4.  Calculate the total Cl present : 1 Cl = 35.45%
    5. Fine the weight compared to NaCl: 35.45/58.44 = 60.66%
* Note: Weight MUST add up to 100%, if not then there is an error in the calculations
Practice Questions:

1.Calculate the percent by weight of each element present in ammonium phosphate [(NH4)3PO4]

2. Bicarbonate of soda (sodium hydrogen carbonate) is used in many commercial preparations. Its formula is NaHCO3. Find the mass percentages (mass %) of Na, H, C, and O in sodium hydrogen carbonate. 

Well! It is pretty straight forward. This is pretty much what we learned all class!

Here is a very instructive video on how to calculate percent composition if it still confuses you!

http://www.youtube.com/watch?v=xbEeyT8nK84

Wednesday, November 24, 2010

Mole Conversions

Today we practiced our mole conversions. The first step to this is calculating molar mass.
Add up all the molecular masses of the compound:
Ex. H2O ---->
mass of 1 oxygen atom = 16.0g

mass of 2 hydrogen atoms = 2.0g

16.0+2.0 = 18.0 therefore the molar mass for H2O is 18g/mol

Remember to use Avogadro's number when converting Particles/atoms/molecules

We also did some examples in class from our Mole Conversions Exercise A-C, this involved converting:
Moles to particles/atoms/molecules
Particles/atoms/molecules to moles
Moles to grams

Ex.. Moles to atoms

How many atoms are present in 3 moles of sulphur

3 moles S x 6.022x1023/1 mole = 1.81 x 1025 Atoms


Ex. Atoms to moles

How many moles of Al are present in 2.6 x 1012 atoms of Aluminum

2.6 x 1012 atoms Al x 1 mole/6.022 x 1023 atoms = 4.32 x 10-12 Moles


Ex. Moles to grams

What is the mass in grams of 12 moles of Fluorine gas

12 moles Fl  x 19g/1mole = 228g of Fluorine

Tuesday, November 23, 2010

Harder Mole Conversions

We learned how to do harder mole conversions.

This means converting from moles to number of atoms for example, where there are multiple steps involved.
One helpful way to do this is to use a mole map.

Here is an example:

Grams----------> Moles----------------> Number of Particles------> #  of Atoms in a Particle



To do this you need to know the operations involved in conversion.

Grams to moles = 1mole/ MMG
Moles to Number of Particles= 1 mole/ 6.022 x 10^23 particles
Number of Particles to Number of atoms = Number of atoms in a particle / 1 molecule of that substance

Going the other way, everything is flipped.

Number of atoms to number of particles= 1 molecule of that substance/ number of atoms in one particle
Number of particles to moles= 1 mole/6.022 x 10^23 particles
Moles to grams = MMG/ 1 mole


Here are 2 examples

Convert 20 g of CO2 into number of atoms of O

First find the molar mass which equals

1 C = 12.0 g/mol
2 O = 32.0 g/mol

so...

20g CO2 x 1 mole/ 44.0g x 6.022 x 10^23/ 1 mole x 2 atoms O/1 molecule CO2=

= 5.47 x 10^ 23 atoms of O



2) Convert 2.34 x 10^ 23 molecules of CaCO3 into grams

= figure out Molar mass of  CaCO3
= 1 Ca = 40.1 g/mol
=1 C =12.0g/mol
=3 O = 16.0 g/mol

= 100.1 g/mol

so ..

2.34 x 10^23 molecules of CaCO3 x 1 mole/6.022 x 10^23 particles x 100.1 g/ 1 mole =

= 38.9 g CaCO3

Sunday, November 21, 2010

Thursday was rough.

There were numerous reasons why Thursday was a rough day for students. Especially for those in grade 11 Chemistry. First of all, we had school. There was a mighty blizzard outside that was life-threatening to many, yet we still had to bust our behinds to school. Not only that, we had a Moles quiz. If you think that is bad, which I know it is bad, without even thinking, we also had a sub. I mean, what is Chemistry 11 without Ms. Chen? Don't even answer. To top it all off, we got 2 worksheets. And that was why Thursday was rough.

Moley Mole

This previous Wednesday's class was a great class. It was such a good class because we saw a video with a great theme song. The song was "Happy Mole Day To You". That was all I learned. Just kidding.
On Wednesday our class was introduced to Amadeo Avogadro's Theory/Hypothesis. Avogadro's theory was that the same amount of different gases at the same pressure and temperature, should contain the same number of particles.
A few examples 2 particles to look at are:
Oxygen : Hydrogen = 16:1
Carbon Dioxide : Hydrogen =  22:1
Carbon Dioxide : Oxygen = 18:1
^^ If the number of particles is the same, the mass ratio equals the mass of the particles. This principal is used for relative masses of all atoms on the periodic table of elements.

There are a few ways to express the measurements of atomic mass. The units possible include amu, u, and daltons.

Molar Mass is the mass of 1 mole; in particles of course. The Molar Mass of an element is the mass labelled on the periodic table shown in grams. Some examples include: **abbreviation for mole = mol!
1 mole of Oxygen is 16.0 g/mol
1 mole of Potassium is 39.1 g/mol
1 mole of Carbon is 12.0 g/mol
^^ all listed above have the same amount of particles!

Lastly, the formula of molar mass is 6.022 X 10 to the power of 23 in grams per mole. {g/mol}

http://www.youtube.com/watch?v=ReMe348Im2w that is the beautiful video we watched.
http://www.animalpictures1.com/r-mole-84-mole-1434.htm and that is a picture of a mole, the animal.

Thursday, November 11, 2010

Drawing and Interpreting Graphs Using Open Office

On November 9, 2010 we started reviewing for the big chapter test coming up on Monday November 15.
We were given a review sheet by  Mrs. Chen which covered accuracy and precision, absolute and relative uncertainty, adding, subtracting, multiplying and dividing using significant figures, rounding, scientific notation and unit conversions.

Later we headed down to the computer lab to do some more graph work using Open Office.


First we graphed the volume( mL) of a sample of gas under increasing temperature(K)

We were asked to use a trend line for our scatter plot

We discovered that the slope of our graph was f(x) =0.55x -0.31.

The point (521, 273) was uncertain on this graph, meaning that when temperature was 521 K, and the volume 273 mL, the point was not directly on the trend line like the other points.

In number 2, we graphed the density(g/mL)  of a water sample under increasing temperature(degrees Celsius)

We determined that the density of water increased from 0.99987 g/mL to 1.00000 g/mL when heated from 1 degree to 4 degrees Celsius.
After that though, the temperature continued to increase but the density decreased.

The graph did not illustrate a consistent unitary rate because it was a parabola and was not constant. Density increases then decreases while the temperature consistently rises.

We then graphed the relationship between volume and the mass of several pieces of glass.

We graphed it, and used a trend line. By determining its slope, we determined the density of the glass.

Friday, November 5, 2010

Density of hot and cold water

Hm... what i remember about last class was that we had a spectacular mini quiz on the lab report!!! Oh goodie... more quizzes to bring down my mark! The lab was about Determining Aluminum Foil Thickness. I will not get into any details because the process was thoroughly described in the previous blog.

*PS, I only wrote that it would bring down my mark because i didnt study for the quiz! See!! I'm a very honest person!

The more interesting work that we did was the graphing of the Density of Water! We graphed 2 charts: One for Cold Water and One for Hot Water.

What we did was on Microsoft Excel and we had to make a graph using the calculations given. With hard work and determination, and imagination like a 6 year old, we had to make it as pretty and colorful as possible, pretty much changing everything.

After all the fun stuff, we had to explain what is different between the 2 graphs. After graphing it, we compared the density of cold water to the density of hot water. Cold water had a density of 1.34g/cm3, and hot water had a density of 1.04g/cm3. During class, we discussed why hot water is less dense than cold water, this is because the particles are more spread out. The hot water has more energy,bouncing off each other, therefore being more spread apart, and causes them to be less dense.




















We didn't cover a whole lot this class, but this video should help your understanding of the density of water, and the difference it'll make if the density is different.



Thursday, November 4, 2010

Lab 2E: Determining Aluminum Foil Thickness

Lab 2E was done in class on November the second. It involved find the thickness of aluminum foil by ways of mathematical formulas.


Equipment:

3 Rectangular pieces of aluminum foil
Centigram balance
Ruler


Procedure:

Obtain 3 pieces of aluminum foil. The foil was then measured twice for both width and length on each end to determine to average length of the side. This was done to minimize experimental error.A centigram balance was then used to find the mass of each piece of aluminum foil.


Results:

The thickness of the aluminum foil was determined by two algebraic equations.
D (density) = m/V (mass/volume)
V = LWH ( length x Width x Height)

The mass, length and width were all known from previous measurements taken in the lab. The mathematical procedure was as follows:


D = m/V

2.7 = 1.04/(16.8)(14.55)(h)
2.7 = 1.04/244.44(h)
(244.44h)(2.7)= 244.44h(1.04/244.44h)

The 244.44h's then cancel on the right side.

687.204h/687.204= 1.04/687.204
h = .001513378
h = 1.51 x 10^-3cm

This same operation was performed for all three pieces of aluminum foil because each had a different length and width then the other two.

Tuesday, November 2, 2010

New Month, New Mentality.

What? MCJC writing their blog the night of Chemistry Class?! No Late Cramming?! I'm not going to lie, I think I have a fever this is so weird. Anyway, today was the first day of November, and we were brutally welcomed to the 11th month of the year to a thrilling quiz that was so much fun I almost felt like writing it again! Okay, it is 6 months until April fools, why not throw in a joke? Moving on, after our spantastic quiz on Significant Figures, we took some beautiful written notes on Density. Now even though I am one of the brightest scientists you will ever meet, I will tell you about these notes we were given.

The Density of a substance is also know as its mass over volume. To calculate an object's density, you quite simply divide mass over volume. D = m/v

Now for most non-scientists, this is where it gets tricky. Well, not really. It just sounds cooler when I say that and when I hear/read that, more often then not I actually decide to listen rather than just pretend to.

For a solid substance, the most commonly used measurement is g/cm3.
For a liquid substance, it is usually g//mL. *** 1 cm cubed of Water = 1 mL

So we know how to calculate the Density, and the units, but what next?

Well, if the density is greater than the liquid's, than the object will sink.
If the density is less than the liquid's, the object will float.

Object > Liquid = Sink
Object < Liquid = Float.

You want an example? I don't. But here is one.
Eg. Calculate the density of an Iron Bar if the bar weighs 1200 grams, with a volume of 1.25L.

D= m/v D = 1200g/1.25L
D = 960 g.
** REMEMBER the units, grams and Litres.

That is all!!! YAY!!!


Here are a couple of videos that we found helpful....

http://www.youtube.com/watch?v=rxb_6UANXqU

http://www.youtube.com/watch?v=Q4EBOE4pJyw&feature=related



ENJOY!

Sunday, October 31, 2010

Oct. 28, 2010 Relative + Absolute Uncertainty

PRECISION is how many times a certain measurement figure appears compared to other measurements taken. (e.g. how many times 2.3 is the result)

ACCURACY is how close your measurement is to the real or accepted number or figure. ( 2.3 ---real value is 2.4)

ABSOLUTE UNCERTAINTY:

-measured in the units, not expressed as a ratio

-2 methods of calculation
-1. Find the average of your measurements(take at least 3). The greatest difference between the measurement amounts is the absolute uncertainty.
****Discard the value that is either much larger or smaller than others.

Ex. Values: 12.2, 12.4, 12.4, 12.2, 11.7

1. Discard 11.7.
2. Find the average (12.2+ 12.4 + 12.4 +12.2/4 = 12.3
3. Find largest difference ( 12.4-12.2 = 0.2)
4. Absolute uncertainty is 12.3 +-0.2

-2. The uncertainty of instruments of measurement
Find the smallest unit/segment of measurement and estimate to fraction of 0.1.
(e.g. 1 mL = 0.1 mL)

RELATIVE UNCERTAINTY

absolute uncertainty / estimated measurement

Absolute uncertainty can be expressed as %

The number of significant figures reveals the relative uncertainty

Wednesday, October 27, 2010

Significant Figures

Significant Figures are important to help find out which numbers are precise, and how to make number's more accurate. In every measurement, the last digit are usually uncertain. 

However, some quantities do not require you to round. For example, you cannot determine that there are 4.1 pairs of jeans, or 2.3 pencils.

There are some rules which you will have to remember in order to make the significant figures precise.

Keep in mind these rules on how to count significant figures:

1) Numbers that are greater than zero are ALWAYS counted as a sig dig. 

For example: 456793 . For this number here, there are 6 sig digs.

2) Zeros that come before a number>0 ARE NOT counted

For example: 0.000087 --> Although there are so many zeros before 87,  those zero's are not counted, therefore there are only 2 sig digs.

3) Zeros that go after the decimal points ARE counted


For example: 56.09  --> This number has 4 sig digs


4) Zeros that go after numbers>0, but ARE NOT before or after a decimal, ARE NOT counted

For example: 4120000000 --> This number has so many zeros, but there are only 3 sig digs


Go to this site for practice on counting sig figs, and see what you get! 

http://science.widener.edu/svb/tutorial/sigfigures.html 

Now on to rounding...

Memorize the following rules:

1) Look to the the number on the right of the digit in which you wish to round


2) If the number>5 round the number up

3) If the number<5 the number stays the same


4) If the number = 5 and there are number>0 after it, then round the number up

5) If the number IS 5 and ends in 5, make sure you round to make the last digit even, (0,2,4,6,8).




Here are some MATH RULES WHEN ADDING AND SUBTRACTING TO REMEMBER:

When you add or subtract you always look at the number that has the fewest number of decimal places, then you round the new number according to that.

For this equation you see that 1.6 has the fewest decimal places, so that means the new answer has to have only one decimal place as well. thus, instead of 23.934m, the answer is 23.9m
 
When you are multiplying or dividing, you find the number that has the fewest number of significant digits. Then you round by the fewest number of significant digits.

 
For more practice on rounding, adding, subtracting, multiplying, and dividing significant figures, visit this site:

 http://www.teacherbridge.org/public/bhs/teachers/Dana/SigFigOperations.html





Wednesday, October 20, 2010

Lab 3B: Separation of a mixture by paper chromatography

Lab 3B was performed in class on October 19. The objective of the lab was to identify the components of mixtures by means of there Rf values. It was fun to see how the food dyes separated into many colours.

Equipment:

3 Large test tubes
3 Erlenmeyer flasks
Ruler
3 pieces of chromatography paper
Food colouring ( yellow, red, or blue)
Green food colouring
Unknown mixture of food colouring
Water
Glass stirring rod


Procedure:


Part 1:


Students in groups of two, acquired 3 Erlenmeyer's flasks, and 3 Large test tubes. The test tubes were placed in the Erlenmeyer's flasks with two centimeters of water in each. A line was drawn across the chromatography paper, four centimeters from the end. The paper was then cut into a point from the pencil line.



Part 2:

 Students received food colouring on there chromatography paper. One, of there choice, (red, yellow, or blue) one green, and one unknown mixture. The food colouring was placed on each paper via glass stirring rod, in spots. The first paper, the colour of your choice, was lowered into the first test tube so that about two centimeters were submerged. As the water climbed the chromatography paper, students made observations on the change in colour, the solute front, and the solvent front. After twenty minutes had elapsed, the paper was taken out, and a line was quickly drawn where the water had stopped climbing. After recording the solvent and solute fronts, the Rf values were recorded, and written on the board. All students copied down this information.

Part 3:

The unknown and the green food colouring was also placed in the test tubes. The same procedures as part 2 were then carried out up until the recording of the Rf values. Data was copied into table three in our notebooks.

 Results:

Thursday, October 14, 2010

Acids and Bases Oct.13 2010

An acid is a substance that tastes sour, has a pH of less than 7 and reacts with metals.  They are formed when a compound made up of hydrogen ions and a negatively charged ions(anions) are dissolved in water(aqueous)
A base is the chemical opposite of an acid, and can be oxides or hydroxides of metals. They are aqueous solutions that can take hydronium ions.

Naming  Simple Acids

1. Use prefix "hydro" at beginning
2 The last syllable of the non-metal is dropped and replace with -ic
3. Add the word acid at the end

ex. HCl


Hydrogen Chloride---------- hydrochloride which then becomes hydrochloric because you drop the ending. Then simply add the word acid at the end

HCl= hydrochloric acid

Naming Complex Acids

If the compound ends with -ate you replace it with -ic
If the compound ends with-ite you replace it with -ous

Then add the word acid at the end

ex. 1
hydrogen nitrite (HNO2) =
1. drop the word hydrogen
2. nitrate= ends with -ite so you replace -ite with -ous
3. then add word "acid".

= nitrous acid


 ex.2

Hydrogen Phosphate(H3PO4)

1. drop the word "hydrogen"
2. Phosphate ends with-ate so you replace it -ate with -ic
3. =Phosphoric
4. Add the word "acid"

= Phosphoric Acid


***** For most acids you would drop the ending completely (e.g. chlorate=chloric and nitrate=nitric) This is not the case for sulphur and phosphorus.

Phosphate becomes phosphoric (not phosphic)
Sulphate becomes sulphuric ( not sulphic)

Monday, October 11, 2010

Ionic and Covalent Compounds


Ionic compounds consist of two or more ions which are held together by ionic bonds. These particles are oppositely charged; the particles which are positively charged are including metal cations, and the particles that are negatively charged are includes anions, or the polyatomic ion. Through the oppositely charged particles, they stick together by electrostatic force. The transfer of electrons always occurs between a metal to a non-metal. Generally, ionic compounds have similar properties; a high melting and boiling point, and being hard and brittle.

Ex. W (+6) & F (-1)
The charges of the positive and negative ion have to be able to cancel out. Therefore, you need 6 F’s in order to satisfy the equation. So the result is:

Ex. WF (Tungsten fluoride)

Also, when you name an ionic compound, the metal always comes before the non-metal.

Ex. Cu (+2 or +1) & S (-2)

Cu2S (Copper (I) Sulphide)

Sometimes there are more than one charge on the ion, and depending on the formula given and the situation, (I), (II) etc… show us what the charge is on the ion.

What are complex anions? Complex anions actually aren’t complex at all; they are just a group of atom that are already combined, which acts as one atom. They are negative particles that are made up of combined gases. For example, chlorate (Clo3-).

Ex. Combine Potassium and Acetate

K(+1) is potassium and CH3COO (-) is Acetate, and since the charges will cancel, the result is simply:

KCH3COO

Now, on to covalent compounds; covalent compounds are essentially the opposite of an ionic compound, instead of transferring electrons, they share them. The two elements in the covalent compound should be both non-metal’s.

To name or to write formula’s of covalent compounds, you need to know your diatomic molecules. They consist of two atoms, and to name it you need to know your prefixes.

The diatomic molecules are: H2, 02, F2, Br2, I2, N2, and Cl2. An easier way to memorize this is to remember the HOFBrINCl, or the Magnificent Seven (the elements are in a shape of a 7 on the periodic table).

The prefixes you should have memorized are :

mono- 1 tetra- 4 hepta- 7 deca- 10
di- 2 penta- 5 octa- 8
tri- 3 hexa- 6 nona- 9

When you name covalent compounds, the first element stays the same. However, with the second element, you add the prefix to the beginning, then change the last part of the element to "ide". Also, if the prefix and the element both have a vowel, cancel out the vowel from the prefix. For example, instead of it being tetraoxide, it'd be tetroxide.

Ex. NI3

N being Nitrogen, and I is Iodide. The prefix for 3 is tri, therefore the result would be:

Nitrogen triodide

For more information on this , go to these links :
http://misterguch.brinkster.net/ionic.html
http://misterguch.brinkster.net/covalentcompounds.html
http://en.wikipedia.org/wiki/Diatomic_molecule

Wednesday, October 6, 2010

Lab 2B: Heating and Cooling Curves of a Pure Substance

Lab 2B was was done in class on October the 4th. It involved the measuring of the freezing and melting points of Dodecanoic cid. It was interesting to see how quickly change of state occurred during the experiment.

Equipment:

Ring stand
Buret clamp
Hot plate
beaker with water (hot and cold)
2 Thermometers
Safety goggles
1 Test Tube
Dodecanoic Acid
Paper to record observations

Procedure:

(Part 2 was done first)

Part 2:
We started off by putting on saftey goggles. Followed by gathering materials needed to conduct the experiment. Students in groups of 2, were given liquid Dodecanoic acid in a test tube. A thermometer was placed in both the acid and the water beaker. The test tube was then suspended inside a beaker filled with cold water by a buret clamp attached to the ring stand. Students then observed the acid as it slowly reached its freezing point. Temperature change was recorded in 30 second intervals, along with observations. Once the acid reached 25 Degrees Celsius, observations and recordings of time intervals were stopped, and students moved on to part 1.

Part 1:
Once the acid had reached 25 Degrees Celsius, students turned on there hot plates. The hot plate's temperature was turned to 2. The cold water from the beaker was replaced by hot water, and placed on the hot plate. The acid was once again suspended inside the beaker of water. Time intervals of 30 seconds were recorded along with observations while the acid reached its melting point. Once the acid had reached 50 Degrees Celsius, observations were stopped and the experiment was complete.

Result:

After preforming the experiment, the melting and freezing points were recorded. The melting point was recorded at 40 - 45 Degrees Celsius, and the Freezing point at 30 - 35 Degrees Celsius. The recording of the temperatures came out very unexpectedly. There were some time intervals where there was no change in temperature, this is because all the energy being transferred from the water to the acid was changed into kinetic energy and used to help break the attractions between the molecules in the acid instead of raising the temperature. Thus creating plateaus in the temperature change graphs made. It was interesting to do this experiment. What made it interesting was knowing that the temperature in which Dodecanoic acid melts and freezes is so close together, and can be changed very quickly.

Sunday, October 3, 2010

October 1st Class was one to remember.

Friday's class was full of fun and learning! Now my job on this Sunday evening is to re-word my knowledge from last class onto this blog. I love blogs, they are a hobby of mine, that is for sure!

Anyway, last class we learned about a few basic laws, including the Law of Multiple Proportion and the Law of Definite Composition. The Law of Multiple Proportion is when at least 2 compounds with different amounts of the same element can be created. The example from the class was Co2->C2O4->C3O6... and so on. The Law of Definite Composition is when a compound will always remain a compound, no matter what the circumstances are. The class example was H2O, it will always be H2O no matter if it is a solid, liquid, or gas.

Moving on, we learned about the heating/cooling curve of a pure substance. Now I could be fancy, and make a chart or grid of some sorts, but lets go old school here, and use our imagination. Now imagine a grid, with the X Axis as Time/Minute, and the Y Axis as Temperature in Celsius. Now picture the letters A to F, resembling points in a stair like fashion, towards the top right corner of the grid. Got it? Great. Now point A is on the Y Axis, and it is a solid state at any temperature below the melting point. A solid is packed very tightly together and does not move a whole lot. Now between points A and B is a line, and that line shows how the Kinetic energy increases, the molecules move faster, and the temperature increases. At point B, the substance remains a solid, but is beginning to melt. The line B-C is a mixture of both a solid and a liquid. It can be either Freezing or Melting, depending on whether the substance is being heated or freezing. The temperature remains constant but it is melting into a liquid at the same time. The substance's particles are becoming more spread out and moving at a faster rate. Now point C. Point C is officially what humans call, a liquid. If it is H2O, then it is water. Cool stuff eh? I thought so. Moving on, C-D is the liquid being heated and the molecules again acquire more heat and speed via Kinetic Energy. Point D exists in a Liquid State. The particles in the substance have enough Kinetic Energy to man up and overcome the adversity of the attraction of particles, so it is not a solid. In fact it is on its way towards being a gas. Now line D-E can either be Evaporation (liquid into gas), or Condensation (Gas to Liquid). Even though the line is in between a liquid and a gas, the temperature remains unchanged. In fact the heat energy is absorbed is not to change the temp, but rather the speed of the molecules. This consistent temperature is known as the boiling point. Now point E is strictly a Gas. Plain and simple. Finallllllllllly, is the line E-F. E-F shows that the gas particles continue to absorb energy and move faster, leading to another increase in temperature. After that, the gas may transform into a solid (Sublimation). If it were Solid to Gas, it would be Deposition. Now for you uncool, mindless, unimaginative people, here is a graph to show all that I explained beautifully.




Textbook Pages 25-34

-The temperature at which matter changes from a liquid to gas is called its boiling point
- A mixture is two or more kinds of matter that have separate identities. They can be divided into different components that are different from each other.
-By classifying matter, you develop a better understanding of it
-Mixtures that are uniform(the same) throughout are called solutions.
Distillation is a procedure used to separate solutions into separate components.
-Changes that produce new substances with different properties are called chemical changes
-Physical changes are changes that can be reversed to obtain the original substance easily. They don't produce new substances.
-Decomposition is the type of change when one kind of matter splits into two or more kinds of matter.
-Electrolysis is a procedure where an electric current is passed through a substance, causing it to decompose into new substances.
-Decomposition and distillation appear similar, but study reveals that they are actually fundamentally different processes.
-Pure substances that can be broken down into new matter are called compounds.
- Compounds are composed of pure substances called elements which cannot be broken down.
-109 known elements
-All matter on Earth made up of 85 elements
-Just 8 of these make up 99% of earth's crust
Law of Definite Composition:
Mixtures can have almost any composition that is desired but compounds will have a definite composition.
e.g. The volume of H2 obtained from water is always twice the volume of O2 obtained.

Law of Multiple Proportions:
Two or more compounds with different proportions of the same elements can be made.
- A particular compound = not more than 1 proportion like a mixture; same elements can form different compounds though, that have different make up but having different composition from the others.

Pg 36-39

-Matter is made up of atoms
-Macroscopic observations are observations you can make by seeing, feeling or smelling.
-The word atom means "smallest piece of something"
- Temperature increases- movement becomes faster and stronger and breaks forces that hold them together as a solid, and become a liquid, and the particles then flow past each other.
-Temp. of a liquid reaches boiling point, the liquid becomes a gas b/c particles move with such energy and force they break forces of attraction and move freely
- Compounds can be solids, liquids or gases
- One atom of an element*----One molecule of a compound
-Molecules are two or more kinds of atoms combined.
-Ions are particles that have an electric charge

Wednesday, September 29, 2010

Sept. 29 Lab 2C

Lab 2C was carried out and it was interesting and fun to do.
All students were asked to create flow charts of the lab beforehand to demonstrate understanding of the the procedure.

Equipment:

Safety goggles
4 Small Test Tubes
Test Tube Rack
Glass Square
2 Eye Droppers
Paper to record observations

Procedure:

Students were given four unknown substances by Mrs. Chen and were instructed to combine the different substances and record observations. Of the four substances, one was a dark greenish colour and the other three were colourless and clear.
There were a total of six possible combinations and therefore 6 observations were made. Students were directed to place one drop of the unknown solution into the given "slot" and then add a drop of another solution to see the reaction take place. This lab was focusing on studying the differences between chemical and physical changes and to see these changes occurring.

Result:

After doing the experiment, one solution remained greenish as the colourless substance did not seem to have any effect on it. Two other solutions remained clear and colourless while the other two solutions turned blue and yellow. Although a vigorous reaction did not take place, the change in colour was fascinating to see and was a good example of the purpose of the lab, which was to observe changes.

Tuesday, September 28, 2010

Different Properties of Matter

What is matter? You can say that matter is anything that has mass, or anything that takes up space, which is volume. Matter has many different properties; it can be broken up into two different and broad categories, pure substances, and mixtures.
Pure Substances has only one set of properties, as well as only one type of particle. Even if there are physical or chemical processes, a pure substance cannot be separated into other types of matter.
Elements cannot be decomposed because it is already in its simplest form. Although one element may have similar properties to other elements, it is still impossible for two different elements to be exactly alike. Elements are made of atoms which can be made into 3 more categories, metal, non-metal, and metalloids. Metal is a chemical element, and it conducts electricity and heat well. Metal forms cations and ionic bonds with non-metals. Non-metals do not conduct electricity or heat very well. Unlike metals, non-metals are very brittle. Metalloids have both properties similar to both metals and non-metals; some metalloids carry an electrical charge under special conditions.
Compounds are another category of pure substances. In a chemical compound, there are two or more elements that are combined chemically. In a compound, the molecule is the smallest particle; it contains multiple atoms of an element. Molecules can be split into 2 categories, ionic, and covalent. An ionic bond bonds a metal and non-metal ion together by electrostatic attraction. The ions have opposite charges, so that is what causes the attraction. A covalent bond bonds by sharing the pairs of electrons. They share the electrons between atoms and other covalent bonds. The bond is stable and balances because of the attractive and repulsive forces.
Mixtures have more than one set of properties and substances. The chemicals are not chemically united, but a mixture still is a combination of two ore more substances. Mixtures can be broken down into homogeneous and heterogeneous. The prefix “homo” means sameness, so a homogeneous mixture is uniform; it has the same appearance and composition throughout the processes. Many people call homogeneous mixtures a solution. The particles in homogeneous mixtures are too tiny to be seen. Also, in a homogeneous mixture, it seems that there is only one component. The prefix “hetero” indicates difference. A heterogeneous mixture has substances or phases that are visibly different. The particles that are in a heterogeneous mixture are larger and more visible. Also, in the heterogeneous mixture it seems to have one or more component.
In physical changes, no new substances are formed, and the chemical composition does not change. This kind of change is not reversible. Unlike physical changes, chemical changes are irreversible. New substance can be produced as a result of a chemical change

This video can help you understand the different states of matter.

http://www.youtube.com/watch?v=tBQcpF_j5Xg&feature=related

This Website can give you a general idea of what Matter really is.

http://www.chem4kids.com/files/matter_intro.html

Sunday, September 26, 2010

Update Sept. 26

Unit conversion and scientific notation quiz is tomorrow, Monday September 27.
Read previous post to figure out what you need to know and study your notes. Good luck to everyone...

Wednesday, September 22, 2010

Chemistry 11 Sept. 21, 2010 Sci Notation + Unitary Rates

Scientific Notation is a method of writing numbers that are either very large or very small. This is done using powers of 10.  For example, 2.0 x 10 to the 3rd power;  the 1st number must be a number between 1 and 10(not including 10)

Examples:
340000000 = 3.4 x 10 to the 8th power
7800000000 = 7.8 x 10 to the 9th power
0.00000023 = 2.3 x 10 to the -7 power.
1) Identify the number
2) Move the decimal place over as many times to make the number between one and 10 in front of the decimal. You can either move it right or move it left depending on whether the number is large or small.
3) The number of spaces that the decimal was moved is the exponent which goes in the top right corner after the 10.
http://www.youtube.com/watch?v=H578qUeoBC0

Unitary Rates:

Start with one unit and convert it by using the conversion factor.
If it requires more than one step  go back to 10 to the power of 0 grams or litres.
Ex.  76.6 mL into L
= 1000 mL = 1 L
= 76.6 mL/ 1000 =(y) L
= 0.0766 L