Tuesday, November 16

Today, after we were all seated, Mr. Henderson started off the day by showing us the fantastic job Tim did on his blog. He briefly reviewed what Tim covered in his blog, which is what we learned yesterday. Then we jumped right into working in our packets and working our way through stoichiometry grade school.

Today we learned about using the mole island to help us solve stoichiometry problems. We started on problem 6 of page 2 in our packets. We had to find how much H2 we would be needed to produce 22.8 g of NH3. First we found that the amount of moles in 22.8 g of NH3, which is 1.34 moles. Then we set the equation up like this-

22.8 g/mol NH3 x 1 mol NH3/17 g NH3 x 3 mol H2/2 mol NH3 x 20 g H2/ 1 mol H2 = 4.02 g H2

This shows that it requires 4.02 g of H2 to produce 22.8 G of NH3.
Mole island shows us how to get to a certain value. To get to the value you follow the arrows, and that's how we completed number 6. We went from C island, to moles C, to moles A, to A island. And the number of moles was the coefficient to the substance.

We then worked on the problems in page 3, which I have the answers to here.

1. Ti: 47.9 g/mol N2: 28.01 g/mol Ti3N4: 199.6 g/mol

2. 3.15 mol N2 x 1 mol Ti3N4/ 2 mol N2 = 1.58 mol Ti3N4

4. 2.85 mol Ti x 1 mol Ti2N4/ 3 mol Ti x 199.6 g Ti3N4/ 1 mol Ti3N4 = 189.6 g Ti3N4

And then as well on page 4.

6. 4.91 mol Ti3N4 x 2 mol N2/ 1 mol Ti3N4 x 28.01 g N2/ 1 mol N2 = 275 g N2

7. 26.3 g Ti x 1 mol Ti/ 27.9 g Ti x 1 mol Ti3N4/3 mol Ti x 199.6 g Ti3N4/1 mole Ti3N4= 36.5 g Ti3N4

8. 93.6 g N2 x 1 mol N2/ 28.01 g N2 x 1 mol Ti3N4/ 2 mol N2 x 199.6 g Ti3N4/ 1 mol Ti3N2 = 334 g Ti3N4

If you are able to do these, then you are are in the fifth grade at stoichiometry school!

To finish off the day we had a demo. The demo consisted of Mr. Henderson adding aluminum (Al) to copper chloride (CuCl2) in different amount then that which would make the entire solution balanced. The equation for Al + CuCl2 is balanced to this-

2Al(s) + 3CuCl2(aq) ---> 3Cu + 2AlCl3

Now by adding .02 moles of Al to .03 moles of CuCl2, our end result would theoretically have had all of the atoms reacting in the single replacement reaction. All we would have left is Cu and AlCl3. But Mr. Henderson added only .01 moles of Al in the first reaction. The .o1 was not enough, and we would have leftover CuCl2. In the second reaction, Mr. Henderson added .03 moles of Al to the CuCl2, and then we have leftover Al. In the first reaction, the aqueous solution left in the flask was a darker blue than the pure CuCl2, but it was not as dark as the second reaction. In the second reaction the flask had turned almost entirely brown in color. This was because of the combination of the brownish Cu and the grey AlCl3. This demo showed us that if we do not have substances in the proper amounts in proportion to the coefficients of the equation, then we will have leftover substances.

So in conclusion, today we learned about using mole island for setting up unit conversions, and we also learned about having leftover substances from a reaction.

Monay, November 15

Today we started class with Mr.H handing out our grades. He assured us that even though some of us may not have done well, it is still the beginning of a quarter and there will be more tests to do well on.

After that was over with, Mr.H showed us the Academic Stimulus Program. This gave students an incentive to do WebAssign assignments early to receive extra credit. Here is a chart that shows how much extra credit is rewarded for doing the WebAssign assignments.

Assignment	Stimulus?	Due Date	Conditions of Bonus Points
Stoichiometry - Set 1	No	Tues, 11/16	--
Stoichiometry - Set 2	Yes	Thurs, 11/18	Xtra 20% if 24 hrs in advance
Stoichiometry - Set 3	Yes	Mon, 11/22	Xtra 15% if 72 hrs in advance
Stoichiometry - Set 4	Yes	Tues, 11/23	Xtra 15% if 24 hrs in advance
Stoichiometry - Set 5	Yes	Tues, 11/30	Xtra 15% if 24 hrs in advance
Stoichiometry - Set 6	Yes	Wed, 12/1	Xtra 15% if 24 hrs in advance

We then moved on the the first page to learn the basics of Stoichiometry. We learned that the coefficients in a balanced chemical equation provided information about the relative amounts of reactants and products involved in the reaction. In the equation

N2 + 3 H2 → 2 NH3 we noticed that it had coefficients of 1, 3, and 2 respectively, so for this equation there has to be 1 molecule of N2 and 3 molecules of H2 to create 2 molecules of NH3. The equation must always keep a ratio of 1:3:2. Here are the answers to problem 1.

1 molecule 3 molecules 2 molecules

2 molecules 6 molecules 4 molecules

3 molecules 9 molecules 6 molecules

5 molecules 15 molecules 10 molecules

1 mole 3 moles 2 moles

2 mole 6 moles 4 moles

10 moles 30 moles 20 moles

5.4 moles 16.2 moles 10.8 moles

1.63 moles 4.90 moles 3.27 moles

We then moved on to problem two, which got a little more difficult. In this problem, we determined the relative masses by which reactants combine to form a given mass of product using molar mass. In the same equation N2 + 3 H2 → 2 NH3 N2 had a molar mass of 28g/mole, H2 had a molar mass of 2g/mol, and NH3 had a molar mass of 17g/mol. If there was 28 grams of N2 there would have to be 6 grams of H2 to create 34 grams of NH3. Notice that the sum of the masses of N2 and H2 equals NH3. Here are the answers to problem 2.

28g 6g 34g

56g 12g 68g

2800g 600g 3400g

45g 9.64g 54.54g

32.1g 6.88g 39.98g

41.53g 8.9g 50.43g

November 10, 2010 Wednesday

Today we started out the class by reviewing for the chemistry test on Friday. Mr. H gave the class ways to review for the test. He told us about a review worksheet on the gbschemphys website. After this we turned to page 13 in our Unit 4 packets and reviewed what we had done the day before.

We know that when an ionic compound dissolves in water, it dissociates into ions. These are the balanced chemical equations which show the dissociation of ionic compounds in water.

a. sodium chloride

NaCl(s) ---> Na+(aq) + Cl-(aq)

b. aluminum nitrate

Al(NO3)3 (s) -----> Al3+(aq) +3 NO3-(aq)

c. barium chloride

BaCl2 ----> Ba2-(aq) +2 Cl-(aq)

d. sodium sulfate

NaSo4(s) ----> 2 Na+(aq) + So4 2-(aq)

We also know from our lab that that Na and NO3 are soluble, but there are more things that are soluble in water. On page 13 of the packet there is a table of solubility rules. The rules for compounds that have high solubility are...

High Solubility

1.All alkali metal and ammonium compounds

2. Hydrogen compounds containing H+

3. All nitrates, acetates, and chlorate's (AgCH3COO only moderately)

4. All chlorides, bromides, and iodides EXCEPT those containing Ag+, Pb2+, and Hg2 2+ (moderately soluble in warm water)

5. All sulfates EXCEPT those containing Pb2+ or the 4 heavier ions of Group IIA ( Ca2+, Sr2+, Ba2+, Ra2+)

Low solubility

6. All sulfides EXCEPT those containing alkaline earths and rules 1 and 2

7. All hydroxides EXCEPT those containing Sr2+ and Ba2+ and rules 1 and 2

8. all phosphates, carbonates, and sulfites EXCEPT for rules 1 and 2

We then turned to page 14 and finished up problem number 3. This was about writing net ionic equations. First you have to present the ions that were present in each solution, and then you identify the solid that are being formed, and then indicate the ions remaining in the solution. And then at the end you write the net equation of the solution.

example a.

Na2SO4 + BaCl2

-ions: Na+ Cl+

-solid: Ba SO4

net ionic equation: Ba2+(aq) + SO4 2-(aq) ---> BaSO4(s)

example b.

AgNO3 + NaCl

-ions: Ag+ Cl+

-solid: Ag+ Cl+

net ionic equation: Ag+(aq) + Cl-(aq) ----> AgCl(s)

We then went on to finish the rest of the page. In problems 5-8 you had to analyze the major species in the solution and write the net ionic equations for the solutions.

5. Pb(ClO4)2(aq) + NaI (aq) ---> PbI2(s) + NaClO4(aq)

Pb2+(aq) + 2I-(aq) ----> PbI2(s)

6. Ca(OH)2(aq) + H3PO4(aq) ---> H2O(l) + Ca3(PO4)2(aq)

Ca2+(aq) + PO4 3-(aq) ----> Ca2(PO4)2

7. CdBr2 + Na2S ---> CdS(s) + 2 NaBr(aq)

Cd2+(aq) + S2+ (aq) ----> CdS(s)

8. NaOH + Fe(NO3)3 ----> Fe(OH)3 + NaNO3 (s)

Fe3+(aq) + 3H9aq) ---> Fe(OH)3 (s)

For problem 9, you are given a sentence which has the names of the solutions and you have to write them out and then write the net ionic equation.

9. Silver nitrate is mixed with sodium chloride solution

AgNO3(aq) + NaCl(aq) ----> AgCl(s) + NaNO3(aq)

net ionic equation: Ag+(aq) + Cl-(aq) ---> AgCl(s)

Some notes to keep in mind for the test....

-Aqueous solution of ionic compounds contain the compound in ion form.

-Ion charges can be found from....

.....the location on the periodic table for ions of the main group elements..

....the Roman numeral used in the name for ions of transition metal elements..

* when two aqueous solutions of ionic compounds are mixed, a precipitate can be formed; an ion from one of the solutions reacts with an ion from the other solution to form a solid precipitate.

Monday, November 8

Mr. Henderson started class today by examining what the rest of this week looks like. We will be learning a new topic for the next couple of days preceding the test, followed by the test Friday. He told us that the test Friday will be composed of: 14 multiple choice questions, 1 double-sided page of writing and balancing equations, no math work, and a section on identifying the reaction type. Mr. H said that what is going to make this test difficult is that it requires a lot of unit 2 work -- writing equations and such. But, we have a very resourceful tool to help us review, this blog! Mr. H walked us through Brooke's blog and talked about how helpful it will be to study for the test. To review, we can also use the Delicious websites created by the class (which, reminder, are due on Thursday!).

We then moved on to the new topic of today, Dissociation and Hydration. He begun this by doing page 11 in our packets. The first small section of this page was identifying a compound as being either ionic or molecular. An ionic compound is one which consists of a cation (metal) and anion (nonmetal), while a molecular compound consist of, normally, two anions. A good, common example of a molecular compound are those hydrocarbons Mr. H keeps talking about. These are the answers to the first part of page 11 ("i" for ionic and "m" for molecular):

1.

a. NaCl I

b. CO2 M

c. NaNO3 I

d. D2H5OH M

e. NH4Cl I

f. C6H12O6 M

If you need extra help distinguishing between the two types of compounds, this is very helpful: http://www.buzzle.com/articles/ionic-compounds-vs-molecular-compounds.html

The following section of the packet introduces a new topic to us, dissociation/hydration. We were supposed to find what the compound looked like after it had dissolved in water (after being broken up into individual ions by the water). The first few problems/answers are:

a. KF(s) -> K+(aq) + F-(aq)

b. CaCl2(s) -> Ca2+(aq) + 2Cl(aq)

c. Na2SO4(s) -> 2Na+(aq) + SO42-(aq)

One of the most difficult things to remember is that Uncle HONClBrIF does not apply to these ions because they're not considered to be "by themselves", but "with" water.

As should be obvious by now, all we are doing is splitting the compound into the ions of which it is made up. But then it gets a little more trickey; we are asked to perform the same type of equation, they give us the name and expect us to derive the formula from that. For example the first question asks what happens when you put lithium phosphate in water:

First, we figure out that the formula for lithium phosphate is Li3PO4

Next, we figure out that the ions of which it is mad up are Li+ and PO43-

Then, we balance the equation to get something which looks like this -- Li3PO4(s) -> 3Li+ + PO43-

After doing a couple more problems like this one, we moved onto yet another new topic. This is the topic of electrolytic (E) and non-electrolytic (NE) solutions. What makes a solution electrolytic is if it is created by ionic compounds, a cation and an anion.

Answers to the last section:

a. NaCl E

b. SO3 NE

c. C2H5OH NE

d. KF E

e. BaCl2 E

f. C12H22O11 NE

He demonstrated this concept through a conductivity test. If, when the wires were placed in the solution, it caused the lightbulb to light, then the solution is electrolytic. This also implies that it contains an ionic compound. He used a video (from the blog) to demonstrate this:

http://g.web.umkc.edu/gounevt/Animations/Animations211/StrongWeakNonElelytes.swf

After this, we moved on to Lab CR2, the Chemical Dropout Lab. We mixed various substances, observing whether or not they each created a precipitate. If a solid developed, or the mixture became cloudy, we knew that a precipitate had been formed.

This picture shows perfectly what precipitates look like.

Overall, today was a very intense day. We were introduced to many new concepts, but luckily Mr. Henderson is explaining them very well and we have many review tools at our fingertips.

Tuesday, November 9

As always, Mr. Henderson started off with a recap of what we did yesterday, and a preview of what will happen in the class. He told us to get out our lab notebooks and page 12 in our unit packets. We first went over what PPT meant in our lab from yesterday, called Chemical Drop Out. PPT means precipitate.

He then went and showed us a video on the Chemistry website, found here. It corresponds with question 5 on page 12. In the video, the first substance that was dissolved in the water was NaCl, sodium chloride. Mr. H explained that the "Mickey Mouses" (H2O) were connecting with sodium chloride. The positive of H2O, H, forms a bond with the negative Cl ion. Then, the O, the negative ion in H2O, bonds with the positive Na. This is an example of dissociated ions.




That is why the light bulb lighted in the video; it is an example of a strong electrolyte. He then explained the next two examples. The second example is called a weak electrolyte because there are more associated ions (ions that did not break off and make bonds with the water) than dissociated ions. This causes the light bulb to light very dimly. When methanol is added to the water, it does not dissociate into ions. This is called a nonelectrolyte.

For Question 6, we had to decide what type of electrolyte it was by the picture. The first picture is SE, strong electrolyte. Someone in the class explained that since there were no bonds and just molecules in the picture, it means that it completely dissociated and therefore is a strong electrolyte. The second picture is a non-electrolyte because there are no floating ions that are separated. Finally, the last one, you guessed it, is weak electrolyte. This is because there are some associated bonds, but there are other dissociated ions, too.

We moved onto question 7, now. We had to find the major species that would be in water by the compound given. He went through A, B, C, D, and F together with us. The answers are as follows:

A.) Compound: NH4Cl
Major Species: NH4+, Cl-, H2O
B.) Compound: C2H5OH
Major Species: C2H5OH, H2O
C.) Compound: H2SO4
Major Species: H+, SO4 2-, H2O
D.) Compound: NaC2H3O2
Major Species: Na+, C2H3O2, H2O
F.) Compound: HF
Major Species: HF, H2O
Minor Species: H+, F-

Note:

• All soluble ionic solids are strong electrolytes.
  
• Certain acids are strong electrolytes and goes as follows: HCl, HNO3, H2SO4, HBr, HI and HClO4
• Other acids are referred to as weak electrolytes.
• Molecular compounds held together by covalent bonds do NOT dissociate and are considered non-electrolytes.

Mr. H told us to refer to this for the WebAssign, due Wednesday, November 10.

PAGE 13

For 1a and 1b, refer to Hannah's blog on how to do them. For C and D, the steps are as follows to solve:

1. Get the formula of reactants
2. Write formula of product AND charges
3. The state that it is in (solid, gas, liquid, or aqueous)
   
4. Balance

The rules of solubility is written in the box on top of page 13.

LAB FROM YESTERDAY/PAGE 13

Mr. H made us take out our little data sheets from the lab, Chemical Drop Outs. He explained that each row (horizontal) had sodium (Na) and the columns (vertical) had nitrates (NO3). We had page 13 out, to refer to the box on the top. It has the solubility rules. For the rows, focus on the ANION, not cation (Na). Similarly, focus on the CATION, not anion (NO3). Mr. H had an example on the board, but he said that most carbonates are insoluble. Insoluble, meaning that there are more PPT's written in the boxes in the row/column. For the lab Conclusion/discussion, we had to write a sentence about each row and column. For example, for the row of Na2CO3, most of my group's boxes are written with PPT. So, this means that carbonate is insoluble. So, my sentence for that row would be carbonate is insoluble with exceptions of aluminum, ammonium, and calcium. The EXCEPT is there because not all of the reactants had a precipitate. Those would be the NR in the boxes. If a row or column has more NR's than PPT's, you would write that that row/column is soluble with exceptions of the boxes with PPT in it. Every group is different, which would result in different sentences.

Remember that the homework for tonight is a WebAssign and a Delicious bookmark!