Blog Week 32

On Monday, we talked a little more about the Nail lab. We wanted to know the reasoning for the rise in temperature of the test tube. We learned that breaking bonds requires energy and when energy increases, so does the temperature. The energy to break the bonds comes from the chemical energy. An exothermic reaction is when energy of the reactants were greater than the products, therefore extra leftover energy exited the system. An Endothermic reaction is when extra energy would need to enter the system before the reaction could take place. On Tuesday, we tried to do some worksheets in class, my group got number three. The picture below is our work:
On Wednesday, we had a sub and we were to do a worksheet. I didn't understand how to do any of it really. I did attempt a couple  for one I got PE + KE = U. Then I attempted number five and got a) increase b) increase c) increase d) decrease e) decrease. I need help understanding the LOL diagrams. On Thursday, we had a tutorial, because Mr. Abud won teacher of the year. Congratulations Mr. Abud! On Friday, we reviewed everything we've learned this year to prepare for the final.

Blog Week 31

On Monday, we started out lab. Our lab was seeing what happens when an iron nail is put in a mixture of Copper Chloride and distilled water. We first massed our nail which came out to be 15.8g. Then we massed our distilled water to be 50mL. after that we weighed the plastic container that would hold the Copper Chloride and that came out to be 3.1g; then we added the Copper Chloride and the mass for the Copper Chloride minus the container came to 6.5g. When we poured the solution into a test tube and the test tube was too small, so we had 24mL of the solution was left over. So, our measurements will probably be off. We did observe what happened to the nail once it was in the solution. We noticed that the nail started turning red and the test tube became warm. This is the reaction equation I have so far for this experiment: Fe + CuCl2 = FeCl2 + Cu. On Tuesday, I was absent. On Wednesday, I picked up with the lab. We measured the nail after it soaked in the solution and it weighed 14.5g. Then we weighed the copper by weighing an empty beaker then subtracting that from the beaker holding the copper. The amount of copper was 2.5g.  We also found the amount of iron by subtracting the first nail mass minus the last nail mass and it came to 1.3g. After that we did out calculations to turn the grams into mols. The equation is mass x 1mol/molar mass= mols. So, for Cu we did this: 2.5g x 1mol/63.5g = .0393 mols of Cu. Then we did Fe: 1.3g x 1mol/55.85g = .0232 mols of Fe. We then found the ratio .0393/.0232 = 1.69. Which is close to a 3 to 2 ratio. From here we went wanted to find out if our balanced equation was right. We learned that Fe could have a charge of +2 or +3. And since the ratio is very close to 3 to 2, we decided that in this situatuion Fe had the charge of +3. So, our balanced equation was 3CuCl2 + 2Fe --> 2FeCl3 + 3Cu. On Thursday, we talked about how next week we are doing the silver coke lab! Where we will turn silver+ to a simple silver. We then went on to discuss more about our nail lab. We came to the conclusion the iron had a charge of +3. We then learned that there were mini reactions with in the bigger reaction. They are called coupled reactants and they were these: Cu+2(aq) + 2e- --> Cu(s) and Fe(s) --> Fe3+(aq) + 3e-. (aq) stand for dissolved in water. The first one gains an electron and it's called a reduction. The second one loses an electron and it's called an oxidation. In this reaction is a redox, it's a coupling of a reduction and oxidation together. On Friday, we had an assessment, I'm definitely going to reassess.

Blog Week 30

On Monday, each group brought their cookies made with a different recipe. My recipe ended up being the real recipe. Other groups had too much of one ingredient or too little of an ingredient. This project was done, so we could understand what limited reactants and an excess reactant was. A limited reactant is a substance that is completely used up in a chemical reaction. An excess reactant is a substance that is partly used, after the chemical reaction there is still some of the reactant left over. On Tuesday, practiced with some problems on stoichiometry. First we balanced the equation, then we decided which reactant was limited and which one was excess. From there we figured out the ratio and continued with the problem doing the same steps from last week's lessons. Here's an example below:

               CH4 + 2O2 --> 2H2O + CO2  <-- need to balance equation

Before:   2.5xs   2.5LR        Ø           Ø     <--- Ø because it's being produced 

Change:  -1.25    -2.5        +2.5      +1.25 <-- are mols  

________________________________  <-- -1.25 bc it was a XS, and it was a 1:2 ratio, only 1/2 of 2.5 

After:        1.25      0          +2.4     +1.25

We also drew particle diagrams. I shows this chemical reaction in the picture below.

On Wednesday, we had a sub and we did more practice on limited and excess reactants. We also worked on converting moles into grams and grams into moles. 

Here are some examples:

3.

               2Al    +  3I2 --> 2AlI3  <-- need to balance equation

Before:   .50 xs  .72 LR      Ø       <--- Ø because it's being produced 

Change:  -.48    -.72        +.48      <-- are mols  

___________________________<-- -.48 bc it was a XS, and it was a 2:3 ratio, only 2/3 of .72

After:       .02       0          +.48     

4. Na2SO4 --> g    .20 mol--> g  Na2SO4 = .20 mols and molar mass = 142 

 .20 mol x 142g/1 mol = 28.4g 

On Thursday, we learned about yield percent. the equation for it is actual/theoretical. An example is 35g NaCl/50g NaCl = .7 = 70% 

We also worked on more examples like on Wednesday. Here is another example:

               2KCIO3  --> 2KCI + 3O2  <-- need to balance equation

Before:       1.2                0          Ø      <--- Ø because it's being produced 

Change:    -1.2           +1.8       +1.8    <-- are mols  

_____________________________  <-- -1.2 bc it's a 3:2 ratio, 3/2 of 1.8

After:          0              +1.8      +1.8

On Friday, I was absent. We were to take an assessment. 

Blog Week 29

On Monday, I was absent. On Tuesday, we worked on reactant equations. We watched Ethanol react with something and oxygen. Then we tested to see if it would combust and it did. A combustion reaction makes carbon dioxide and water.

Combustion -> CO2 + H2O. We had an equation for our experiment it was: C2H6O + 3O2, we found the molar mass for C2H6O = 46.08 g/mol and the molar mass for 96 g/mol. So, 46.08 + 96 = 142.08 g/mol. Then we found the reaction equation C2H6O + 3O2 -> 2CO2 + 3H2O. The molar mass for 2CO2 is 88.02 g/mol and the molar mass for 3H20 is 54.06 g/mol. So, 88.02 + 54.06 = 142.08 g/mol. So, the reaction equation is right because the molar masses are equal from before and after the reaction. C2H6O + 3O2 -> 2CO2 + 3H2O, so for every 1 C2H6O there's three H20. Then for every 50 C2H6O there's 15o H2O. We also, went over the equation for g<--> mol, they are g x 1mol/__g = __mol and mol x __g/1mol =__g. On Wednesday, we learned how to do BCA tables. We did a practice:
               2H2S + 3O2 --> SO2 + 2H2O <-- need 3O2 to get rid of 2H2S.
Before:   +2.4       XS          Ø           Ø     <--- Ø because it's being produced 
Change:  -2.4       -3.6        +2.4      +2.4  <-- are mols  
________________________________  <-- 3.6 because it's a 2:3 ratio, (3/2) x 2.4 = 3.6 
After:        0           XS        +2.4      +2.4

After that we wanted to find out how much (g) of H20 did we produce?
mol x molar mass/1 mol = g
2.4mol H2O x 18g/1 mol = 43.2 g.
    ^                      ^
in equation      H2=2 and O = 16 H2O = 18

On Thursday, we did more examples:

2.            C3H8 + 5O2 --> 3CO2 + 4H2O <-- needed to balance equation

Before:       +4       XS          Ø           Ø       <--- Ø because it's being produced 

Change:      -4       -20        +12        +16     <-- are mols  

________________________________    <-- 4 bc it's a 1:3 ratio, (1/3) x 12 = 4; 20 bc 1:5, (1/5) x 4 = 20

After:           0         XS        +12      +16          16 bc 1:4, (1/4) x 4 = 16

6.           3H2 + N2 --> 2NH3 <-- need 3H2 to get rid of N2.

Before:   .45     .15           Ø     <--- Ø because it's being produced 

Change:  -.45    -.15       +.30  <-- are mols  

________________________<-- .45 bc it's a 3:1, (3/1) x .15 = .45 and .30 bc it's a 2:1, (2/1) x .15 = .30   

After:       0         0          +.30

Then we wanted to find how much (g) of 2NH3 we produced?

molar masses N=14 H3=3 

17g/1mol x .30 = 5.1 g of 3NH3.

On Friday, we took an assessment. I think I did okay, but I didn't know the type of chemical reaction. Then I took the assessment from last week. I understood everything, except the chemical reactions.