Pre Test Tutoring

For anyone who wants to take advantage of it, I will be at the school at 7:30 Monday morning. If the door is locked, I will leave a note telling you which room to go to. Have questions ready since we might not have much time to review the material.

Hope to see some of you guys there!

Stoichiometry Solutions

Practice Problems Sheet

To view the practice problems, click on the link above.

!Solutions!

Below are the problems' solutions. If you have any questions, simply post a comment and I'll try to answer them as soon as possible. Click on the images to zoom in.

**Pay attention to significant figures. If the problem starts with three, then the final answer can only have three. This might require rounding.**

1) look at photo.
2) 125gFeCl3 -->  124g NH4Cl,  127gNH4OH --> 194gNH4Cl FeCl3 is the limiting reactant
3) There are 45.9g NH4OH left over.
4) 44.4% yield

1) Look at photo
2) 12.5mol CuNO3 --> 1.12 x 10^3g Fe(NO3)2, 9.50 x 10^28 molecules Fe --> 2.80 x 10^7g Fe(NO3)2

CuNO3 is the limiting reactant

3) 8.7 x 10^6g Fe(NO3)2
4)88% yield

1) Look at photo
2)125mol Al --> 6370g Al2O3, unlimited O2 --> ∞Al2O3  Al is the limiting reactant
3)O2 is unlimited so the amount left over is unlimited
4)100%

 

1) Look at photo

2)125gAl --> 14.0g H2, 129ml H2SO4 --> .521gH2  H2SO4 is the limiting reactant

3)120g Al are left over after the reaction

I Want 'em Quantum Numbers

Lets look back on what we discussed Friday.

As we discussed, quantum numbers are a method to pinpointing the probable location of a specific electron. Remember that my analogy, useful or not, likened quantum numbers to the address label on an envelope. Both increase in level of specificity until an exact location is pinpointed. With quantum numbers, there are 4 locations used to pinpoint the electron. Going from most general to most specific, they would be listed n, l, m, and s. So, lets look back at some rules governing how we use these.

n:

(n) applies to the principle energy level. In simplest terms, n corresponds with the period in which you're working. For example, if we were to look at Titanium, the period in which it is found would be period 4, so n would equal 4.  For Zirconium, the period is 5, so n=5. So on and So on. Using n, we can find the general roaming grounds of an electron, but we're nowhere near its exact location. For that, we must find l.

l:

(l) represents the orbital in which an electron can be found. This is also very simple as long as you remember the following chart.

s=0

p=1

d=2

f=3

This can narrow our search, but as we discussed Friday, there are many "configurations" that the electron can be found in. We showed these configurations by drawing and labeling slots. These slots lead into determining (m).

m:

(m) is the most difficult to find, but if you remember how to label correctly and how to place electrons in orbitals correctly, you will have it made. First, you must remember that each orbital has a different number of ways that electrons can move around the nucleus. s has only 1 while p=3, d=5, and f=7. To determine in which configuration you can find an electron, you must first have a method of labeling each of those configurations. So what we do first is draw a slot for each configuration. Then we take the (l) we determined earlier and place that number under the far right slot. At this point, all you need to be able to do is count. Move from the far right to the far left decreasing by one integer with each slot. For example, lets continue to use Titanium. We know that it is found in the d-orbital, so (l)=2. Now we draw out the number of configurations that can be taken, which is 5. This is what it would look like. 

_  _  _  _  _ next, we must label them by placing (l) at the far right and moving leftwards.

-2  -1   0   1   2  

You MUST remember that one electron must be in each slot before any can pair. This means that all 5 slots will have a single ↑ in each before you will see any with ↑↓. Knowing this, we can see that Titanium will have an m value of -1 because it has two electrons in the d orbital, the last of which is located in the second slot ↑. If we were to look at Cobalt, it would also have an m value of -1 because it too has its outer electron in the second slot, but it would look like ↑↓. Finally, we must look at S.

S:

S represents the spin of the electron. Fortunately for us, there are only 2 ways an electron can spin. To denote this we use +1/2(↑)and -1/2(↓). It's that simple. Back to our example. Titanium's outer electron is located in the -1 orbital and since it is the first, it is written as an ↑. For this reason we apply an s value of +1/2 to it. Now look at Cobalt, we know that at this point, electrons have started pairing. We know that the outer electron is also located in the -1 orbital, but it is paired and looks like this ↑↓. Seeing that it points downward, we can apply an s value of -1/2 to it.

To wrap up, lets look at a couple of elements and determine their quantum numbers.

Titanium:            n=4,  l=2,  m=-1, s=1/2

Cobalt:                 n=4,  l=2,  m=-1, s=-1/2

Zirconium:         n=5,  l=2,   m=-1,  s=1/2

Welcome Students!

Welcome students and parents alike! With the many projects going on in my life right now, I decided that supplying my students with a location for supplemental information would help everyone. The ultimate goal is to provide additional reference material that supports the lessons learned in class each day. In addition, if students have any questions, you can post it in the comment section of each new Blog post. I hope to check the Blog each night, so hopefully I can address any questions in a timely manner.