Unit 10, Day 3 (aka Friday, March 18)

We entered the classroom on Friday and immediately went over pages 19 and 20 in our packets. The answers to the 4.1/10.1 Reading Sheet are as follows?:
1. c, d
2. 2.0 M
3. a
4. b
5. d
6. 0.220 mol
7. 0.0209 L
8. c
9. (ask Mr. H)
10. d
11. a - 0.50 M
b - 0.250 mol
c - 0.050 M
12. Mole fraction: a
Mass %: e
Molarity: b

We were then reminded of the WebAssigns due on Monday and Tuesday. Mr. H proceeded to go over the blogs from the last two days. Kendall's encompassed the concept of reading a graph of the solubility curve. Kevin's blog slightly offended Mr. H because it talked about the special St. Patrick's day demo we had the day before. Kevin claimed it was "no biggie" while Mr. Henderson argued that it was indeed "a biggie." Kevin's blog also included information about how to determine molarity as well as a recap of the lab(s) we did that day.
Mr. H then instructed us to turn to page 3 in our packets where we used the equation moles/liters = molarity to figure out the problems. He told us that before we calculated, we had to establish the numerator as moles (which is often converted from grams) as well as the denominator as liters (which is often converted from milliliters). For number 8, we had to convert 35.5 g of HC2H3O2 to moles. Our answer was 0.591 moles. Then we took 88.0 mL and converted it to 0.088 L. With the moles over the denominator, our answer for molarity was 6.72. Mr. H also taught us a snazzy way to write the symbol of molarity for that equation:

[HC2H302] = 6.72 M

Pretty simple. Number 4 was done in a similar manner. 93g of KCl were converted to 1.249 mol and .60 L was already workable. With the moles over liters, the molarity came out to be about 2.1 M.

Page four included some concepts about ions. In order to grasp it, we had to turn to page 19, where the marginal notes disclosed information about dissolving molarity. For number 14, Mr. H told us that we didn't need a calculator for part a. The concentration of the ions is 0.60 M. How? I honestly don't know, ask Mr. H. Then for part b we were reminded to get used to the idea that the number of moles = L x mol/1L. So for part b, .1500L x (.60 mol/1L) = 0.090 mol Fe3+.

Number 18 on page 4 is basically impossible. I would recommend you go see Mr. H for that one. I'm clueless. I'm gonna go see him too.

Then, Mr. H handed out some sheets of paper and told us this was the demo. It was a dilution problem that he was prepared to show us. I'm a faliure at scribing cuz I just realized that I left my compsition book with the demo information at school. Well, I'll have it in class tomorrow so if you need to see it, come see me (yes, it is right for once). Then the bell rang...LOL!

Sorry about my mediocre scribing today, Mr. H. I'll do better next time.