Friday April 29th

Class began with collecting the answers from a reading sheet webassign which had previously been for homework. This can be found on page #25 of your packet.

CHAPTER 13.2-3 pH, pOH and the Ion Product of Water
1. b able to both accpet and donate a proton.

2. a H+ and b OH-

3. d 1.0x10e-14 = [H+] x [OH-]

4. a the dissociation of water into its ions

5. neutral
acidic
basic
basic

6. a increases

7. c 2e-7 M

8. b decreases

9. b basic

10. a acidic

11. b lower, acidic

12. basic
acidic
neutral

13. c 11.0

14. d the conjugate acid of water in an acid is ionization

15.d HNO3(aq) + H2O --> [H3O+](aq) + [NO3-](aq)

16. b pH=1

17. a [Na+]=0.50 M, [OH-]=0.50 M, and [H+]=2.0e-14 M

18. d Use a stoichiometric calculation

*Mr. Henderson pointed out that we should write down the equations/formulas on the right side of the page on a notecard*

equations/formulas:

• [H3O+] x [OH-] = 1.0e-14


• [H+] x [OH-] = 1.0e-14


• pH= -log[H3O+]


• pH= -log[H+]


• [H3O+]=[H+]=10^(-pH)


• pOH= -log[OH-]


• [OH-]= 10^(-pOH)


• pH + pOH = 14 (at 25 degrees C)

We then turned to page #4 where we worked on #9 part a.

a. the first step is to write the equation

H2O + HCN <---> [H3O+] + [CN-]

I 2.0 0 0

C -x +x +x

E 2-x x x

Ka expression= ([CN-] x [H3O+])/[HCN]

Ka=6.2e-10=(x^2)/(2-x)=(x^2)/(2)-->(x^2)=1.24e-9

x=3.5e-5

This is when the 5% approximation comes into play. This says that since the value of x is so small that it can just be ignored in the 2-x part of the equation.

3.5e-5 M=[CN-]=[H3O+]

We then turned to page #6 to finish the table in probblem 12 and work on question #13

12. [H3O+] [OH-] pH pOH acidic or basic?

1.0e-2 1.0e-13 2.00 12.00 acidic

1.0e-11 1.0e-3 11.00 3.00 basic

1.0e-5 1.0e-9 5.00 9.00 acidic

1.0e-3 1.0e-11 3.00 11.00 acidic

1.0e-7 1.0e-7 7.00 7.00 neutral

2.0e-14 5.0e-2 13.70 0.30 basic

6.0 1.67e-15 -0.78 14.78 acidic

2.5e-15 4.0 14.60 -0.60 basic

13. a. [H+]= 2.0 M

pH= -log(2.0)=-0.30

b. [OH-]= 2.0 M

pOH= -log(2.0)= -0.30

pH= 14.30

c. [OH-]= 0.100 M

pOH= -log(0.100)=1.00-->pH=13.00

Review of the six strong basses:Li, Na, K, Ca, Sr, Ba

Light and Energy/Bohr Model of the atom TUESDAY, FEBRUARY 10, 2011

Mr Henderson started out the class by having us take out our packets and turn to page#13-14 to give us the correct answers to one of the webassigns due the previous day

(Pages 13-14)The answers to this webassign are:
1)a. Its predictions were inaccurate for atoms having more than one electron.
2)c. The best that one can know of an electron's location is to know the probability that it is in a given region of space.
3)d. Complex differential equations
4)b. Where an electron has a 90% probability of being located
5)b. The value of n cannot be negative AND c. The larger the n number, the higher the energy of the electron.
6)a. The value of l reveals information about the shape of the orbital housing the electron AND c. The value of l cannot be negative AND d. The value of l must be a whole number
7)b. The value of ml indicated the directional orientation of the orbital AND The value of ml must be a whole number
8)b. The value of ms indicated the spin direction
9)b. an l value of 1
10)a. TRUE b. TRUE c. TRUE d. TRUE e. TRUE
11)TRUE
12)b. 2
13)a. For every n value, there are just as many sublevels.
14)a. INVALID b. VALID c. INVALID d. INVALID
15)c. The 3s orbitals are larger than the 2s orbitals.

The class then looked at the blog from the previous day done by Dmitriy and Mr. Henderson had the class watch the video he added of the plane crashing into 12 feet of concrete.

Next the class took out their calculators and turned to page #1 to work on problem 4a. and 4b.
These problems will probably be the only time calculators will be needed to use for the rest of this unit.
The equation needed to remember for these equations is E=h*f=h*c/wavelength. f stands for frequency.

4a...red photon (f=4.80*10^14 Hz)

E=h*f=(6.62*10^-34)(4.80*10^14 Hz)=3.18*10^-19 Joules

4b...green photon (wavelength=5.43*10^-7)

E=h*c/wavelength=(6.62*10^-34)(2.998*10^8)/5.43*10^-7=3.66*10^-19 Joules

We continued the class period by turning to page #2 and started the problems in the chart at the bottom of the page.
Photon(Energy in Joules) use the equation from page #1
a. 3.03*10^-19
b. 4.09*10^-19
c. 4.55*10^-19
d. 4.85*10^-19
e. 5.00*10^-19
f. 1.64*10^-18
g. 1.93*10^-18
h. 1.06*10^-18
i. 1.55*10^-18

For these next problems use the equation En=Rh/n^2

• Rh is the constant called the Rydberg constant and its value is -2.180*10^-18 Joules.
• replace n with the number given in the problem.

E Initial(in Joules)
a.-2.42*10^-19
b. -1.36*10^-19

E Final(in Joules)
a. -5.45*10^-19
b. -5.45*10^-19

Change in E(in Joules)
a. -3.03*10^-19
b. -4.09*10^-19

The class was then given glasses that when put on revealed the true colors from white light included in roygbiv(red,orange,yellow,green,blue,indigo,violet)

We then also looked at different types of lighting which revealed different strengths of the colors from roygbiv.

At the end of the period the class ended the day by beginning a lab in the back of the room. The lab will be continued tomorrow

HW:next webassign not due till the 14th.

Friday, December 10, 2010

The class period started out looking at the blog from the previous night. Mr Henderson then had the class take out the unit 6 packets.

To answer the following problems one of five equations must be used.

1)P*V=n*R*T
(R=0.08206 L * atm/mol * k)
P=pressure=force/area, force=cumulative force from all collisions of particles with container walls.(pounds/inch2=psi, atmosphere=atm, millimeters of mercury=mmHg, torr)

V=volume(liters,mL)

n=number of moles

T=temperature= measure of average kinetic energy or KE (C or kelvin). Kelvin(K)=C + 273

2)V1/T1=V2/T2
(when n and P are constant)

3)P1* V1= P2* V2
(when n and T are constant)

4)P1/T1= P2/T2
(when n and V are constant)

5)[P1*V1]/T1=[P2*V2]/T2

Page#9 number 1

1) V1= 16.9 V2=?
T1=25C (25+273=298), 298K T2=268K

Since the information given contains temperature and volume the next step is to look at which of the five equations contains temperature and volume and the equation is V1/T1=V2/T2

16.9/298=V2/268

V2=15.198 Liters

Page #10 number 7

7)V1=412 mL V2=663
P1=1.00 atm P2=?

Since the information given contains volume and pressure the equation for this problem is P1*V1=P2*V2

412*1=663*P2

P2=0.621 atm

Page #11 number 15

15) P1=2.50atm P2=?
T1=5C(5+273=278), 278K T2=650C(650+273=923), 923K

P1/T1=P2/T2

2.5/278=P2/923

P2=8.30 atm

Page #13 number 26 and 32

26) V1=18.5 L V2=?
P1=85.5 kPa P2=1atm, 1atm=101.3kPa
T1=296 K T2=273 K

[P1*V1]/T1=[P2*V2]/T2

[25.5kPa*18.5L]/296 K=[101.3kPa*V2]/273 K

V2=14.4L

32)V1=5.25L
T1=23C(23+273=296), 296K
P1=8.00 atm
n=?
m=?

P*V=n*R*T n=[P*V]/[R*T]

n=[8.00 atm*5.25L]/[0.08206L*atm/mol*K *296 K]=1.73 mol

m=1.73 mol CO2*[44 gCO2/1 mole CO2]=76.1 g

The class period ended with Mr. Henderson making ice cream for the class.
HW:2 webassigns due Monday and test Friday

Friday, November 5

Mr. Henderson started out the period by looking at the scribe from the previous day by Katie then began to work on problems #33, 34, and 35 on page #10.

*when working on these problems and those similar to them...
1) write the reactants
2) write the products
3) balance the equation

PAGE #10

33.Butane gas (C4H10) is burned in a gas lighter.

C4H10 + 13/2O2 ---> 4CO2 + 5H2O ... need to have coefficients that are whole numbers so multiply all coefficients by 2

C4H10 + 13O2 ---> 8CO2 + 10H2O

34.Zinc reacts with silver(I) nitrate in a single replacement reaction

Zn + 2AgNO3 ---> 2Ag + Zn(NO3)2

35.Fluorine reacts with sodium bromide in a single replacement reaction.

* UNCLE HONClBrIF*
F2 + 2NaBr ---> 2NaF + Br2

NOTE:
~Whenever something is burned, combustion occurs and the result is the oxide of the element.
~Hydrocarbons are most commonly burned.
>octane(car or bus ride to school)
>methane CH4(to heat home or cook food)
>propane C3H8(grill)
-result in carbon dioxide CO2 and H20(water)
~CO2 causes environmental problems
(greenhouse gas -> greenhouse affect -> global warming)
-sunlight can get through and heat the earth but cannot escape well

DEMO:
SOLUBLE SOLID-
Potassium iodide in beaker(white solid)
solution=solute (present in least amount)
water=solvent(present in greatest amount)
-end result is the solid is gone(dissolved in water)
KI(s) ---> KI(aq)

TO BE ABLE TO TELL IF A SOLUTE DISSOLVES...
1. boil water(about 30 minutes)
~result would be KI in solid state (white)
2. taste it(shouldn't just taste like water)

Copper Chloride
-dissolves in water(changes color to a greenish blue)

Lead Nitrate(heavy)
-dissolves in water(whitish liquid)
Pb(NO3)2(s) ---> Pb(NO3)2(aq)

ALL LIQUIDS MAKE THE LIGHT BULB LIGHT UP BECAUSE THE SOLIDS BROKE DOWN INTO CHARGED IONS.

SUGAR DOES NOT LIGHT THE LIGHT BULB BECAUSE NO IONS ARE PRESENT.

HW:
*TEST MOVED TO NEXT FRIDAY
*webassign due Monday
*delicious assignment probably due Wednesday(tags- hcp3y1011 unit4 first name, last initial. do not use commas to separate the tags just separate by spaces and keep the tags in this order.
*lab notebooks will be collected Friday

Monday, September 20

Today Mr. Henderson started the day out by signing up the remainder of people for blog days if they have not already done the blogging. He then goes on to tell our class that today we will be doing a lab test. The lab test will be done throughout today and tomorrow.

Homework: Labs #5, 6, and 7 are due Tuesday (tomorrow)

webassign is due Wednesday

TEST THURSDAY

FOR THE TEST, STUDY FROM THE BLOG, TEXTBOOK, AND OBJECTIVE SHEET FOUND IN THE UNIT PACKET.

A sheet was then handed out explaining what should be included in the labs which will be turned in tomorrow.

Mr. Henderson then began to explain what our class is going to be doing for the rest of the week. Tomorrow we will continue working on the sugar lab and then have a lab test. Then we will review for the test Thursday. Wednesday we will be starting to learn about the next unit. Thursday is the test and on Friday first someone from the TLC will be coming to talk to our class and then we will be going to the computer lab.

Next Mr. Henderson told our class to take out our Chemistry Basics packet while he opened Anthony's blog from the previous day (Friday). After looking at the blog we turned to page 13 in our packets and worked on problem #5, which said if a piece of aluminum has a mass of 14.0 g, what volume does it occupy?(the density of aluminum=2.70 g/cm cubed). To find the answer, Mr. Henderson set up the equation volume=mass/density=14.0g/2.70g per cm cubed. The answer when put into the calculator came out to be 5.185185185185. Using the significant digits rule the answer is 5.19 cm cubed. We also learned the answer could be 5.19 mL because cm cubed equals mL.

The next problem we did was #11 on page #14. The question was several beads of an unknown metal are placed into a partially-filled graduated cylinder. The water level in the graduated rises from 24.58 mL to 49.12 mL. The mass of the beads was determined to be 127.88 g. Determine the density of the metal. To solve the problem Mr. Henderson set up the equation density=mass/volume=127.88g/24.54mL. The volume was found by subtracting 49.12-24.58=24.54 mL. 127.88g/24.54mL=5.21108. Using the significant figures rule the answer is 5.211g per mL.

Once we finished the 2 problems Mr. Henderson went on to tell us that for our test on Thursday that we will have to know how to do the percent difference(difference between the 2 measures/average of the 2 measure x 100) and find the density of a substance(density=mass/volume).

Then it was time to start the lab for the day. Mr.Henderson explained that the purpose of the lab is to determine the percent sugar in each of the beverages. What percent of the beverage is sugar? The reason why we would want to know this is because sugar contributes to the mass. The beverages used are made mostly of water and sugar but also flavorings and dies.

Today we did the first part of the lab which was finding the mass and volume of water that had either 0%, 5%, 10%, 15% or 20% of sugar in it. These measurements would be used to determine the amount of sugar in the beverages which would be tested tomorrow. Mr. Henderson then reminded our class to wear our safety goggles and to dispose of the sugar water in the sink. Also he then explained how to use the pipets by turning the wheel.

In the lab, as expected, it was found that the more sugar the water contained, the higher its mass. The mass of the 0% sugar water was 4.97g, 5% was 5.03g, 10% was 5.07g, 15% was 5.16g, and the 20% was 5.25g. The volume of the 5 substances was a constant 5.00 mL.

This is a picture of my lab group’s data table from lab #8

This is a picture of the rubric which will used to grade the labs.