Check Your Understanding

1. Change the velocity or direction of the
2. Acceleration is the rate of change of velocity. Deceleration is also a rate of change of velocity, but where the velocity is lowered.
3. You use the formula of final velocity minus initial velocity to find the rate of change of the velocity.
4. 20 m/s
5. They're both the same direction because acceleration is a change of velocity.
6.

Activity 9.1 B

1.


2.
a)   Δv = vf - vi -- 16 m/s f - 11 m/s f = +5 m/s
b) Δv = vf - vi -- 18 m/s f - 16 m/s f = +2 m/s
c) Δv = vf - vi -- 18 m/s f - 18 m/s f = 0 m/s
d) Δv = vf - vi -- 14 m/s f - 18 m/s f = -4 m/s
e) Δv = vf - vi -- 11 m/s f - 14 m/s f = -3 m/s

What did you find out?

It was gaining speed in the 0-20 second interval and the 20-40 second interval, it was slowing down in the 60-80 second and the 80-100 second intervals, and it was in velocity zero in the 40-60 second interval.

8.1 Check Your Understanding

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8.1 C and 8.1 D

8.1C


The line did not go through all the dots. This tells me that the lawnmower's motion accelerates and decelerates.


8.1 D Part1



Through 0-6 seconds, there is uniform motion. From 6 - 10 seconds, there is no movement. From 10-12 seconds, there is new uniform motion. Then from 12-16 seconds, there is negative uniform motion before becoming stationary from 16-18 seconds. It then concludes with negative uniform motion once again.

The displacements are:
2mN, 0m, 4mN, 6mS, 0m, 2mS

The slopes are:
Positive, zero, positive, negative, zero, and negative

The total distance is:
16m

8.1 D Part2



(a) The object starts at the origin and travels south
with uniform motion. - 4
(b) The object starts 2 m [S] and travels north
with uniform motion. - 2
(c) The object starts at the origin and travels north
with uniform motion. - 1
(d) The object starts 2 m [N] and travels south
with uniform motion. - 3

Reaction Rates Lab

Factors Affecting Reaction Rates

Part 1: -Two 250mL beakers -Ice water -Hot water -Thermometer -3 Effervescent Tablets -Stopwatch Part 2: -3 Small Test Tubes -Test Tube Rack -Dish Soap -3% Hydrogen Peroxide Solution -Pottasium Iodide -Copper(II) Chloride -Sodium Chloride Part 3: -Mortar and Pestle -Sodium Carbonate -4 Small Test Tubes -Test Tube Rack -Dilute Hydrochloric Acid Solution

Purpose: To determine how temperature, surface area, and the presence of a catalyst affect the rate of a chemical reaction.

Materials:  

               

Procedure:  The procedure used was from the BC Science 10 textbook on page 278

Results:

Part 1:

	Hot	Cold	Warm
Temperature	69­OC	6OC	32OC
Prediction	N/A	N/A	4-5 minutes
Time to dissolve the effervescent tablets	2 minutes 17 seconds	Stopped at 5 minutes (40% Dissolved)	Stopped at 5 minutes (70% Dissolve)

Part 2:

	KI	CuCl2	NaCl
Observations	Turned yellow and bubbled steadily. After 2 and a half minutes, it overflowed out of the test tube.	Turned blue and reacted quickly for about 30 seconds before settling half way up the test tub	Not very much happened in the reaction.

Part 3:

Less Powder	More Powder
Not a lot seemed to happen. When tested out with a stir stick, the powder was slightly solid.	Again, not a lot was going on. But it solidified quicker and stronger.

Discussion

Analyze:

-There is a very noticeable relationship between temperature and rate of reaction. The higher the temperature, the faster the molecules will move, and the more collisions you will get.

-My prediction was incorrect. I assumed that the warmth of the water would be enough to dissolve the tablet, but in fact, it did not do very much better than the cold water.

-I think that of the additives to the hydrogen peroxide solution, the KI, and CuCl­₂ worked well as catalysts. It was obvious to us though that the KI sped up the reaction the best as it continuously bubbled for well over 5 minutes.

-When you’re looking at surface area, a powder (in the case of the question 5g of sodium carbonate) would have more exposure to everything else than a lump. Powders will have thousands of faces exposed; lumps may only have a few.

-In step 3, the more exposed area to the reaction helped the decomposition out by speeding up the process in which it happened.

Conclude and Apply:

If there was a reaction with finely ground powder and a concentrated acid, there would be a few ways that you could slow it down. These include using a less concentrated acid and not using a powdered form of a reactant. Another way could be cooling down the temperature of the reactants prior to combining them for the reaction.

Conclusion

This lab proved that temperature, surface area, and catalysts are all successful in speeding up the process of a chemical reaction. Whether it’s temperature, surface area, concentration or a catalyst, it was made obvious to us that you can drastically increase or decrease the amount of time it can take to produce a product through a chemical reaction.

Reaction Rates Assignment

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Equations Problem Set

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5-3 A

https://my43.sd43.bc.ca/schools/Riverside/Classes/fbdc2b5b-182c-4090-953b-9d5a0b1bc809-defaultaspx/science10DI/Mr%20Taylors%20Documents/Organic%20Compounds%20Activity.pdf

1) Molecular Formula - CH4
    Structural Formula -











2) a) Ethane - C2H6, Propane C3H8
    b)




3)






4)









What Did you Find Out?

1)

a) 1
b) 1
c) 1
d) 2
e) 3

2)
The Hydrogen atoms are always the Carbon times 2 plus 2. This is because each "middle" carbon will have 2 Hydrogens on it, and the two end carbons will have those 2 plus an extra.

Balancing chemical Equations

1.          2H2 + O2 = 2H2O
2.          N2 + 3H2 = 2NH3
3.          S8 + 12O2 = 8SO3
4.          2N2 + O2 = 2N2O
5.          2HgO = 2Hg + O2
6.          6CO2 + 6H2O = C6H12O6 + 6O2
7.          Zn + 2HCl = ZnCl2 + H2
8.          SiCl4 + 4H2O = H4SiO4 + 4HCl
9.          2Na + 2H2O = 2NaOH + H2
10.      2H3PO4 = H4P2O7 + H2O
11.      C10H16 + 8Cl2 = 10C + 16HCl
12.      2CO2 + 4NH3 = OC(NH2)2 + 2H2O
13.      4Si2H3 + 17O2 = 8SiO2 + 6H2O3
14.      2Al(OH)3 + 3H2SO4 = Al2(SO4)3 + 6H2O
15.      4Fe + 3O2 = 2Fe2O3
16.      Fe2(SO4)3 + 6KOH = 3K2SO4 + 2Fe(OH)3
17.      2C7H6O2 + 15O2 = 14CO2 + 6H2O
18.      H2SO4 + 8HI = H2S + 4I2 + 4H2O
19.      4FeS2 + 11O2 = 2Fe2O3 + 8SO2
20.      2Al + 3FeO = Al2O3 + 3Fe
21.      1Fe2O3 + 3H2 = 2Fe + 3H2O
22.      Na2CO3 + 2HCl = 2NaCl + H2O + CO2
23.      2K + Br2 = 2KBr
24.      C7H16 + 11O2 = 7CO2 + 8H2O
25.      P4 + 5O2 = 2P2O5

26.      Dicarbon dihydride + Oxygen = Carbon dioxide + Water
2C2H2 + 5O2 = 4CO2 + 2H2O
27.      Potassium oxide + Water = Potassium hydroxide
K2O + H2O = 2KOH
28.      Hydrogen peroxide = Water + Oxygen
2H2O2 = 2H2O + O2
29.      Aluminum + Oxygen = Aluminum oxide
4Al + 3O2 = 2Al2O3
30.      Sodium peroxide + Water = Sodium hydroxide + oxygen
2Na2O2 + 2H2O = 4NaOH + O2
31.      Silicon dioxide + Hydrogen fluoride = Silicon tetrafluoride + Water
SiO2 + 4HF = SiF4 + 2H2O
32.      Carbon + waterè Carbon monoxide + Hydrogen
C + H2O à CO + H2
33.      Potassium chlorateè Potassium chloride + Oxygen
2KClO3 = 2KCl + 3O2
34.      Potassium chlorate =  Potassium perchlorate + Potassium chloride
4KClO3 = 3KClO4 + KCl
35.      Aluminum sulfate + Calcium hydroxide = Aluminum hydroxide + Calcium sulfate
Al2(SO4)3 + 3Ca(OH)2 = 2Al(OH)3 + 3CaSO4
36.      Tetraphosphorus decoxide + Waterè Hydrogen phosphate
P4O10 + 6H2O = 4H3PO4
37.      Iron III chloride + Ammonium hydroxide = Iron III hydroxide + Ammonium chloride
FeCl3 + 3NH4OH = Fe(OH)3 + 3NH4Cl
38.      Antimony + Oxygen = Tetrantimony Hexoxide
4Sb + 3O2 = Sb4O6
39.      Tricarbon octahydride + Oxygen = Carbon dioxide + water
C3H8 + 5O2 = 3CO2 + 4H2O
40.      Dinitrogen pentoxide + Water = Hydrogen nitrate
N2O5 + H2O = 2HNO3
41.      Nitrogen trihydride + Nitrogen monoxide = Nitrogen + Water
4NH3 + 6NO = 5N2 + 6H2O
42.      Aluminum + Hydrogen chloride = Aluminum chloride + Hydrogen
2Al + 6HCl = 2AlCl3 + 3H2
43.      Phosphorus pentachloride + water = Hydrogen chloride + Hydrogen phosphate
2PCl5 + 8H2O = 10HCl + 2H3PO4
44.      Magnesium + Nitrogenè Magnesium nitride
3Mg + N2 = Mg3N2
45.      Iron + Waterè Iron III oxide + Hydrogen
2Fe + 3H2O = Fe2O3 + 3H2
46.      Sodium hydroxide + Chlorine = Sodium chloride + Sodium hypochlorite + water
2NaOH + 2Cl = NaCl + NaClO + H2O
47.      Lithium oxide + Water = Lithium hydroxide
Li2O + H2O = 2LiOH
48.      Ammonium nitrate = Dinitrogen monoxide + water
2NH4NO3 = 2N2O + 4H2O
49.      Lead II nitrate = Lead II oxide + Nitrogen dioxide + Oxygen
2Pb(NO3)2 = 2PbO + 4NO2 + O2
50.      Calcium chlorate = Calcium chloride + Oxygen
Ca(ClO3)2 = CaCl2 + 3O2

Chapter 5 Review

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Combining Solutions of Ionic Compounds Lab Report

Combining Solutions of Ionic Compounds

Purpose:  To find out how the properties of acids and bases be used to classify solutions.

Materials: -4x6 spot plate

-Masking tape

-Solutions A,B,C, and D (some are corrosive)

-4 Pieces of Mg ribbon

-4 Pieces of red litmus paper

-4 Pieces of blue litmus paper

-Bromothymol blue solution

-Indigo carmine solution

-Methyl orange solution 

          

Procedure:  The procedure used was from the BC Science 10 textbook on page 230

Results:

Mg Ribbon	Red Litmus	Blue Ribbon	Bromothymol Blue	Indigo Carmine	Methyl Orange	Unknown
Began to bubble somewhat quickly, bubbles formed on the bottom.	Red paper, but the solution looks blue	Nothing	The solution turned yellow	The solution turned blue	The solution turned red	A
Nothing	The paper was blue	Nothing	The solution turned royal blue	The solution turned dark green	The solution turned yellow orange	B
A few small bubbles	The paper was red	Nothing	The solution turned green	The solution turned blue	The solution turned yellow orange	C
A few small bubbles	The paper was blue	Nothing	The solution turned light blue	The solution turned  blue	The solution turned yellow orange	D

Discussion

Analyze:

From most acidic to most basic I think is A,C,B,D.  I am not certain as I really am not confident with my results. Even with this, I believe that the neutral solution was C.  I believe that the most alkaline solution was solution B. My reasoning is that it was the only solution to not react at all with the Mg ribbon and it reacted with the liquid indicators as well. The ribbon was useful for testing the pH as acids would react with it to produce a salt and hydrogen gas. The gas then was able to bubble in the solution.

Conclude and Apply:

If there was a pH of 3 in a solution, the indicators would be red. If there was a pH of 10, the indicators would be blue. If I needed 3 tests to see if a sample was acidic, neutral, or basic, I would use red Litmus paper, blue litmus paper, and an alkaline earth metal. My reasoning is pretty obvious for the litmus paper as it would give a pretty instant reading. The metal would be good confirmation if the sample was acidic as it would react. For the colour of the Roccella tincoria being dipped in vinegar, my educated guess would be that it turns red. This is because vinegar is acidic. To find out the colour of seawater with Bromothymol blue added to it, I needed to find out the pH of seawater. I learned that is a little bit basic. And when Bromothymol blue reacts with bases, you get a dark blue colour.

Conclusion

This experiment was interesting. I think that the results we got however were off. They were different than the other group we worked with. However, we were able to clearly identify the acid thanks to the liquid indicators, litmus, and the magnesium.  Another problem we had was the beakers with solution were not properly labelled, so we weren’t 100% sure which solutions were A,B,C, and D.

Acids and Bases Poster

Lab Write Up

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Bohr vs Lewis Diagrams

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Science Lab Safety Poster

Made by Jacob.

Periodic Table Scavenger Hunt

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Building an Atom

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http://www.mediafire.com/view/?ydk2fxd7qbfnn01

WHMIS Symbols

Workplace Hazardous Materials Information System

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Scientific Method Worksheet

https://my43.sd43.bc.ca/schools/Riverside/Classes/fbdc2b5b-182c-4090-953b-9d5a0b1bc809-defaultaspx/science10DI/Mr%20Taylors%20Documents/Scientific%20Method%20WS.pdf


1) The scientists assumed that there was no life on Mars because the broths did not produce as much carbon dioxide as it would have on Earth.

2) The scientists got soil from different parts of the Martian surface so they would be able to not only compared results with what happens on Earth, but the other part of Mars as well.

3) Micro organisms were the characteristic of life being tested for in this experiment.

4) Scientists found the evidence from this experiment to be inconclusive. The only way to be sure that life exists on Mars would be to find living proof of it.