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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