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Sagot :
### Solution:
#### (a) Diagram of the Apparatus
To carry out the experiments for the decomposition of hydrogen peroxide using manganese(IV) oxide as a catalyst, you can use the following apparatus:
1. A conical flask to hold the hydrogen peroxide solution.
2. A measuring cylinder to measure the volume of gas produced.
3. A delivery tube to transfer the gas from the conical flask to the measuring cylinder.
4. A stopper with a hole to fit the delivery tube in the conical flask.
5. A stopwatch to record the time.
Diagram:
```
________
| | <- Conical Flask
Stopper -| |
| | <--- Hydrogen Peroxide solution with manganese(IV) oxide
| |
|________|
|
| Delivery tube
_________ /
| |/ <- Measuring Cylinder inverted in water to collect gas
|_________|
```
#### (b) Plot a Graph of the Results
To plot the graph, you will use the time (in seconds) on the x-axis and the volume of gas produced (in cm³) on the y-axis. You will have two plots on the same graph:
1. For 0.3 g/cm³ of manganese(IV) oxide:
[tex]\[ \{(0, 0), (30, 29), (60, 55), (90, 79), (120, 98), (150, 118), (180, 133), (210, 146)\} \][/tex]
2. For 0.5 g/cm³ of manganese(IV) oxide:
[tex]\[ \{(0, 0), (30, 45), (60, 84), (90, 118), (120, 145), (150, 162), (180, 174), (210, 182)\} \][/tex]
#### (c) Is the Manganese(IV) oxide acting as a catalyst in this reaction? Explain your answer. [2]
Yes, the manganese(IV) oxide is acting as a catalyst in this reaction. This can be inferred because the manganese(IV) oxide speeds up the decomposition of hydrogen peroxide without being consumed in the reaction. The results show that different amounts of manganese(IV) oxide affect the rate at which gas is produced, but the manganese(IV) oxide itself does not undergo any permanent chemical change.
#### (d)
##### (i) At which stage does the reaction proceed most quickly?
The reaction proceeds most quickly at the beginning of the experiment.
##### (ii) How can you tell this from your graph?
You can tell this from the graph by observing that the slope (rate of change) of the volume of gas produced is steepest at the beginning (initial segment) of the experiment. This steep slope indicates a high rate of reaction.
##### (iii) In terms of particles, explain why the reaction is quickest at the point you have chosen in (i).
In terms of particles, the reaction is quickest at the beginning because the concentration of hydrogen peroxide is highest at this point. More hydrogen peroxide molecules are available to collide with the surface of the manganese(IV) oxide catalyst, leading to more frequent collisions and thus a higher rate of reaction.
#### (e) Why does the slope of the graph become less steep as the reaction proceeds? [2]
The slope of the graph becomes less steep as the reaction proceeds because the concentration of hydrogen peroxide decreases over time. As the hydrogen peroxide gets decomposed into water and oxygen, there are fewer reactant molecules available to collide with the catalyst. This results in fewer effective collisions per unit time and thus a slower rate of reaction.
#### (f) What volume of gas has been produced when using 0.3 g of manganese(IV) oxide?
The volume of gas produced when using 0.3 g of manganese(IV) oxide is 146 cm³.
#### (a) Diagram of the Apparatus
To carry out the experiments for the decomposition of hydrogen peroxide using manganese(IV) oxide as a catalyst, you can use the following apparatus:
1. A conical flask to hold the hydrogen peroxide solution.
2. A measuring cylinder to measure the volume of gas produced.
3. A delivery tube to transfer the gas from the conical flask to the measuring cylinder.
4. A stopper with a hole to fit the delivery tube in the conical flask.
5. A stopwatch to record the time.
Diagram:
```
________
| | <- Conical Flask
Stopper -| |
| | <--- Hydrogen Peroxide solution with manganese(IV) oxide
| |
|________|
|
| Delivery tube
_________ /
| |/ <- Measuring Cylinder inverted in water to collect gas
|_________|
```
#### (b) Plot a Graph of the Results
To plot the graph, you will use the time (in seconds) on the x-axis and the volume of gas produced (in cm³) on the y-axis. You will have two plots on the same graph:
1. For 0.3 g/cm³ of manganese(IV) oxide:
[tex]\[ \{(0, 0), (30, 29), (60, 55), (90, 79), (120, 98), (150, 118), (180, 133), (210, 146)\} \][/tex]
2. For 0.5 g/cm³ of manganese(IV) oxide:
[tex]\[ \{(0, 0), (30, 45), (60, 84), (90, 118), (120, 145), (150, 162), (180, 174), (210, 182)\} \][/tex]
#### (c) Is the Manganese(IV) oxide acting as a catalyst in this reaction? Explain your answer. [2]
Yes, the manganese(IV) oxide is acting as a catalyst in this reaction. This can be inferred because the manganese(IV) oxide speeds up the decomposition of hydrogen peroxide without being consumed in the reaction. The results show that different amounts of manganese(IV) oxide affect the rate at which gas is produced, but the manganese(IV) oxide itself does not undergo any permanent chemical change.
#### (d)
##### (i) At which stage does the reaction proceed most quickly?
The reaction proceeds most quickly at the beginning of the experiment.
##### (ii) How can you tell this from your graph?
You can tell this from the graph by observing that the slope (rate of change) of the volume of gas produced is steepest at the beginning (initial segment) of the experiment. This steep slope indicates a high rate of reaction.
##### (iii) In terms of particles, explain why the reaction is quickest at the point you have chosen in (i).
In terms of particles, the reaction is quickest at the beginning because the concentration of hydrogen peroxide is highest at this point. More hydrogen peroxide molecules are available to collide with the surface of the manganese(IV) oxide catalyst, leading to more frequent collisions and thus a higher rate of reaction.
#### (e) Why does the slope of the graph become less steep as the reaction proceeds? [2]
The slope of the graph becomes less steep as the reaction proceeds because the concentration of hydrogen peroxide decreases over time. As the hydrogen peroxide gets decomposed into water and oxygen, there are fewer reactant molecules available to collide with the catalyst. This results in fewer effective collisions per unit time and thus a slower rate of reaction.
#### (f) What volume of gas has been produced when using 0.3 g of manganese(IV) oxide?
The volume of gas produced when using 0.3 g of manganese(IV) oxide is 146 cm³.
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