GRAPHING EXERCISES CHEMISTRY - Free Printable

Problem Analysis and Solution

The provided document describes an experiment where hydrogen peroxide (\( \text{H}_2\text{O}_2 \)) is decomposed using manganese dioxide (\( \text{MnO}_2 \)) as a catalyst. The reaction produces water (\( \text{H}_2\text{O} \)) and oxygen gas (\( \text{O}_2 \)). The data from the experiment shows the volume of \( \text{O}_2 \) produced in each test tube over five minutes, with varying masses of \( \text{MnO}_2 \).

We will solve each question step by step.

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1. What volume of \( \text{O}_2 \) did tube #3 produce between the second and fourth minutes?

#### Solution:
- From the table, for tube #3:
- Volume of \( \text{O}_2 \) at 2 minutes: \( 8.8 \, \text{ml} \)
- Volume of \( \text{O}_2 \) at 4 minutes: \( 10.2 \, \text{ml} \)
- The volume of \( \text{O}_2 \) produced between the second and fourth minutes is the difference:
\[
\text{Volume produced} = 10.2 \, \text{ml} - 8.8 \, \text{ml} = 1.4 \, \text{ml}
\]

#### Answer:
\[
\boxed{1.4 \, \text{ml}}
\]

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2. How much \( \text{O}_2 \) is produced in tube #5 during the first two minutes?

#### Solution:
- From the table, for tube #5:
- Volume of \( \text{O}_2 \) at 1 minute: \( 12.4 \, \text{ml} \)
- Volume of \( \text{O}_2 \) at 2 minutes: \( 14.4 \, \text{ml} \)
- The total volume of \( \text{O}_2 \) produced in the first two minutes is the sum of the volumes at 1 minute and the additional volume produced in the second minute:
\[
\text{Total volume} = 12.4 \, \text{ml} + (14.4 \, \text{ml} - 12.4 \, \text{ml}) = 14.4 \, \text{ml}
\]

#### Answer:
\[
\boxed{14.4 \, \text{ml}}
\]

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3. How much oxygen did tube #7 and #8 produce together during the third minute?

#### Solution:
- From the table:
- For tube #7 at 3 minutes: \( 20.2 \, \text{ml} \)
- For tube #8 at 3 minutes: \( 22.1 \, \text{ml} \)
- The total volume of \( \text{O}_2 \) produced by both tubes during the third minute is:
\[
\text{Total volume} = 20.2 \, \text{ml} + 22.1 \, \text{ml} = 42.3 \, \text{ml}
\]

#### Answer:
\[
\boxed{42.3 \, \text{ml}}
\]

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4. What volume of oxygen gas, in litres, was produced during this procedure?

#### Solution:
- To find the total volume of \( \text{O}_2 \) produced, we need to sum the volumes produced by all nine tubes at the end of 5 minutes.
- From the table, the volumes at 5 minutes are:
\[
\begin{aligned}
&\text{Tube 1: } 5.1 \, \text{ml}, \\
&\text{Tube 2: } 7.6 \, \text{ml}, \\
&\text{Tube 3: } 11.3 \, \text{ml}, \\
&\text{Tube 4: } 13.3 \, \text{ml}, \\
&\text{Tube 5: } 17.1 \, \text{ml}, \\
&\text{Tube 6: } 21.8 \, \text{ml}, \\
&\text{Tube 7: } 24.8 \, \text{ml}, \\
&\text{Tube 8: } 27.3 \, \text{ml}, \\
&\text{Tube 9: } 30.4 \, \text{ml}.
\end{aligned}
\]
- Summing these volumes:
\[
\text{Total volume} = 5.1 + 7.6 + 11.3 + 13.3 + 17.1 + 21.8 + 24.8 + 27.3 + 30.4 = 157.7 \, \text{ml}
\]
- Convert milliliters to liters:
\[
157.7 \, \text{ml} = 0.1577 \, \text{L}
\]

#### Answer:
\[
\boxed{0.1577 \, \text{L}}
\]

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5. Graph the amount of oxygen produced each minute in tubes #2, #4, #6.

#### Solution:
- This requires plotting the data for tubes #2, #4, and #6 over the five-minute period. The data points are:
- Tube #2:
\[
\begin{aligned}
&\text{1 min: } 2.8 \, \text{ml}, \\
&\text{2 min: } 4.6 \, \text{ml}, \\
&\text{3 min: } 7.1 \, \text{ml}, \\
&\text{4 min: } 7.1 \, \text{ml}, \\
&\text{5 min: } 7.6 \, \text{ml}.
\end{aligned}
\]
- Tube #4:
\[
\begin{aligned}
&\text{1 min: } 5.9 \, \text{ml}, \\
&\text{2 min: } 8.5 \, \text{ml}, \\
&\text{3 min: } 10.4 \, \text{ml}, \\
&\text{4 min: } 11.8 \, \text{ml}, \\
&\text{5 min: } 13.3 \, \text{ml}.
\end{aligned}
\]
- Tube #6:
\[
\begin{aligned}
&\text{1 min: } 11.0 \, \text{ml}, \\
&\text{2 min: } 14.8 \, \text{ml}, \\
&\text{3 min: } 17.5 \, \text{ml}, \\
&\text{4 min: } 19.8 \, \text{ml}, \\
&\text{5 min: } 21.8 \, \text{ml}.
\end{aligned}
\]
- Plot these points on a graph with time (minutes) on the x-axis and volume of \( \text{O}_2 \) (ml) on the y-axis.

#### Answer:
Graphing is visual and cannot be represented textually. However, the data points are provided above for plotting.

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6. By comparing the slope of the graph curves, which tube was producing oxygen at the fastest rate between minutes four and five?

#### Solution:
- The rate of oxygen production is given by the slope of the graph between minutes 4 and 5. We calculate the change in volume for each tube:
- Tube #2:
\[
\text{Slope} = \frac{\text{Volume at 5 min} - \text{Volume at 4 min}}{5 - 4} = \frac{7.6 - 7.1}{1} = 0.5 \, \text{ml/min}
\]
- Tube #4:
\[
\text{Slope} = \frac{\text{Volume at 5 min} - \text{Volume at 4 min}}{5 - 4} = \frac{13.3 - 11.8}{1} = 1.5 \, \text{ml/min}
\]
- Tube #6:
\[
\text{Slope} = \frac{\text{Volume at 5 min} - \text{Volume at 4 min}}{5 - 4} = \frac{21.8 - 19.8}{1} = 2.0 \, \text{ml/min}
\]
- The highest slope corresponds to the fastest rate of oxygen production. Tube #6 has the highest slope (\( 2.0 \, \text{ml/min} \)).

#### Answer:
\[
\boxed{\text{Tube #6}}
\]

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7. Make a graph using the mass of manganese dioxide and the volume of oxygen for all tubes at five minutes.

#### Solution:
- This requires plotting the mass of \( \text{MnO}_2 \) (in grams) on the x-axis and the volume of \( \text{O}_2 \) produced at 5 minutes (in ml) on the y-axis. The data points are:
\[
\begin{aligned}
&\text{Mass of } \text{MnO}_2: 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9 \, \text{g}, \\
&\text{Volume of } \text{O}_2 \text{ at 5 min}: 5.1, 7.6, 11.3, 13.3, 17.1, 21.8, 24.8, 27.3, 30.4 \, \text{ml}.
\end{aligned}
\]
- Plot these points on a graph.

#### Answer:
Graphing is visual and cannot be represented textually. However, the data points are provided above for plotting.

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Final Answers:

1. \(\boxed{1.4 \, \text{ml}}\)
2. \(\boxed{14.4 \, \text{ml}}\)
3. \(\boxed{42.3 \, \text{ml}}\)
4. \(\boxed{0.1577 \, \text{L}}\)
5. Graph the data points for tubes #2, #4, and #6.
6. \(\boxed{\text{Tube #6}}\)
7. Graph the mass of \( \text{MnO}_2 \) vs. the volume of \( \text{O}_2 \) at 5 minutes.