### 1. Balanced Reaction for the Decomposition of Hydrogen Peroxide with Catalase
The decomposition of hydrogen peroxide ([tex]\( \text{H}_2\text{O}_2 \)[/tex]) into water ([tex]\( \text{H}_2\text{O} \)[/tex]) and oxygen gas ([tex]\( \text{O}_2 \)[/tex]) in the presence of the catalase enzyme as a catalyst can be represented by the following balanced chemical equation:
[tex]\[ \text{2 H}_2\text{O}_2 \,(\text{aq}) \rightarrow \text{2 H}_2\text{O} \,(\text{l}) + \text{O}_2 \,(\text{g}) \][/tex]
This equation indicates that two molecules of hydrogen peroxide decompose to produce two molecules of water and one molecule of oxygen gas.
### 2. Calculation of [tex]\( \Delta V \)[/tex] for Trial 5
To find the change in volume of the gas ([tex]\( \Delta V \)[/tex]) produced during the reaction in Trial 5, we need to compute the difference between the final volume and the initial volume of gas produced. The steps for the calculation are as follows:
1. Initial Volume of Gas for Trial 5:
[tex]\[
\text{Initial Volume} = 0 \, \text{mL}
\][/tex]
2. Final Volume of Gas for Trial 5:
[tex]\[
\text{Final Volume} = 10 \, \text{mL}
\][/tex]
3. Change in Volume ([tex]\( \Delta V \)[/tex]):
[tex]\[
\Delta V = \text{Final Volume} - \text{Initial Volume}
\][/tex]
[tex]\[
\Delta V = 10 \, \text{mL} - 0 \, \text{mL} = 10 \, \text{mL}
\][/tex]
Thus, the change in volume of gas ([tex]\( \Delta V \)[/tex]) for Trial 5 is [tex]\( 10 \, \text{mL} \)[/tex].
