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Sagot :
Let's break down the solution step by step for the given questions.
### Part (g)
How long did it take for [tex]\(60 \, \text{cm}^3\)[/tex] of gas to be produced when the experiment was carried out using [tex]\(0.5 \, \text{g}\)[/tex] of the manganese(IV) oxide?
1. Given Data:
- Time (s): [tex]\(0, 30, 60, 90, 120, 150, 180, 210\)[/tex]
- Volume for [tex]\(0.5 \, \text{g} / \text{cm}^3\)[/tex] manganese(IV) oxide (cm³): [tex]\(0, 45, 84, 118, 145, 162, 174, 182\)[/tex]
2. Determine the relevant time interval:
- The target volume is [tex]\(60 \, \text{cm}^3\)[/tex].
- From the given data, we can see [tex]\(45\)[/tex] cm³ at [tex]\(30\)[/tex] seconds and [tex]\(84\)[/tex] cm³ at [tex]\(60\)[/tex] seconds.
3. Interpolate to find the exact time:
- The volume [tex]\(60 \, \text{cm}^3\)[/tex] lies between [tex]\(45 \, \text{cm}^3\)[/tex] (at [tex]\(30 \, \text{s}\)[/tex]) and [tex]\(84 \, \text{cm}^3\)[/tex] (at [tex]\(60 \, \text{s}\)[/tex]).
- We can use linear interpolation to find the exact time [tex]\(t\)[/tex] it took to reach [tex]\(60 \, \text{cm}^3\)[/tex]:
[tex]\[ \text{Time taken} = 30 + \left(\frac{60 - 45}{84 - 45}\right) \times (60 - 30) \][/tex]
4. Calculate the interpolated time:
[tex]\[ \text{Time taken} = 30 + \left(\frac{15}{39}\right) \times 30 \][/tex]
[tex]\[ \text{Time taken} = 30 + \left(\frac{15 \times 30}{39}\right) \][/tex]
[tex]\[ \text{Time taken} = 30 + \frac{450}{39} \][/tex]
[tex]\[ \text{Time taken} \approx 30 + 11.538 \][/tex]
[tex]\[ \text{Time taken} = 41.538 \, \text{s} \][/tex]
So, it took approximately [tex]\(41.538\)[/tex] seconds for [tex]\(60 \, \text{cm}^3\)[/tex] of gas to be produced with [tex]\(0.5 \, \text{g}\)[/tex] of manganese(IV) oxide.
### Part (h)
Write a balanced chemical equation for the decomposition of hydrogen peroxide.
The balanced chemical equation for the decomposition of hydrogen peroxide is:
[tex]\[ 2 \, \text{H}_2\text{O}_2 \rightarrow 2 \, \text{H}_2\text{O} + \text{O}_2 \][/tex]
In this reaction:
- [tex]\( \text{H}_2\text{O}_2 \)[/tex] represents hydrogen peroxide.
- [tex]\( \text{H}_2\text{O} \)[/tex] represents water.
- [tex]\( \text{O}_2 \)[/tex] represents oxygen gas.
### Part (g)
How long did it take for [tex]\(60 \, \text{cm}^3\)[/tex] of gas to be produced when the experiment was carried out using [tex]\(0.5 \, \text{g}\)[/tex] of the manganese(IV) oxide?
1. Given Data:
- Time (s): [tex]\(0, 30, 60, 90, 120, 150, 180, 210\)[/tex]
- Volume for [tex]\(0.5 \, \text{g} / \text{cm}^3\)[/tex] manganese(IV) oxide (cm³): [tex]\(0, 45, 84, 118, 145, 162, 174, 182\)[/tex]
2. Determine the relevant time interval:
- The target volume is [tex]\(60 \, \text{cm}^3\)[/tex].
- From the given data, we can see [tex]\(45\)[/tex] cm³ at [tex]\(30\)[/tex] seconds and [tex]\(84\)[/tex] cm³ at [tex]\(60\)[/tex] seconds.
3. Interpolate to find the exact time:
- The volume [tex]\(60 \, \text{cm}^3\)[/tex] lies between [tex]\(45 \, \text{cm}^3\)[/tex] (at [tex]\(30 \, \text{s}\)[/tex]) and [tex]\(84 \, \text{cm}^3\)[/tex] (at [tex]\(60 \, \text{s}\)[/tex]).
- We can use linear interpolation to find the exact time [tex]\(t\)[/tex] it took to reach [tex]\(60 \, \text{cm}^3\)[/tex]:
[tex]\[ \text{Time taken} = 30 + \left(\frac{60 - 45}{84 - 45}\right) \times (60 - 30) \][/tex]
4. Calculate the interpolated time:
[tex]\[ \text{Time taken} = 30 + \left(\frac{15}{39}\right) \times 30 \][/tex]
[tex]\[ \text{Time taken} = 30 + \left(\frac{15 \times 30}{39}\right) \][/tex]
[tex]\[ \text{Time taken} = 30 + \frac{450}{39} \][/tex]
[tex]\[ \text{Time taken} \approx 30 + 11.538 \][/tex]
[tex]\[ \text{Time taken} = 41.538 \, \text{s} \][/tex]
So, it took approximately [tex]\(41.538\)[/tex] seconds for [tex]\(60 \, \text{cm}^3\)[/tex] of gas to be produced with [tex]\(0.5 \, \text{g}\)[/tex] of manganese(IV) oxide.
### Part (h)
Write a balanced chemical equation for the decomposition of hydrogen peroxide.
The balanced chemical equation for the decomposition of hydrogen peroxide is:
[tex]\[ 2 \, \text{H}_2\text{O}_2 \rightarrow 2 \, \text{H}_2\text{O} + \text{O}_2 \][/tex]
In this reaction:
- [tex]\( \text{H}_2\text{O}_2 \)[/tex] represents hydrogen peroxide.
- [tex]\( \text{H}_2\text{O} \)[/tex] represents water.
- [tex]\( \text{O}_2 \)[/tex] represents oxygen gas.
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