Question 5.4.3: Additional Answer Options

2022 - November - Paper 2 - Question 5

Or

Option 2

1\mathrm{mol}=32\mathrm{g}

\mathrm{xmol}=0.9\mathrm{g}

x=0,0275\mathrm{~mol}

n\left(\mathrm{H}_2\mathrm{O}_2\right)=2\mathrm{n}\left(\mathrm{O}_2\right)

=2(0,0275)

=0,056

\text{ rate }/\text{ tempo }=\frac{\Delta\mathrm{n}}{\Delta\mathrm{t}}

2,1\times10^{-3}=\frac{0,056-0}{\Delta t}

=2(0,0275)

\Delta\mathrm{t}=26,67(\mathrm{~s})

Or

Option 3

n\left(O_2\right)=\frac{m}{M}

=\frac{0,9}{32}

=0,028\mathrm{~mol}(0,0281)

\operatorname{Rate}\left(\mathrm{O}_2\right)=1/2\text{ rate }\left(\mathrm{H}_2\mathrm{O}_2\right)

=1/2\left(2,1\times10^{-3}\right)

=1,05\times10^{-3}

\text{ rate }/\text{ tempo }=\frac{\Delta\mathrm{n}}{\Delta\mathrm{t}}

1,05\times10^{-3}\stackrel{}{=}\frac{0,028}{\Delta t}

\Delta t=26,67(\mathrm{~s})

Or

Option 4

\text{ Rate }\mathrm{H}_2\mathrm{O}_2=2,1\times10^{-3}\mathrm{~mol}\cdot\mathrm{s}^{-1}

\operatorname{Rate}\left(\mathrm{O}_2\right)=1/2\operatorname{rate}\left(\mathrm{H}_2\mathrm{O}_2\right)

=1/2\left(2,1\times10^{-3}\right)

=1,05\times10^{-3}

In one second:

\mathrm{n}\left(\mathrm{O}_2\right)=\frac{\mathrm{m}}{\mathrm{M}}

1,05\times10^{-3}=\frac{m}{32}

\mathrm{m}\left(\mathrm{O}_2\right)=0,0336\mathrm{~g}

\text{ rate }=0,0336\mathrm{~g}\cdot\mathrm{s}^{-1}

\text{ rate }=\frac{\Delta\mathrm{m}}{\Delta\mathrm{t}}

0,0336=\frac{0,9-0}{\Delta t}

\Delta\mathrm{t}=26,79(\mathrm{~s})

Related subjects & topics
Grade 12 Physical Sciences
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