Question 9.4: Additional Answer Options

2021 - November - Paper 2 - Question 9

Or

Option 2

\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}

=\frac{2}{52}

=0,038\mathrm{~mol}\quad(0,04\mathrm{~mol})

=0,038\times6,02\times10^{23}

=2,315\times10^{22}

\text{ Number }\left(\mathrm{e}^{-}\right)=3\mathrm{~N}(\mathrm{Cr})

=3\left(2,315\times10^{22}\right)

=6,946\times10^{22}

Q=6,95\times10^{22}\times1,6\times10^{-19}

=11113,85\mathrm{C}

Or

Option 3

\mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}}

=\frac{2}{52}

=0,038\mathrm{~mol}

n\left(e^{-}\right)=3n(Cr)

=3(0,038)

=0,115\mathrm{~mol}

1\mathrm{mol}\ldots96500\mathrm{C}

0,115\mathrm{~mol}\quad11134,62\mathrm{C}

Related subjects & topics
Grade 12 Physical Sciences
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