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3 years ago

Determine the number of atoms in 3.62 mL of mercury. The density of mercury is 13.5g/mL

1 answer:

vladimir1956 [14]3 years ago

3 0

=3.62 * 13.5
then you'll get mass.
then devide by molar mass of Hg
then multiply by avegadro number.(L)

get the answer .
hope this would help u

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3 years ago

Write packing efficiency of fcc ,BCC, SCC and formula also . ?

Minchanka [31]

❖ Packing Efficiency ❖

➪ The percentage of total space occupied by particles is called packing efficiency.

$\\ \qquad{\rule{200pt}{3pt}}$

$\bigstar \boxed{ \sf \: \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}$

❖ Packing efficiency of simple cubic structure (SCC).❖

$\rm \: {Let \: } \bf{r } \: \rm{ be \: the \: radius \: of \: a \: sphere }\: \\ \rm and \: \bf{a} \: \rm {be \: the \: edge \: of \: unit \: cell.} \\ \\ \; \bigstar \boxed{ \sf{Volume_{(sphere)} = \dfrac{4}{3} \pi r^3}} \bigstar$

❒ Since, simple cubic unit cell contain 1 atom (sphere). So, the total volume of sphere will be :

$: : \implies \sf \: 1 \times \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi r^3$

Volume of unit cell = a³

Volume of unit cell = (2r)³

Volume of unit cell = 8r³

❒ Now, we know that,❒

$\bigstar \boxed{ \sf \: \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}$

➪ Substituting the known values in the formula, we get the following results:

$: : \implies \rm \: {\dfrac{\dfrac{4}{3}\pi r^3}{8r^3} \times 100} \\ \\ : : \implies \dfrac{\pi}{6} \times 100 \\\\\ : : \implies \bf {52.4 \%}$

❒ Hence, the packing efficiency of simple cubic structure is 52.4%.

$\\ \qquad{\rule{200pt}{3pt}}$

➪ Packing efficiency of cubic close packing (SCC)/face centred cubic structure (FCC).

❒ Since, simple cubic unit cell contain 2 atom (sphere). So, the total volume of sphere will be:

$: : \implies \rm 4 \times \dfrac{4}{3}\pi r^3 = \dfrac{16}{3}\pi r^3$

Volume of unit cell = a³

Volume of unit cell = (2√2r)³

Volume of unit cell = 16√2r³

❒ Now, we know that,❒

$\bigstar \boxed{ \sf \: \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}$

➪Substituting the known values in the formula, we get the following results:

$: : \implies \rm {\dfrac{\dfrac{16}{3}\pi r^3}{16\sqrt{2}r^3} \times 100 }\\\\ \implies \rm \dfrac{\pi}{3\sqrt{2}} \times 100 \\\\\ \implies\bf52.4 \%$

❖ Hence, the packing efficiency of face centred cubic structure is 74%.

$\\ \qquad{\rule{200pt}{3pt}}$

❖ Packing efficiency of body cubic structure (BCC).

❒ Since, simple cubic unit cell contain 2 atom (sphere). So, the total volume of sphere will be:

$: : \implies \rm \: 2 \times \dfrac{4}{3}\pi r^3 = \dfrac{8}{3}\pi r^3$

Volume of unit cell = a³

Volume of unit cell = (4r/√3)³

Volume of unit cell = 64r³/3√3

❒ Now, we know that,❒

$\bigstar \boxed{ \sf \: \bigg(Packing \; effeciency = \dfrac{Total \; value \; of \; sphere}{Volume \; of \; unit \; cell} \times 100 \bigg)}$

➪Substituting the known values in the formula, we get the following results:

$: : \implies \rm {\dfrac{\dfrac{8}{3}\pi r^3}{\dfrac{64r^3}{3\sqrt{3}}} \times 100} \\\\ \implies \dfrac{3\pi}{8} \times 100 \\\ \\ \bf \implies \: 68 \%$