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ioda
3 years ago
13

The complete combustion of 1.00 kg of coal releases about 3.0 x 10E7 J of energy. The conversion of 1.00 kg of mass into energy

is equivalent to the burning of how many metric tonnes of coal?
Chemistry
2 answers:
zalisa [80]3 years ago
3 0

Answer: 0.001 metric tonnes

Explanation:

According to Einstein equation, Energy and mass are inter convertible.

E=mc^2

E=  Energy

m = mass

c= speed of light

Given : Complete combustion of 1.00 kg of coal releases about 3.0\times 10^7 J of energy.

Given mass: 1 kg

Converting kg to metric ton using the conversion factor:

1 kg=0.001 metric tonnes

Thus 1 kg of coal would also be equal to 0.001 metric tonnes.

yKpoI14uk [10]3 years ago
3 0

Answer:

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Explanation:

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A 1.525g sample of a compound between nitrogen and hydrogen contains 1.333 g of nitrogen. Calculate its empirical formula. The e
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Which of these elements has two electrons in its outer p orbitals?
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Read 2 more answers
Write packing efficiency of fcc ,BCC, SCC and formula also . ?​
Minchanka [31]

❖ <u>Packing Efficiency</u> ❖

➪ The percentage of total space occupied by particles is called <u>packing efficiency</u>.

\\ \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

❒ <u>Now, we know that</u>,❒

\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).

\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 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

❒ <u>Now, we know that</u>,❒

\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).

\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 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

❒ <u>Now, we know that</u>,❒

\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 \%

❒ Hence, the packing efficiency of body centred cubic structure is 68%.

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

6 0
2 years ago
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