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4 years ago

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The energy of colored light increases in the order red, yellow, green, blue, violet.List the metallic elements used in the flame

tests in increasing order of the energy of the light emitted

1 answer:

Vinil7 [7]4 years ago

3 0

Answer:

Explanation:

Flame test:

The metals ions can be detected through the flame test. Different ions gives different colors when heated on flame. Tom perform the flame test following steps should follow:

1. Dip a wire loop in the solution of compound which is going to be tested.

2. After dipping put the loop of wire on bunsen burner flame.

3. Observe the color of flame.

4. Record the flame color produce by compound

Color produce by metals:

Red = Lithium, zirconium, strontium, mercury, Rubidium (red violet)

Orange-red = calcium

Yellow = sodium, iron (brownish yellow)

Green = green

Blue = cesium. arsenic, copper, tantalum, indium, lead

Violet = potassium (lilac)

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There are two isotopes of an unknown element, X-19 and X-21. The abundance of X-19 is 14.55%. A weighted average uses the percen

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Putting the value in expression for area,

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

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Since, $1 nm=10^{-7}cm$

Thus, $0.4961 nm=4.961\times 10^{-8} cm$

Putting the value,

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Putting the value,

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

now, number of atom in unit cell can be calculated by using the following formula:

$n=\frac{\rho N_{A}V_{c}}{A}$

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Putting all the values,

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Thus, there will be 18 $Cr_{2}O_{3}$ units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of $Cr^{3+}$ and $O^{2-}$ is 62 pm and 140 pm respectively.

Converting them into cm:

$1 pm=10^{-10}cm$

Thus,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

Volume of sphere will be sum of volume of total number of cations and anions thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, volume of sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Putting the values,

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The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, atomic packing factor is 0.758.

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