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3 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]3 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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Answer:

The answer to your question is below

Explanation:

Covalent bonds are bonds between to atoms that share a pair of electrons, there are three kinds of covalent bonds but I'll describe only two:

Covalent non polar bond: is a covalent bond between two elements of the same element. Ex two hydrogens, two chlorine, two oxygenes, etc.

Covalent polar bond: is a covalent bond between 2 elements of different elements, for example: hydrogen and chlorine or nitrogen, they are polar because on of the element that form it is smaller than the other one, then a partial positive and a partial negative charge is formed.

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3 years ago

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2 years ago

A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of

Bad White [126]

Answer:

a) $T_{2} = 360.955\,K$ , $P_{2} = 138569.171\,Pa\,(1.386\,bar)$ , b) $T_{2} = 347.348\,K$ , $V_{2} = 0.14\,m^{3}$

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

$\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}$

The number of moles of the ideal gas is:

$n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}$

$n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}$

$n = 5.541\,mol$

The final temperature is:

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}$

$T_{2} = 360.955\,K$

The final pressure is:

$P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}$

$P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)$

$P_{2} = 138569.171\,Pa\,(1.386\,bar)$

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

$\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}$

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}$

$T_{2} = 347.348\,K$

The final volume is:

$V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}$

$V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})$

$V_{2} = 0.14\,m^{3}$

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3 years ago