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EastWind [94]
3 years ago
7

How many different flame colors were you able to detect? How many different elements must have been exposed to the flame to prod

uce this number of colors?
Chemistry
1 answer:
Bogdan [553]3 years ago
5 0

Answer:

This question is incomplete

Explanation:

This question is incomplete, however, there are some basic things that be used to answer the completed question on your own. Flame test is a test that is used to identify metal ions in a compound. Although, not all metal ions produce a colour in a flame test.

In a flame test, a "clean wire loop" is dipped in an unknown solid/mixture of solids, the loop where the solids must have attached to is then placed in the tip of a blue flame (perhaps of a bunsen burner). A colour change/changes is then observed during the course of this process. Some popular metal ions and there colour in flame test are listed below

Lithium ion ⇒ red

Sodium ion ⇒ yellow

Potassium ion ⇒ lilac

Calcium ion ⇒ orange-red

Barium ion ⇒ pale-green

Copper ion ⇒ blue-green

rubidium ion ⇒ red-violet

Lead ion ⇒ gray white

The number of different colours observed will ultimately determine the number of elements exposed to the flame

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If a sample of oxygen occupies a volume of 85.0 L at a pressure of 67.4 kPa and a temperature of 245K , what volume would the ga
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Answer: V2 = 35.54L

Explanation:

Applying

P1= 67.4, V1= 85, T1= 245, P2= 179.6, V2= ?,. T2=273

P1V1/ T1= P2V2/T2

Substitute and simplify

(67.4*85)/245 = (179.6*V2)/273

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3 years ago
The surface area of an object to be gold plated is 49.8 cm2 and the density of gold is 19.3 g/cm-3. A current of 3.15. A is appl
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Answer: Time required to deposit an even layer of gold with given thickness is 5.3 \times 10^{2} sec.

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The given data is as follows.

     Surface area = 49.8 cm^{2},

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     Current = 3.15 A,       thickness of gold layer = 1.2 \times 10^{-3} cm

It is known that relation between volume, area and thickness is as follows.

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               = 49.8 \times 1.2 \times 10^{-3} cm

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Therefore, we will calculate the time required to deposit an even layer of gold with given thickness is calculated as follows.

  0.05988 \times cm^{3} \times \frac{19.3 g Au}{1 cm^{3}} \times \frac{1 mol Au}{197 g Au} \times \frac{3 mol e^{-}}{1 mol Au} \times \frac{96485}{1 mol e^{-}} \times \frac{1 As}{1 C} \times \frac{1}{3.20 A}

        = 5.3 \times 10^{2} sec

Thus, we can conclude that time required to deposit an even layer of gold with given thickness is 5.3 \times 10^{2} sec.

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