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Len [333]
3 years ago
8

Which of the following statements are TRUE concerning this laboratory exercise - The systematic identification of an organic com

pound ?
True False Your compound , when you receive it , may be impure , ie. a component of a mixture.
True False Your unknown may be a solid, liquid, or gas.
True False Your compound will contain at least C , H and oxygen and one or more of several other possible elements such as N , one or more halogens, S and Se.
True False In a "real world" situation, the compound could be one of several million possible known organic compounds.
True False You can use spectroscopic results to indicate the appropriate chemical tests to perform.
Chemistry
1 answer:
emmainna [20.7K]3 years ago
5 0

Answer:

True;

True;

False;

True;

True.

Explanation:

Organic compounds can be found pure, but generally, they are in mixtures, such as the petroleum, so it is possible to the compound received be impure;

Organic compounds can be found in the three states: solid, liquid or gas. The state will depend on the molecular forces in it and its molar mass. For example, gasoline is liquid in the ambient temperature, natural gas is a gas, and the glucose is solid;

The main characteristic of the organic compounds is the presence of C and H, the other elements can or can't be in the structure: O, N, halogens, and S;

Because of the carbon can form chains, there are several million possible known organic compounds;

The spectroscopic would give some results such as composition, molar mass, and diffraction, and with these results, it would be possible to identify some chemical properties and the tests needs more identification.

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The energy range expected is 6.6 × 10^-19 J < E < 7.33 × 10^-19 J

The energy of the photon is given by;

E = hc/λ

E = energy of the photon

h = Plank's constant

c = speed of light

λ = wavelength of light

For the upper boundary range;

E = ?

h = 6.6 × 10^-34 Js

c = 3  × 10^8 m/s

λ = 270 × 10^-9

E =  6.6 × 10^-34 Js × 3  × 10^8 m/s / 270 × 10^-9

E = 7.33 × 10^-19 J

For the lower range;

E = ?

h = 6.6 × 10^-34 Js

c = 3  × 10^8 m/s

λ =300 × 10^-9

E =  6.6 × 10^-34 Js × 3  × 10^8 m/s / 300 × 10^-9

E = 6.6 × 10^-19 J

Hence, the energy range 6.6 × 10^-19 J < E <  7.33 × 10^-19 J

Learn more: brainly.com/question/24857760

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PLEASE GIVE ME A PIC FOR THIS ANSWER

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Answer:

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