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kolbaska11 [484]
3 years ago
12

The products obtained by cracking an alkane, X, are methane, ethene and propene. The mole fraction of ethene in the products is

0.5. What is the identity of X?. A C6H14. B C8H18. C C9H20. D C11H24.
Chemistry
1 answer:
MrMuchimi3 years ago
6 0

<span>The identity of X based from the given problem is C9H20. </span>The correct answer between all the choices given is the third choice or letter C. I am hoping that this answer has satisfied your query about this specific question.

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How much mass defect is required to release 5.5 x 1020 J of energy?
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Answer : The mass defect required to release energy is 6111.111 kg

Explanation :

To calculate the mass defect for given energy released, we use Einstein's equation:

E=\Delta mc^2

E = Energy released = 5.5\times 10^{20}J

\Delta m = mass change = ?

c = speed of light = 3\times 10^8m/s

Now put all the given values in above equation, we get:

5.5\times 10^{20}Kgm^2/s^2=\Delta m\times (3\times 10^8m/s)^2

\Delta m=6111.111kg

Therefore, the mass defect required to release energy is 6111.111 kg

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Which subatomic particle gives off visible light when it drops back down to a lower energy state?
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Answer:

Option C = electron

Explanation:

Electrons are responsible for the production of colored light.

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When the energy is provided to the atom the electrons by absorbing the energy jump to the higher energy levels. This process is called excitation. The amount of energy absorbed by the electron is exactly equal to the energy difference of orbits.

De-excitation:

When the excited electron fall back to the lower energy levels the energy is released in the form of radiations. this energy is exactly equal to the energy difference between the orbits. The characteristics bright colors are due to the these emitted radiations. These emitted radiations can be seen if they are fall in the visible region of spectrum.

Other process may involve,

Fluorescence:

In fluorescence the energy is absorbed by the electron having shorter wavelength and high energy usually of U.V region. The process of absorbing the light occur in a very short period of time i.e. 10 ∧-15 sec. During the fluorescence the spin of electron not changed.

The electron is then de-excited by emitting the light in visible and IR region. This process of de-excitation occur in a time period of 10∧-9 sec.

Phosphorescence:

In phosphorescence the electron also goes to the excitation to the higher level by absorbing the U.V radiations. In case of Phosphorescence the transition back to the lower energy level occur very slowly and the spin pf electron also change.

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Answer:

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