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3 years ago

One drawback of x-ray scattering experinment.

Chemistry

1 answer:

Kitty [74]3 years ago

5 0

Answer:

Rutherford passed beams of alpha particles through a thin gold foil to observe the atom and noted how the alpha particles scattered from the foil.

Observations of Rutherford's alpha ray scattering experiment:

1.Most of the α-particles passed straight through the gold foil without any deviation.

2. Some of the α-particles were deflected by the foil by some angles.

3.Interestingly one out of every 12,000 alpha particles appeared to rebound

Conclusion of Rutherford's scattering experiment:

1.Most of the space inside the atom is empty because most of the α-particles passed through the gold foil without getting deflected.

2. Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space.

3. A very small fraction of α-particles were deflected by very large angles, indicating that all the positive charge and mass of the gold atom were concentrated in a very small volume within the atom.

<h2 /><h2>Hopefully u will satisfy with my answer of ur question..!!</h2>

<h2>Mark on brainleast please..!!</h2>

<h2>Have a nice day ahead dear..!!</h2>

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3 years ago

Help. chemistry...........

nasty-shy [4]

<span>The </span>equilibrium<span> will </span>shift<span> to favor the side of the reaction that involves fewer moles of gas.
Its C

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3 years ago

Read 2 more answers

While ethanol (CH3CH2OH) is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by

grin007 [14]

Answer:

$n_{C2H_5OH}^{eq}=14.234mol$

Explanation:

Hello,

In this case, the reaction is:

$C_2H_4+H_2O\rightleftharpoons CH_3CH_2OH$

Thus, the law of mass action turns out:

$Kc=\frac{[CH_3CH_2OH]_{eq}}{[H_2O]_{eq}[CH_2CH_2]_{eq}}$

Thus, since at the beginning there are 29 moles of ethylene and once the equilibrium is reached, there are 16 moles of ethylene, the change $x$ result:

$[CH_2CH_2]_{eq}=29mol-x=16mol\\x=29-16=13mol$

In such a way, the equilibrium constant is then:

$Kc=\frac{\frac{x}{V} }{\frac{16mol}{V}* \frac{3mol}{V}} =\frac{\frac{13mol}{75.0L} }{\frac{16mol}{75.0L}* \frac{3mol}{75.0L}} =20.31$

Thereby, the initial moles for the second equilibrium are modified as shown on the denominator in the modified law of mass action by considering the added 15 moles of ethylene:

$Kc=\frac{\frac{13+x_2}{V} }{\frac{16+15-x_2}{V}* \frac{3-x_2}{V}} =20.31$

Thus, the second change, $x_2$ finally result (solving by solver or quadratic equation):

$x_2=1.234mol$

Finally, such second change equals the moles of ethanol after equilibrium based on the stoichiometry:

$n_{C2H_5OH}^{eq}=x+x_2=13mol+1.234mol\\n_{C2H_5OH}^{eq}=14.234mol$

Best regards.

5 0

3 years ago

One drawback of x-ray scattering experinment.​

One drawback of x-ray scattering experinment.