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Vlad1618 [11]
3 years ago
13

In 1911, Ernest Rutherford tested the atomic model existing at the time by shooting a beam of alpha particles (42He, helium nucl

ei) at a very thin sheet of gold foil. He found that while most particles went straight through the foil undeflected, a very few were deflected at great angles as they passed through the foil. Why was this discovery a reason to change the atomic model
Chemistry
1 answer:
STatiana [176]3 years ago
5 0

Answer:

At the time of Rutherford's experiment, the accepted model for the atom was the Thomson plum-pudding model of the atom, in which the atom consists of a "sphere" of positive charge distributed all over the sphere, with tiny negative particles (the electrons) inside this sphere.

In his experiment, Rutherford shot alpha particles towards a very thin sheet of gold foil. He observed the following things:

1- Most of the alpha particles went undeflected, but

2- Some of them were scattered at very large angles

3- A few of them were even reflected back to their original directions

Observations 2) and 3) were incompatible with Thomson model of the atom: in fact, if this model was true, all the alpha particle should have gone undeflected, or scattered at very small angles. Instead, due to observations 2) and 3), it was clear that:

- The positive charge of the atom was all concentred in a tiny nucleus

- Most of the mass of the atom was also concentrated in the nucleus

So, Rutherford experiment lead to a change in the atomic model of the atom, as it was clear that the plum-pudding model was no longer adequate to describe the results of Rutherford's experiment.

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Answer: A

Explanation:

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1 year ago
Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100 °C is ________ atm. (a
svetoff [14.1K]

Answer:

The answer to your question is        P = 1.357 atm

Explanation:

Data

Volume = 22.4 L

1 mol

temperature = 100°C

a = 0.211 L² atm

b = 0.0171 L/mol

R = 0.082 atmL/mol°K

Convert temperature to °K

Temperature = 100 + 273

                      = 373°K

Formula

               (P + \frac{a}{v^{2}} )(v - b) = RT

Substitution

               (P + \frac{0.211}{22.4})(22.4 - 0.0171) = (0.082)(373)

Simplify

               (P + 0.0094)(22.3829) = 30.586

Solve for P

                           P + 0.0094 = \frac{30.586}{22.3829}

                           P + 0.0094 = 1.366

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7 0
3 years ago
Calculate the pH of a solution created by placing 2.0 grams of yttrium hydroxide, Y(OH)3, in 2.0 L of H2O. Ksp for Y(OH)3 is 6.0
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Answer:

pH = 8.314

Explanation:

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equil:   S               S         3S

∴ Ksp = [ Y+ ] * [ OH- ]³ = 6.0 E-24

⇒ 6.0 E-24 = ( S )*( 3S )³

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⇒ pOH = - Log [ OH- ]

⇒ pOH = - Log ( 2.06 E-6 )

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What net ionic equation describes the reaction when these solutions are mixed?
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Hope this answers the question. Have a nice day.
7 0
3 years ago
Read 2 more answers
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