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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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Evaporation!
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What is the oxidation number of nitrogen in nitrogen gas?
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During mitosis, this structure moves individual chromosomes?
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Anaphase

Explanation:

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3 years ago
Excess Ca(OH)2 is shaken with water to produce a saturated solution. The solution is filtered, and a 50.00 mL sample titrated wi
julia-pushkina [17]

<u>Answer:</u> The K_{sp} for calcium hydroxide is 5.324\times 10^{-6}

<u>Explanation:</u>

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is Ca(OH)_2

We are given:

n_1=1\\M_1=0.0983M\\V_1=11.22mL\\n_2=2\\M_2=?M\\V_2=50mL

Putting values in above equation, we get:

1\times 0.0983\times 11.22=2\times M_2\times 50\\\\M_2=0.011M

The concentration of Ca(OH)_2 comes out to be 0.011 M.

The balanced equilibrium reaction for the ionization of calcium hydroxide follows:

Ca(OH)_2\rightleftharpoons Ca^{2+}+2OH^-

The expression for solubility constant for this reaction follows:

K_{sp}=[Ca^{2+}][OH^-]^2

Putting the values in above equation, we get:

K_{sp}=(0.011)\times (2\times 0.11)^2

K_{sp}=5.324\times 10^{-6}

Hence, the K_{sp} for calcium hydroxide is 5.324\times 10^{-6}

6 0
3 years ago
Read 2 more answers
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Anika [276]

Answer:

That reminds me of the avengers logo

8 0
3 years ago
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