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Ne4ueva [31]
3 years ago
7

Which atomic models does Rutherford’s experimental evidence support? Explain why these models (Dalton, Bohr, Thomson models) are

compatible with the experimental results.
Chemistry
2 answers:
ipn [44]3 years ago
5 0

Answer:

Rutherford was the first scientist who proposed the nuclear model of the atom. According to his atomic model, most of the space of an atom is empty, while the nucleus containing protons and neutrons lie at the center of the atom while electrons revolve around nucleus in definite orbits.

If we talk about studies of some other scientists like Dalton, Neil Bohr and JJ Thomson, they all are compatible with Rutherford's results to a large extent.

For example: Dalton's atomic model assumed that atoms of any substance are similar in size and atoms react to form compounds. Rutherford's concept indicated that atoms contain electrons and they are in a specific number which can be shared to form compounds.

If we talk about Bohr's model, it states that electrons revolve around nucleus in specific shells, this again is compatible with Rutherford's results which gave the concept of shells.

If we talk about Thomson's Plum pudding model, that describe atom as negative particles floating within a soup of diffuse positive charge. This is also compatible with the results of Rutherford that state that negative electrons surround positive nucleus.

Rutherford's model was best atomic model but still it took help from many previous studies and therefore was compatible with the results of old models.


Hope it help!

Gnesinka [82]3 years ago
5 0

Answer:

Rutherford was the first scientist who proposed the nuclear model of the atom. According to his atomic model, most of the space of an atom is empty, while the nucleus containing protons and neutrons lie at the center of the atom while electrons revolve around nucleus in definite orbits.

Explanation:

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Answer:

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Explanation:

Step 1: Given data

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Step 2: Write the dissociation equation

NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)

Step 3: Calculate the concentration of OH⁻

We will use the following expression.

[OH^{-} ]=\sqrt{Kb \times Cb } = \sqrt{1.8  \times 10^{-5} \times 0.300 } = 2.3 \times 10^{-3} M

Step 4: Calculate the pOH

We will use the following expression.

pOH =-log[OH^{-} ]= -log(2.3 \times 10^{-3} M) = 2.6

Step 5: Calculate the pH

We will use the following expression.

pH+pOH=14\\pH = 14-pOH = 14-2.6 = 11.4

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