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Dafna1 [17]
3 years ago
8

Which atomic model proposed that electrons move in specific orbits around the nucleus of an atom?

Chemistry
2 answers:
Tatiana [17]3 years ago
8 0

<u>Answer:</u> The correct answer is Bohr's atomic model.

<u>Explanation:</u>

For the given options:

<u>Dalton's atomic model:</u> This model states that every matter is made up of smallest unit known as atom.

<u>Thomson's atomic model:</u> He proposed a model known as plum pudding model. He considered atom to be a pudding of positive charge in which negative particles are embedded such as plum.

<u>Rutherford's atomic model:</u> He gave an experiment known as gold foil experiment. In his model, he concluded that in an atom, there exist a small positive charge in the center.

<u>Bohr's atomic model:</u> This model states that electron revolve around the nucleus in discrete orbits in an atom.

<u>Quantum atomic model:</u> This model determines the location of electrons in an atom in a 3-D space.

Hence, the correct answer is Bohr's atomic model.

Brilliant_brown [7]3 years ago
4 0
Hi there ,
The Bohre's atomic model represents movement of electrons in specific orbit around the nucleus of an atom.
Hope it helps.
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For the reaction H2(g) + 12(g) = 2HI(g), Kc = 50.2 at 445°C. If
Studentka2010 [4]

Qc < Kc, the reaction proceeds from left to right to reach equilibrium

<h3>Further explanation </h3>

Given

K = 50.2 at 445°C

[H2] = [I2] = [HI] = 1.75 × 10⁻³ M At 445ºC

Reaction

H2(g) + I2(g) ⇔2HI(g)

Required

Qc

Solution

Qc for the reaction

\tt Qc=\dfrac{[HI]^2}{[I_2][H_2]}\\\\Qc=\dfrac{(1.75.10^{-3})^2}{1.75.10^{-3}\times 1.75\.10^{-3}}=1  

Qc < Kc ⇒ reaction from left(reactants) to right (products) (the reaction will shift on the right) until it reaches equilibrium (Qc = Kc)  

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Before I can answer your question we need to know the different types of chemical reactions that can occur:

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Combustion is an answer that can usually be ruled out unless oxidation is occuring in the chemical equation (hyrdrogen occuring with oxygen to react)

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Hope this helps and if you need anymore clarification feel free to ask.


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