Answer:
Niels Bohr, refined the model of an atom by proposing a quantized shell structure atomic model in order to describe how the electrons are able to maintain stable orbits around the nucleus
Based on the predictions of classical mechanics the electron motion of the Rutherford model was unstable as the electrons where expected to have lost some energy during motion and thus having to come rest in the nucleus
According to the modification by Neils Bohr in 1913, electrons move in shells or orbits of fixed energy and emission of electromagnetic radiation takes place only when electrons changes the orbit in which they move
Explanation:
Answer:
D. Grams liquid x mol/g x delta Hfreezing
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to reason that the stoichiometry used to calculate energy released when a mass of liquid freezes, involves the grams of the liquid, the molar mass of the liquid, as given in all the group choices, and the enthalpy of freezing because that is the process whereby a liquid goes solid.
In such a way, we infer that the correct factor would be D. Grams liquid x mol/g x delta Hfreezing which sometimes is the negative of the enthalpy of fusion as they are contrary processes.
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Answer:
2KClO3 —> 2KCl + 3O2
The coefficients are 2, 2, 3
Explanation:
From the question given above, we obtained the following equation:
KClO3 —> 2KCl + 3O2
The above equation can be balance as follow:
There are 2 atoms of K on the right side and 1 atom on the left side. It can be balance by putting 2 in front of KClO3 as shown below:
2KClO3 —> 2KCl + 3O2
Now, the equation is balanced.
Thus, the coefficients are 2, 2, 3
The molar concentration will be greater than 0.01 M 
.
Since more of the compound was measured out than what was calculated, you can think of the solution as being 'stronger' than what it was calculated to be. Since a 'stronger' concentration results in a number that is higher, the molarity of this solution is going to be greater than 0.01 M.
You should take note that the question is about stability. A compound is stable if it does not easily react with other elements. Hence, its reactivity must be low. As you move down the group, reactivity decreases. So, the halide at the very bottom is the least reactive. It would then be logical that the most stable conjugate base is I⁻ and the least stable conjugate base is the most reactive which is F⁻.
