Answer:
Explanation:
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In this case, since the reaction for the formation of ammonia is:
We can evidence the 1:2 mole ratio of nitrogen gas to ammonia; therefore, the appropriate stoichiometric setup for the calculation of the moles of the latter turns out to be:
And the result is:
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Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
Explanation :
As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.
Let us assume that the radius of Cl⁻ be, (x) pm
So, the radius of Na⁺ =
In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.
Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.
Given:
Distance between Na⁺ nuclei = 566 pm
Thus, the relation will be:
The radius of Cl⁻ ion = (x) pm = 181 pm
The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm
Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.
The angular momentum of an electron in the third Bohr orbit of a hydrogen atom is given by mvr=3h÷2π
<h3>What is momentum?</h3>Momentum is defined as the amount of motion occurring in something that is moving, or the force that drives something forward to keep it moving.
Bohr never assumed stable electronic orbits with the electronic angular momentum quantized as
l=mvr =
Quantization of angular momentum means that the radius of the orbit and the energy will be quantized as well.
Bohr assumed that the discrete lines seen in the spectrum of the hydrogen atom were due to transitions of an electron from one allowed orbit/energy to another.
Learn more about momentum here:
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