Answer:
(a) a = 5.08x10⁻⁸ cm
(b) r = 179.6 pm
Explanation:
(a) The lattice parameter "a" can be calculated using the following equation:
<em>where ρ: is the density of Th = 11.72 g/cm³, N° atoms/cell = 4, m: is the atomic weight of Th = 232 g/mol, Vc: is the unit cell volume = a³, and </em>
<em>: is the Avogadro constant = 6.023x10²³ atoms/mol. </em>
Hence the lattice parameter is:
![a = \sqrt[3]{1.32 \cdot 10^{-22} cm^{3}} = 5.08 \cdot 10^{-8} cm](https://tex.z-dn.net/?f=%20a%20%3D%20%5Csqrt%5B3%5D%7B1.32%20%5Ccdot%2010%5E%7B-22%7D%20cm%5E%7B3%7D%7D%20%3D%205.08%20%5Ccdot%2010%5E%7B-8%7D%20cm%20)
(b) We know that the lattice parameter of a FCC structure is:
<em>where r: is the atomic radius of Th</em>
Hence, the atomic radius of Th is:
I hope it helps you!
International Union of Pure and Applied Chemistry
They are examples of elements because there's only one molecule in both
Answer:
Therefore, the oxidation state of N in NaNO₂ is +3
Explanation:
Problem: calculating the oxidation state of nitrogen N in NaNO₂
let us denote the oxidation number of Nitrogen as N:
We know from the periodic table that Na has an oxidation number of +1 i.e it will readily want to lose 1 electron so as to complete its octet.
Oxygen is known to have an oxidation number of -2
Summation of the oxidation number of each atoms is 0 for neutral compound.
Therefore to calculate the oxidation state of Nitrogen in NaNO₂, we express as:
+1 + N + (-2 x 2) = 0
1 + N = 4
N = 4-1 = +3
Therefore, the oxidation state of N in NaNO₂ is +3
The early atmosphere of the earth would consist of gases like nitrogen and carbon dioxide and about 2.5 billion years ago oxygen started to form and some of these gases would have been expelled by volcanoes and also the atmosphere would stay close to the surface of the earth due to gravity that held it there.
