1) Calculate the number of moles of O2 (g) in 300 cm^3 of gas at 298 k and 1 atm
Ideal gas equation: pV = nRT => n = pV / RT
R = 0.0821 atm*liter/K*mol
V = 300 cm^3 = 0.300 liter
T = 298 K
p = 1 atm
=> n = 1 atm * 0.300 liter / [ (0.0821 atm*liter /K*mol) * 298K] = 0.01226 mol
2) The reaction of a metal with O2(g) to form an ionic compound (with O2- ions) is of the type
X (+) + O2 (g) ---> X2O or
2 X(2+) + O2(g) ----> X2O2 = 2XO or
4X(3+) + 3O2(g) ---> 2X2O3
In the first case, 1 mol of metal react with 1 mol of O2(g); in the second case, 2 moles of metal react with 1 mol of O2(g); in the third, 4 moles of X react with 3 moles of O2(g)
So, lets probe those 3 cases.
3) Case 1: 1 mol of metal X / 1 mol O2(g) = x moles / 0.01226 mol
=> x = 0.01226 moles of metal X
Now you can calculate the atomic mass of the hypotethical metal:
1.15 grams / 0.01226 mol = 93.8 g / mol
That does not correspond to any of the metal with valence 1+
So, now probe the case 2.
4) Case 2:
2moles X metal / 1 mol O2(g) = x / 0.01226 mol
=> x = 2 * 0.01226 = 0.02452 mol
And the atomic mass of the metal is: 1.15 g / 0.02452 mol = 46.9 g/mol
That is similar to the atomic mass of titanium which is 47.9 g / mol and whose valece is 2+.
4) Case 3
4 mol meta X / 3 mol O2 = x / 0.01226 => x = 0.01226 * 4 / 3 = 0.01635
atomic mass = 1.15 g / 0.01635 mol = 70.33 g/mol
That does not correspond to any metal.
Conclusion: the identity of the metallic element could be titanium.
The elements in each group have the same number of electrons in the outer orbital. Or also called valence electrons. Khan academy has a great video online explaining why this happens. (It only happens for main group elements). Here is a link (sorry you can’t click it in Brainly) https://www.khanacademy.org/science/chemistry/periodic-table/copy-of-periodic-table-of-elements/v/periodic-table-valence-electrons. Feel free to message me for a better explanation, I would explain now but I’m not sure how much you know about this. If you know how to write an electron configuration you can see how all the electron configurations for the same group (not the transitional metals only the main groups) have the same number of valence electrons. I hope that helped, sorry I was vague about the explanation :)
Answer:
0. 414
Explanation:
Octahedral interstitial lattice sites.
Octahedral interstitial lattice sites are in a plane parallel to the base plane between two compact planes and project to the center of an elementary triangle of the base plane.
The octahedral sites are located halfway between the two planes. They are vertical to the locations of the spheres of a possible plane. There are, therefore, as many octahedral sites as there are atoms in a compact network.
The Octahedral interstitial void ratio range is 0.414 to 0.732. Thus, the minimum cation-to-anion radius ratio for an octahedral interstitial lattice site is 0. 414.
One of the atoms involved must be a metal and the other a nonmetal
