The answer is 60.3% magnesium, 39.7% oxygen.
Solution:
The chemical equation for the reaction is 2 Mg + O2 → 2 MgO.
Since magnesium reacts completely with oxygen, it is the limiting reactant in the reaction. Hence, we can use the number of moles of magnesium to get the mass of MgO produced:
moles of magnesium = 14.7g / 24.305g mol-1
= 0.6048 mol
mass of MgO = 0.6048mol Mg(2 mol MgO/2mol Mg)(40.3044g MgO/1 mol MgO)
= 24.376g MgO
We can now solve for the percentage of magnesium:
% Mg = (14.7g Mg / 24.376g MgO)*100% = 60.3%
We also use the number of moles of magnesium to get the mass of oxygen consumed in the reaction:
mass of O2 = 0.6048 mol Mg (1mol O2 / 2mol Mg) (31.998g / 1mol O2)
= 9.676g
The percentage of oxygen is therefore
% O2 = (9.676g O2 / 24.376g MgO)*100%
= 39.7%
Notice that we can just subtract the magnesium's percentage from 100% to get
% O2 = 100% - 60.3% = 39.7%
Answer: Bromide is many orders of magnitude better than fluoride in leaving group ability
Explanation:
As Size of an atom Increases, the Basicity Decreases this is because if we move downwards from the top of the periodic table to the bottom of the periodic table, the size of an atom increases. As size increases, basicity will decrease, meaning the element will be less likely to act as a base implying that the element will be less likely to share its electrons.
in the same vein. With an increase in size, basicity decreases, making the ability of the leaving group to leave increase to increase . This can be seen in the halogens going down the group from
F--- worst
Cl----fair
Br ----good
I-----excellent
with fluorine having the worst ability to leave than Bromine which is better in terms of the leaving group ability.
