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%
Because Na has one too many electron to be stable, while Mg has 2. once they loose those electrons, they will be come ion. loosing one electron will give the ion a 1+ charge, and loosing 2 will create a 2+ charge
Answer:
It is mostly water (up to 95% by volume), and contains dissolved proteins (6–8%) (e.g. serum albumins, globulins, and fibrinogen), glucose, clotting factors, electrolytes (Na+, Ca2+, Mg2+, HCO3−, Cl−, etc.), hormones, carbon dioxide (plasma being the main medium for excretory product transportation) and oxygen.
hope this helps :)
can i pls have a brainilist!
Explanation:
Because molarity is classified as moles of solute per liter of water, dilution of the water may result in a reduction of its concentration.
Therefore, because the amount of moles of solute has to be constant for dilution, you will use the molarity and volume of that same target solution to calculate how many moles of solute will be present in the sample of the stock solution that you dilute.
c = 
⇒ n
=
c
⋅ V
= 0.250 M ⋅ 6.00 L =
1.5 moles HCl
Now all you have to do is figure out what volume of 6.0 M stock solution will contain 1.5 moles of hydrochloric acid
c =
V = 
= 
=
0.25 L
Expressed in milliliters, the answer will be
→ rounded to two sig figs
The -COOH is bonded to a long chain of hydrocarbons in a fatty acid.
A hydrocarbon chain is comprised of both carbon (C) and hydrogen (H) atoms. The 'acid' part is made up of the -COOH portion, whereas the 'fatty' part is made up of the long hydrocarbon chain that it is attached to.
