Answer:
3
Explanation:
If oxygen reacts with iron, then both must be reactants and rust the product of that reaction
Answer:
Magnesium is a naturally ubiquitous; (appearing & found evrywhere) element and has three naturally occurring stable isotopes, 24Mg, 25Mg and 26Mg, with relative abundance of 78.99%, 10.00% and 11.01%, respectively.
However, they differ only because a 24Mg atom has 12 neutrons in its nucleus, a 25Mg atom has 13 neutrons, and a 26Mg has 14 neutrons.
Explanation:
4.20 mol Al would react completely with 4.20 x (1/2) = 2.10 mol Fe2O3, but there is not that much Fe2O3 present, so Fe2O3 is the limiting reactant. (1.75 mol Fe2O3) x (2/1) x ( 55.8452 g Fe/mol) = 195 g Fe 3 MgO + 2 H3PO4 → Mg3(PO4)2 + 3 H2O (15.0 g MgO) / (40.3045 g MgO/mol) = 0.37217 mol MgO (18.5 g H3PO4) / (97.9953 g H3PO4/mol) = 0.18878 mol H3PO4 0.18878 mol H3PO4 would react completely with 0.18878 x (3/2) = 0.28317 mole of MgO, but there is more MgO present than that, so MgO is in excess and H3PO4 is the limiting reactant. Now we must consider why the problem tells us "17.6g of Mg3(PO4)2 is obtained". The first possibility is that it's just there for the sake of confusion -- in which case ignore it and proceed this way: ((0.37217 mol MgO initially) - (0.28317 mole MgO reacted)) x (40.3045 g MgO/mol) = 3.59 g MgO left over However, if the amount of magnesium phosphate obtained is given because the reaction was stopped before it was complete, the amount obtained governs the amount reacted and the amount left over, so proceed this way: (17.6g Mg3(PO4)2) / (262.8581 g Mg3(PO4)2/mol) x (3/1) = 0.20087 mol MgO reacted ((0.37217 mol MgO initially) - (0.20087 mole MgO reacted)) x (40.3045 g MgO/mol) = 6.90 MgO left over
Answer :
(A) The number of moles of 
ions per liter is, 0.1 moles/L
(B) The number of molecules of 
ion is, 
(C) The pH of the solution will be, 4
<u>Solution for part A :</u>
First we have to calculate the pOH of the solution.
As we know that,
Now we have to calculate the moles of ion per liter.
![pOH=-\log [OH^-]\\\\1=-\log [OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%20%5BOH%5E-%5D%5C%5C%5C%5C1%3D-%5Clog%20%5BOH%5E-%5D)
![[OH^-]=0.1moles/L](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.1moles%2FL)
<u>Solution for part B :</u>
First we have to calculate the ion concentration.
![pH=-\log [H^+]\\\\13=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D%5C%5C%5C%5C13%3D-%5Clog%20%5BH%5E%2B%5D)
![[H^+]=10^{-13}moles/L](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-13%7Dmoles%2FL)
Now we have to calculate the number of molecules of ion
As, 1 mole contains 
number of molecules of ion
So, 
moles contains 
number of molecules of ion
<u>Solution for part C :</u>
![pH=-\log [H^+]\\\\pH=-\log (1\times 10^{-4})](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D%5C%5C%5C%5CpH%3D-%5Clog%20%281%5Ctimes%2010%5E%7B-4%7D%29)
