W = AB x F x Cos < AB, F
or just W= AB x F for short
Answer:
Explanation:
We have the reactions:
A: 
B: 
Our <u>target reaction</u> is:
We have 
as a reactive in the target reaction and is present in A reaction but in the products side. So we have to<u> flip reaction A</u>.
A: 
Then if we add reactions A and B we can obtain the target reaction, so:
A:
B: 
For the <u>final Kc value</u>, we have to keep in mind that when we have to <u>add chemical reactions</u> the total Kc value would be the <u>multiplication</u> of the Kc values in the previous reactions.
Answer:
2.4 moles of oxygen are needed to react with 87 g of aluminium.
Explanation:
Chemical equation:
4Al(s) + 3O₂(l) → 2AlO₃(s)
Given data:
Mass of aluminium = 87 g
Moles of oxygen needed = ?
Solution:
Moles of aluminium:
Number of moles of aluminium= Mass/ molar mass
Number of moles of aluminium= 87 g/ 27 g/mol
Number of moles of aluminium= 3.2 mol
Now we will compare the moles of aluminium with oxygen.
Al : O₂
4 : 3
3.2 : 3/4×3.2 = 2.4 mol
2.4 moles of oxygen are needed to react with 87 g of aluminium.
<span> Ksp = [Ag+]^2[CO32-]that should be it </span>
