Answer:
how can I solve this ?4Al+3O2 produce 2Al2O3 find a) oxygen atoms needed to react with 5.4 g of aluminium b) grams of oxygen needed to react with 0.6 mol of aluminium?
(A) n=m/M,
n(Al)=5.4/27=0.2 moles
n(O2)=n(Al)*3/4=0.2*3/4=0.15 moles
Number of oxygen atoms= n(O2)*Avogadro's number
=0.15*6.02*10^23=9.03*10^22 oxgyen atoms
(B)
n=m/M
n(Al)=0.6/27=0.02222 moles
n(O2)=n(Al)*3/4=0.016666 moles
m=n*M
m(O2)=0.0166666*32=0.53333 grams
Answer: 5
Explanation: add up all the electrons and it will amount to 23. Arranging by the old model for electronic configuration, we have : 2, 8, 8, 5
The last number being 5 represent its valence electron
Answer:
<h3>601.93 g/mol</h3>
<h3>explanation:</h3>
Problem: The Ba3(PO4)2 (molar mass = 601.93 g/mol) precipitate that formed from a salt mixture has a mass of 0.667 g.
The head of a matchstick has a great deal of chemical energy stored in it, including combustible substances that produce a flame when rubbed against a suitable surface. ... As the combustible materials burn, some of the chemical energy is transformed into heat energy, and some is transformed into light energy. Hope this helps
