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:
Total number of heat absorbed is 4.08kJ
Explanation:
Explanation is contained in the picture attached
Answer:
Empirical formula is C3H3O
Molecular formula C9H9O3
Explanation:
From the question given, we obtained the following data:
Carbon = 63.15%
Hydrogen = 5.30%
Oxygen = 31.55%
We can obtain the empirical and molecular formula by doing the following as illustrated in the attached file. Please see attachment for explanation.