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:
324.18 g/mol
Explanation:
Let the molecular mass of the antimalarial drug, Quinine is x g/mol
According to question,
Nitrogen present in the drug is 8.63% of x
So, mass of nitrogen = 
Also, according to the question,
2 atoms are present in 1 molecule of the drug.
Mass of nitrogen = 14.01 amu = 14.01 g/mol (grams for 1 mole)
So, mass of nitrogen = 14.01×2 = 28.02
These 2 must be equal so,
solving for x, we get:
<u>x = 324.18 g/mol</u>
They like cheese because it is easy to eat and it tases really good
The 28 is the sum of the protons and neutrons in the element silicon.
ALL silicon atoms have 14 protons in the nucleus, so we can turn this into an equation:
#protons + #neutrons = 28
14 + #neutrons = 28
#neutrons = 14
#protons = 14
