important: past hSC question not many ppl got it right (1 Viewer)

[clicker]

an element X forms both a monoxide (XO) and a dioxide (XO2).

conversion of 10.00g of X0 to XO2 produces 15.33g of the dioxide.

what is therelative atomic mass of X?

 

[miss_b]

I got

(74.72/5.33)
=14.02g/mol

Making it nitrogen

 

[clicker]

yeh thats right
howd u get that?

 

[miss_b]

XO+O--->XO2

m(XO)=10.00g
M(XO)=16+[M(X)]

m(XO2)=15.33g
M(XO2)=32+[M(X)]

n(XO):n(XO2)
=1:1
.: 10/(16+[M(X)]) = 15.33/(32+[M(X)])
16+[M(X)] = (320+10*[M(X)] )/15.33
15.33*M(X) = 74.72+10*M(X)
5.33*M(X) = 74.72
M(X)=14.018=14.02g/mol

So, relative atomic mass of X is 14.02g/mol making it nitrogen

 

[clicker]

thanx heeps

 

[clicker]

thanx heeps your a genius