Answer:
19.8% of Nitrogen
Explanation:
In the Al(NO₃)₃ there are:
1 atom of Al
3 atoms of N
And 9 atoms of O
The molar mass of Al(NO₃)₃ is:
1 Al * (26.98g/mol) = 26.98g/mol
3 N * (14g/mol) = 42g/mol
9 O * (16g/mol) = 144g/mol
26.98 + 42 + 144 = 212.98g/mol
We can do a conversion using these molar masses to find the mass of nitrogen is the sample, that is:
2.57g * (42g/mol / 212.98g/mol) =
0.51g N
Percent composition of nitrogen is:
0.51g N / 2.57g * 100
= 19.8% of Nitrogen
Hope this helps
Answer- 1 mole
The equilibrium constant for the reaction is 0.00662
Explanation:
The balanced chemical equation is :
2NO2(g)⇌2NO(g)+O2(g
At t=t 1-2x ⇔ 2x + x moles
The ideal gas law equation will be used here
PV=nRT
here n=
=
= density
P =
density is 0.525g/L, temperature= 608.15 K, P = 0.750 atm
putting the values in reaction
0.75 = 
M = 34.61
to calculate the Kc
Kc=![\frac{ [NO] [O2]}{NO2}](https://tex.z-dn.net/?f=%5Cfrac%7B%20%5BNO%5D%20%5BO2%5D%7D%7BNO2%7D)
x M NO2 +
M NO+
M O2
Putting the values as molecular weight of NO2, NO,O2

34.61= 
x= 0.33
Kc= 
putting the values in the above equation
Kc = 0.00662