Question

11) Determine the percent composition of Ba(NO2)2. Express percentages to 1 place after the decimal. (4 pts)

Ba=
N=_
0=

Answer 1

molar mass of ba(no2)2=229.34

mass of ba=137.27

mass of n=14.007*2=28.014

mass of o=15.999*4=63.996

so %of ba=137.27÷229.34*100=59.85%

% of n=28.014/229.34*100=12.21%

% of o=63.996/229.3*100=27.90%

i hope it helps btw i am from india

Step-by-step explanation: