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HNO2
M=1u+14u+2*16u=47u
47u-100%
14u-x(%)
x=29,78%azotu
2.
H2SO4
M=1*2u+32u+4*16u=98u
98u-100%
64u-x(%)
x=65,3%tlenu
H=1u
N=14u
O=16u
masa tlenu w kwasie=16u×2=32u
masa HNO₂=1u+14u+32u=47u
%N=14u/47u×100%≈29,8%
2)H₂SO₄-kwas siarkowy VI
H=1u
S=32u
O=16u
masa wodoru w kwasie 1u×2=2u
masa tlenu w kwasie=16u×4=64u
masa H₂SO₄=2u+32u+64u=98u
%O=64u/98u×100%≈65,3%