oblicz zawartość procętową fostforu w kwasie fosforowym V
mH3PO4=3*1u+31u+4*16u=3u+31u+64u=98u
mP=31u
98u--------100%
31u----------x
x=31,6% stanowi P w H3PO4
Masa H₃PO₄ = 3·1u + 31u + 4·16u = 98u
98u --------- 100%
31 ------------ x
98x = 3100/98
x= 31.63%
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mH3PO4=3*1u+31u+4*16u=3u+31u+64u=98u
mP=31u
98u--------100%
31u----------x
x=31,6% stanowi P w H3PO4
Masa H₃PO₄ = 3·1u + 31u + 4·16u = 98u
98u --------- 100%
31 ------------ x
98x = 3100/98
x= 31.63%