18.
C+2CI₂=CCI₄
2Na+CI₂=2NaCI
2AI+3CI₂=2AICI₃
Be+CI₂=BeCI₂
19.
CrBr₂
CrBr₃
Mr(CrBr₂)=52+2x80=212
ω(Cr)=52÷212=0,245 ω%(Cr)=0,245×100%=24,5%
Mr(CrBr₃)=52+3x80=292
ω(Cr)=52÷292=0,178 ω%(Cr)=0,178 ×100%=17,8%
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Verified answer
18.
C+2CI₂=CCI₄
2Na+CI₂=2NaCI
2AI+3CI₂=2AICI₃
Be+CI₂=BeCI₂
19.
CrBr₂
CrBr₃
Mr(CrBr₂)=52+2x80=212
ω(Cr)=52÷212=0,245 ω%(Cr)=0,245×100%=24,5%
Mr(CrBr₃)=52+3x80=292
ω(Cr)=52÷292=0,178 ω%(Cr)=0,178 ×100%=17,8%