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Gigica2001
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ofer 15p
Nr de moli de CaO continuti in 500g oxid de calciu de puritate 90 la suta

Răspuns :

Raoull
[tex]m_{CaO}=500g[/tex] (masa impura)
[tex]P=90[/tex]%

[tex]P=\frac{m_{pur}}{m_{impur}}*100=\ \textgreater \ m{pur}=\frac{P*m_{impur}}{100}=\frac{90*500}{100}=90*5=450g[/tex] CaO

[tex]M_{CaO}=40+16=56g/mol[/tex]

[tex]n=\frac{m(pur)}M={450}{56}=8.03[/tex] moli CaO

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