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Submitted by IvanArsic on Sun, 06/15/2014 - 20:13
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Koliko atoma kiseonika ima u 56ml ozona(normalni uslovi)?

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trolololo lo - Sun, 06/15/2014 - 21:08

V=56ml=0.056dm3

V/Vm=n;

0.056dm3/22.4dm3/mol= 0.0025mol

n*Na=N

N=6*10231/mol*0.0025mol=0.015*1023

i sada ovo pomnožiš sa 3 zato što imaš 3 atoma kiseonika u ozonu(O3)

N=0.015*1023*3=0.045*1023=4,5*1021