Answered

89,6 Litre x normal koşullarda kaç gramdır?

Sagot :

Yukselbaysoz378rans

Cevap:

4 mol

Açıklama:

Normal şartlarda 1 mol gaz 22.4L hacim kaplar

o zaman

1 kalıp 22.4L ise

x kalıp 89.6L (İçler dışlar)

89.6/22.4=4 mol gelir

1 mol C3H8 de 8Gram H varsa

4 kalıp xgram vardır

4×8=32gram H içerir

XZerossrans

ѕєℓαм

cevap 4 mol olacak

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