ÇÍÓÈ ÇáÊÑßíÒ ÇáãæáÇÑí áãÍáæá áåíÏÑæßÓíÏ ÇáÕæÏíæã ÊÑßíÒå 50% ææÒäå ÇáäæÚí 1.52
æÇÈí ÞÇäæä ÇáÍá Çæ ÇáØÑíÞÉ
therealnizar
27-04-2012, 11:19
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ÍíË:
C: ÊÑßíÒ ÇáãÍáæá
P: äÞÇæÉ ÇáãÍáæá Ýí ÇáÍÇáÉ ÇáÊí ØÑÍÊåÇ 50% (ÇáÊØÈíÞ ÇáÚÏÏí ÊÃÎÐ 50 ÝÞØ æáÇ ÊÞÓã Úáì 100)
d : ÇáßËÇÝÉ Ãæ ÇáæÒä ÇáäæÚí . Ýí ÍÇáÊß d=1.52
M: ÇáßÊáÉ ÇáãæáíÉ áåíÏÑæßÓíÏ ÇáÕæÏíæã (NaOH) M=23+16+1=40 g/mol
ÇáÊØÈíÞ ÇáÚÏÏí:
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ÇÐä ÇáÊÑßíÒ Çáãæáí ááãÍáæá åæ:
C=19 mol/l