MOL NED\u0130R?
\n Atomlar ve molek\u00fcller en g\u00fc\u00e7l\u00fc mikroskoplarla bile g\u00f6r\u00fclemiyecek kadar k\u00fc\u00e7\u00fck taneciklerdir.Bu taneciklerin olu\u015fturdu\u011fu 6.02 x1023 tanelik k\u00fcmeye 1 mol denir.Burada 6.02 x1023 say\u0131s\u0131n\u0131n ne ifade etti\u011fini belirtmeden ge\u00e7emiyece\u011fim.Bu say\u0131y\u0131 a\u00e7\u0131k bir \u015fekilde yazarsak, 602000000000000000000000 tanedir. Bahsetti\u011fimiz say\u0131n\u0131n ne kadar b\u00fcy\u00fck oldu\u011funu art\u0131k tahmin edebiliyorsunuz.<\/p>\n
6.02 x1023 Dolay\u0131s\u0131yla kimyasal hesaplamalarda 1 mol yerine N tane, N tane yerine 1 mol yaz\u0131labilir.
\n 6.02 x1023 say\u0131s\u0131na Avagadro Say\u0131s\u0131 denir. Bazen sorularda k\u0131saca N harfiyle g\u00f6sterilir.
\n 1 mol demir (Fe) 6.02 x1023 tane demir (Fe) atomu demektir.
\n 1 mol su (H2O) 6.02 x1023 tane su (H2O) molek\u00fcl\u00fc
\n demektir.
\n Bile\u015fkelerin b\u00fcnyesinde farkl\u0131 cinste atomlar vard\u0131r. \u00d6rne\u011fin 1 tane H2O molek\u00fcl\u00fcnde 2 tane H atomu, 1 tane O atomu vard\u0131r.
\n Sonu\u00e7 olarak 1 mol H2O atomu i\u00e7in;<\/p>\n
-6.02 x1023 tane molek\u00fcl
\n -2 mol H atomu
\n -1 mol O atomu
\n -Toplam 3 mol atom
\n -2 x 6.02 x1023 tane H atomu
\n -1 x 6.02 x1023 tane O atomu
\n -Toplam 3 x 6.02 x1023 tane atom i\u00e7erir.\u0130fadeleri kullan\u0131labilir.<\/p>\n
\u00d6RNEK:
\n 1 mol C6H12O6 i\u00e7in a\u015fa\u011f\u0131dakilerden hangisi yanl\u0131\u015ft\u0131r?<\/p>\n
a) 6.02 x1023 tane molek\u00fcl i\u00e7erir.
\n b) 6 mol C atomu i\u00e7erir.
\n c)12 x 6.02 x1023 tane H atomu i\u00e7erir.
\n d)Toplam 24 tane atom i\u00e7erir.
\n e) 6 mol O atomu i\u00e7erir.<\/p>\n
\u00c7\u00d6Z\u00dcM\u00dc:<\/p>\n
A,B,C,E se\u00e7enekleri do\u011frudur. Fakat D se\u00e7ene\u011findeki ifade yanl\u0131\u015ft\u0131r. Ya 24 mol atom i\u00e7erir olmal\u0131yd\u0131, ya da 24 x 6.02 x1023 tane atom i\u00e7erir olmal\u0131yd\u0131. <\/p>\n
Do\u011fru cevap \u201cD\u201d se\u00e7ene\u011fidir.<\/p>\n
ATOM A\u011eIRLI\u011eI<\/p>\n
12C izotopu standart se\u00e7ilerek ( 12 de 1\u2019i , 1akb [atomik k\u00fctle birimi]) di\u011fer elementlerin bu izotopla k\u0131yaslanmas\u0131 sonucu hesaplanan k\u00fctlelerdir. 1akb \u00e7ok k\u00fc\u00e7\u00fck bir birimdir. \u00d6rne\u011fin, H=1 O=16 N=14 S=32…
\n Buna g\u00f6re;
\n 1 tane C atomu 12 x 1akb dir.
\n 1 mol (6.02 x1023 ) C atomu 12 x6.02 x1023 akb dir.
\n 1 mol C atomu 12×1=12 gr.
\n 6.02 x1023 tane Catomu 12 gr.dir.<\/p>\n
MOLEK\u00dcL A\u011eIRLI\u011eI:<\/p>\n
Bile\u015fi\u011fi olu\u015fturan elementlerin gram cinsinden k\u00fctlelerinin toplam\u0131na denir. <\/p>\n
\u00d6rnek : 1 mol C6H12O6 (glikoz) ka\u00e7 gramd\u0131r ? (C=12 O=16 H=1) <\/p>\n
6.12 = 72 gram C
\n 12.1 = 12 gram H
\n 6.16 = 96 gram O <\/p>\n
1 mol C6H12O6 180 gram olur. <\/p>\n
B\u0130R TANE MOLEK\u00dcL\u00dcN K\u00dcTLES\u0130: Bir tek molek\u00fcl\u00fcn k\u00fctlesi, molek\u00fcl a\u011f\u0131rl\u0131\u011f\u0131n\u0131n Avagadro say\u0131s\u0131na b\u00f6l\u00fcm\u00fcyle bulunur. (MA=Molek\u00fcl a\u011f\u0131rl\u0131\u011f\u0131)<\/p>\n
1 tek molek\u00fcl\u00fcn k\u00fctlesi = MA\/6,02.1023<\/p>\n
\u00d6rnek : 1 tane H2O molek\u00fcl\u00fc ka\u00e7 gramd\u0131r ? (H=1 O=16)
\n 1 tane H2O = 18\/ 6.1023 = 3.10-23 gramd\u0131r. <\/p>\n
B\u0130R TANE ATOMUN K\u00dcTLES\u0130: Atom a\u011f\u0131rl\u0131\u011f\u0131n\u0131n Avagadro say\u0131s\u0131na oran\u0131d\u0131r. ( A.A :Atom A\u011f\u0131rl\u0131\u011f\u0131)
\n 1 tek atomun k\u00fctlesi = A.A.\/ 6,02.1023<\/p>\n
\u00d6rnek : 1 tane C atomu ka\u00e7 gramd\u0131r ? (C=12)
\n 1 tane C atomu = 12\/6.1023 = 2.10-23 gramd\u0131r. <\/p>\n
N.\u015e.A HAC\u0130M: B\u00fct\u00fcn gazlar\u0131n bir mol\u00fc N.\u015e.A\u2019da ( 0 0C ve 1 atmosfer de) 22,4 litredir. (N.\u015e.A=Normal \u015eartlar Alt\u0131nda)<\/p>\n
\u00d6rnek : 3,01.1023 tane C2H4 molek\u00fcl\u00fc ka\u00e7 gramd\u0131r ? (C=12 H=1) <\/p>\n
n=3,01.1023\/6,02.1023= 0,5 mol 1 mol C2H4 28 gram ise 0,5 mol C2H4 14 gram olur.
\n n=3,01.1023\/6,02.1023= 0,5 mol 1 mol C2H4 28 gram ise 0,5 mol C2H4 14 gram olur.
\n MOL HESAPLAMA Y\u00d6NTEMLER\u0130:<\/p>\n
\u00d6rnek : 23 gram C2H5OH ka\u00e7 mold\u00fcr ? (C=12 O=16 H =1)
\n MA = 2.12 + 6.1 + 16.1= 46 gram\/mol
\n n=23\/46 = 0,5 mol <\/p>\n
\u00d6rnek : N.\u015e.A\u2019da 2,8 litre olan CO2 gaz\u0131 ka\u00e7 mol ve ka\u00e7 gramd\u0131r ?(C=12 O=16)
\n n=2,8\/22,4 = 0,125 mol.
\n 1 mol CO2 44 gram oldu\u011funa g\u00f6re 0,125 mol CO2 0,125.44= 5,5 gramd\u0131r.
\n 22,4 litre = 1 mol ; 11,2 litre= 0,5 mol ; 5,6 litre = 0,25 mol ;2,8 Litre = 0,125 mol<\/p>\n","protected":false},"excerpt":{"rendered":"
MOL NED\u0130R? Atomlar ve molek\u00fcller en g\u00fc\u00e7l\u00fc mikroskoplarla bile g\u00f6r\u00fclemiyecek kadar k\u00fc\u00e7\u00fck taneciklerdir.Bu taneciklerin olu\u015fturdu\u011fu 6.02 x1023 tanelik k\u00fcmeye 1 mol denir.Burada 6.02 x1023 say\u0131s\u0131n\u0131n ne ifade etti\u011fini belirtmeden ge\u00e7emiyece\u011fim.Bu say\u0131y\u0131 a\u00e7\u0131k bir \u015fekilde yazarsak, 602000000000000000000000 tanedir. Bahsetti\u011fimiz say\u0131n\u0131n ne kadar b\u00fcy\u00fck oldu\u011funu art\u0131k tahmin edebiliyorsunuz. 6.02 x1023 Dolay\u0131s\u0131yla kimyasal hesaplamalarda 1 mol yerine N …<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1407,1403],"tags":[2230,2261,2305,2137,2250,2252,5128,5127],"class_list":["post-2019","post","type-post","status-publish","format-standard","hentry","category-fen-ve-teknoloji-odevleri","category-odevler","tag-atom","tag-element","tag-glikoz","tag-izotop","tag-kimyasal-hesaplamalar","tag-mol","tag-mol-kavarmi","tag-mol-kavarmi-ve-kimyasal-hesaplamalar"],"_links":{"self":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/2019","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/comments?post=2019"}],"version-history":[{"count":0,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/posts\/2019\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/media?parent=2019"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/categories?post=2019"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.islamidavet.com\/kutuphane\/wp-json\/wp\/v2\/tags?post=2019"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}