HOA 8 T 64

      kiÓm tra bµi cò    !"!#$%&'()*+ # !",!#$%&'()*+ )$-./+01233,,456 7!*45608 ,,9 §¸p ¸n: :4$;<! = + 8,,>?@71 ,, A +:BC23D3"C2 v n C m = MC m 33,0 5,1 5,0 == EF + EF 1 1,33 0,75 m C M= = #+)33 400 2,5 160 n mol = = 4 CuSO 7 EF ,+:B233D"2 ):B233D3"2 *4567 moln 4,0 40 16 == EF §¸p ¸n: MC M 025,0 4 5,2 == 0,06 0,04M 1,5 m C M = MC M 8,0 5,0 4,0 == Bµi 43: Pha chÕ dung dÞch G/,,H 1I. A!E/0J?I.@K! L,MMN"OFK! .</ +23,,73P +23,,7= 4 CuSO 4 CuSO 4 CuSO 07 gm H 45550 0 2 =− g mC m m m C DD CT DD CT 5 100 5010 %100 % %100% = ⋅ = ⋅ ⇒⋅= '?IQ8I.N@F! I./ Cân@F21K!? Cân@F)2I.$R)2I.@+ 1B,N,NK!?K!;@FS  (0IQ23,,3P Bµi 43: Pha chÕ dung dÞch 4 CuSO 4 CuSO 4 CuSO 0 Bµi 43: Pha chÕ dung dÞch moln CuSO 05,0105,0 4 =⋅= I./ (TK!?U :B,N,NI.@K!? K 23đế ạ ,  ";@FS "IQừ ạ đề ,,= /ầ 4 CuSO 4 CuSO 4 CuSO gMnm CuSO 816005,0 4 =⋅=⋅= Bµi 43: Pha chÕ dung dÞch A!E/0J?V4"I.@K! ,MMN"OFK!.< / +23,,43P +23,,4= '?IQ8I.N@F! I./ Cân@F#341K!? Cân@F3I.$R3I. @+1B,N,NK!?K!;@FS  (0IQ23,,43P Bµi 43: Pha chÕ dung dÞch mC m m m C DD CT DD CT 30g 100 15020 %100 % %100% = ⋅ = ⋅ ⇒⋅= gm H 12030150 0 2 =− 0 Bµi 43: Pha chÕ dung dÞch I./ (2"T2NaClK!?U :B,N,NI.@K!? K 23đế ạ ,  ";@FS "IQừ ạ đề 23,,4= ầ có moln NaCl 1,0205,0 =⋅= 50ml = 0,05lít Mnm NaCl 58,51,0 = 5,85g⋅=⋅= [...]...luyện t p và củng cố Bài t p : Đun nhẹ 40g dung dịch NaCl cho đến khi nước bay hơi h t, người ta thu được 8g muối NaCl khan T nh nồng độ phần trăm của dung dịch Bài làm: Trong 40g dung dịch NaCl có 8g muối NaCl khan vậy: mCT 8 C% = 100% = 100% = 20% mDD 40 Bài 1,2,3 SGK trang 149 Đọc nội dung bài phần II cách pha loãng m t dung dịch theo nồng độ cho trước . cè A!E/:S)3,,4 ;I.FW"IXIQ T ?4;0/N1V 8 ,, A!!01)3,,47 T ? 4;KEF %20%100 40 8 %100% =⋅=⋅= DD CT m m C A!""#%&'1)Y :Z,!/NGG/O ,,H1I. . 23đế ạ ,  ";@FS "IQừ ạ đề 23,,4= ầ có moln NaCl 1,0205,0 =⋅= 50ml = 0,05l t Mnm NaCl 58, 51,0 = 5 ,85 g⋅=⋅= luyÖn t p vµ cñng cè A!E/:S)3,,4 ;I.FW"IXIQ T ?4;0/N1V 8 ,, A!!01)3,,47 T ? 4;KEF %20%100 40 8 %100% =⋅=⋅= DD CT m m C . !",!#$%&'()*+ )$-./+01233,,456 7!*4560 8  ,,9 §¸p ¸n: :4$;<! = + 8 ,,>?@71 ,, A +:BC23D3"C2 v n C m = MC m 33,0 5,1 5,0 == EF + EF 1 1,33 0,75 m C