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P a d a re ak s i 100 m L l a r u tan P b ( N O 3 ​ ) 2 ​ 0 , 4 M d an 100 m L l a r u tan K 2 ​ C O 3 ​ 0 , 4 M jika Ks p P b C O 3 ​ = 7 , 4.1 0 − 4 , ma ss a z a t y an g m e n g e n d a p a d a l ah .. ( A r P b = 207 , C = 12 , O = 16 , K = 39 )

  1. 56,67 gram

  2. 10,68 gram

  3. 5,34 gram

  4. 2,67 gram

  5. 1,068 gram

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M. Robo

Master Teacher

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Pembahasan

M o l space P b left parenthesis N O subscript 3 right parenthesis subscript 2 equals space 100 space m L space x space 0 comma 4 space M space equals space 40 space m m o l space  M o l space K subscript 2 C O subscript 3 space equals space 100 space m L space x space 0 comma 4 space M space equals space 40 space m m o l  space space space space space space space space space space space space space space space space space space P b left parenthesis N O subscript 3 right parenthesis subscript 2 space space left parenthesis a q right parenthesis space space space plus space space space K subscript 2 C O subscript 3 space left parenthesis a q right parenthesis space space space rightwards arrow space space P b C O subscript 3 space space space plus space 2 space K N O subscript 3  M u l a space space space space space space space space space space space space 40 space m m o l space space space space space space space space space space space space space space space space space space space space space 40 space m m o l  R e a k s i space space space space minus space 40 space m m o l space space space space space space space space space space space space space space space space space space minus 40 space m m o l space space space space space plus space 40 space m m o l space space space space plus 80 space m m o l  S i s a space space space space space space space space space space space space space space space space space space space 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space 0 space space space space space space space space space space space space space space space space space space space space space space 40 space m m o l space space space space space space space space space 80 space m m o l  M a s s a space P b C O subscript 3 space equals space m o l space x space M r space equals 0 comma 04 space m o l space x space 267 space g divided by m o l space equals space 10 comma 68 space g r a m

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Jika K sp ​ CaC 2 ​ O 4 ​ = 2 , 6 × 1 0 − 9 , konsentrasi ion oksalat yang diperlukan untuk membentuk endapan dalam larutan yang mengandung 0,02 M ion kalsium adalah …

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