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S e ban y ak 50 m L l a r u ta n C a ( N O 3 ​ ) 2 ​ 1 0 − 2 M d i c am p u r kan d e n g an 50 m L l a r u tan N a 2 ​ C O 3 ​ 1 0 − 2 M d e n g an re ak s i : C a ( N O 3 ​ ) 2 ​ ( a q ) + N a 2 ​ C O 3 ​ ( a q ) → C a C O 3 ​ ( s ) + 2 N a N O 3 ​ ( a q ) J ika Ks p C a C O 3 ​ = 9.1 0 − 9 . M a ss a y an g m e n g e n d a p se ban y ak .. ( A r C a = 40 ; C = 12 ; O = 16 ; N a = 23 ; N = 14 )

  1. 100 gram

  2. 0,100 gram

  3. 0,050 gram

  4. 0,025 gram

  5. 0,0025 gram

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M. Robo

Master Teacher

Jawaban terverifikasi

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Pembahasan

M o l space C a left parenthesis N O subscript 3 right parenthesis subscript 2 equals space 50 space m L space x space 0 comma 01 space M space equals space 0 comma 5 space m m o l space  M o l space N a subscript 2 C O subscript 3 space equals space 50 space m L space x space 0 comma 01 space M space equals space 0 comma 5 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 C a left parenthesis N O subscript 3 right parenthesis subscript 2 space left parenthesis a q right parenthesis space plus space N a subscript 2 C O subscript 3 space left parenthesis a q right parenthesis space space space rightwards arrow C a C O subscript 3 space left parenthesis s right parenthesis space space space space plus space 2 space N a N O subscript 3  M u l a space space space space space space space space space space space space space space space space 0 comma 5 space m m o l space space space space space space space space space space space space 0 comma 5 space m m o l  R e a k s i space space space space space space space space space space space space minus 0 comma 5 space m m o l space space space space space minus space 0 comma 5 space m m o l space space space space space space space space space space space plus space 0 comma 5 space m m o l space space space space space plus 1 comma 0 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 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 0 space space space space space space space space space space space space space space space space space space space space space space space 0 comma 5 space m m o l space space space space space space space space space space 1 comma 0 space m m o l    M a s s a space C a C O subscript 3 space end subscript equals space m o l space x space M r space equals space 0 comma 0005 space m o l space x space 100 space g divided by m o l space equals space bold 0 bold comma bold 05 bold space bold italic g bold italic r bold italic a bold italic m

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Diketahui K sp ​ X ( OH ) 2 ​ dalam air adalah 3 , 2 × 1 0 − 11 . Larutan jenuh akan mengendap pada pH sebesar ....

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5.0

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