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Sebanyak 11,6 gram senyawa hidrat N a subscript 2 S O subscript 4 space x space H subscript 2 O dipanaskan sampai terbentuk N a subscript 2 S O subscript 4 sebanyak 7,1 gram, menurut reaksi  N a subscript 2 S O subscript 4. space x space H subscript 2 O left parenthesis s right parenthesis rightwards arrow N a subscript 2 S O subscript 4 left parenthesis s right parenthesis plus space x space H subscript 2 O left parenthesis g right parenthesis.  Jika Ar Na=23; S=32; O=16; H=1, rumus senyawa kristal tersebut adalah ....

  1. N a subscript 2 S O subscript 4. space H subscript 2 O

  2. N a subscript 2 S O subscript 4. space 2 space H subscript 2 O

  3. N a subscript 2 S O subscript 4. space 3 space H subscript 2 O

  4. N a subscript 2 S O subscript 4. space 4 space H subscript 2 O

  5. N a subscript 2 S O subscript 4. space 5 space H subscript 2 O

A. Ratna

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Pembahasan

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell M a s s a space h i d r a t space left parenthesis N a subscript 2 S O subscript 4. x H subscript 2 O right parenthesis space a d a l a h space 11 comma 6 space g r a m comma space k e t i k a space d i p a n a s k a n space a d a space m a s s a space h i l a n g comma space m a s s a space y a n g space h i l a n g space i n i space m e r u p a k a n space m a s s a space H subscript 2 O. space end cell row blank blank cell K e m u d i a n space m a s s a space t e r s i s a space a d a l a h space 7 comma 1 space g r a m space i n i space m e r u p a k a n space m a s s a space a n h i d r a t space left parenthesis g a r a m space m u r n i right parenthesis. space space end cell row cell M a s s a space H subscript 2 O space space end cell equals cell space m a s s a space h i d r a t space – space m a s s a space a n h i d r a t space end cell row blank equals cell space 11 comma 6 space g r a m space – space 7 comma 1 space g r a m space end cell row blank equals cell space 4 comma 5 space g r a m end cell end table

K e m u d i a n comma space u n t u k space m e n e n t u k a n space j u m l a h space H subscript 2 O space m a k a space h i t u n g space p e r b a n d i n g a n space m o l space d a r i space H subscript 2 O space d a n space g a r a m.  m o l space N a subscript 2 S O subscript 4 colon M o l space H subscript 2 O equals fraction numerator m a s s a space N a subscript 2 S O subscript 4 over denominator M r space N a subscript 2 S O subscript 4 end fraction colon fraction numerator m a s s a H subscript 2 O over denominator M r H subscript 2 O end fraction  equals fraction numerator 7 comma 1 over denominator 142 end fraction colon fraction numerator 4 comma 5 over denominator 18 end fraction  equals 0 comma 05 colon 0 comma 25  equals 1 colon 5  S e h i n g g a space X space a d a l a h space 5 space  M a k a space r u m u s space h i d r a t n y a equals N a subscript 2 S O subscript 4.5 H subscript 2 O

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Suatu senyawa hidrat sebanyak 67,8 g dipanaskan, lalu massa senyawa tersebut berkurang 27 g. Analisis lebih lanjut menunjukkan bahwa anhidrat senyawa tersebut mengandung 12,00 g Ca; 9,59 g S; dan sisa...

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