Determina la composición centesimal de la sacarosa?
Determina la composición centesimal de la sacarosa.
Determina la composición centesimal de la sacarosa.
En resumen
Determina la composición centesimal de la sacarosa (C12H22O11)1. Calcular Mm de la SacarosaC : 12 x 12 = 144 g / molH : 22 x 1 = 22 g / molO : 11 x 16 = 176 g / mol````````````````````````````````````````````````````````` Mm = 342 g / mol2.
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Determina la composición centesimal de la sacarosa (C12H22O11)1.
Calcular Mm de la SacarosaC : 12 x 12 = 144 g / molH : 22 x 1 = 22 g / molO : 11 x 16 = 176 g / mol````````````````````````````````````````````````````````` Mm = 342 g / mol2.
Calcular % C 342 g - - - - 100 % 144 g - - - - - x x = 42.
11 % C3.
Calcular % H 342 g - - - - 100 % 22 g - - - - - x x = 6.
43 % H4.
Calcular %O 342 g - - - - 100 % 176 g - - - - - x x = 51.
46 % O.
Mm SnO2 Sn : 1 x 118. 7 = 118. 7 g / mol O : 2 x 16 = 32 g / mol `````````````````````````````````````` Mm = 150. 7 g / mol estaño : 100 % . 150. 7 g x - - - - - - - 118. 7 g x = 78. 76 % Sn OXIGENO 100% - - - - - - -…
Mm Al2(SO4)3 = 342 g / mol Al : 2 x 27 = 54 g / mol S : 3 x 32 = 96 g / mol O : 12 x16 = 192 g / mol ¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨¨ Mm = 342 g / mol Al : 342 g - - - - - - 100% 54 g - - - - - x x = 15. 78 % S : 342 g - - - - - -…