3727 2 C C 0299984 C 03 g mL3 10 g de az\u00facar sin refinar se disuelven en agua y

3727 2 c c 0299984 c 03 g ml3 10 g de azúcar sin

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° = 37.27 ° 2 C C = 0.299984 C≈ 0.3 g / mL 3) 10 g de azúcar sin refinar se disuelven en agua y se diluyen a volumen total de 100 mL . Usando un tubo de 200 mm a 25 °C , esta solución da una lectura de 12.648 ° . Después de la inversión, llevada a cabo de acuerdo a ICUMSA, la solución produce una lectura de 3.922 ° . Determinar el % de sacarosa en el azúcar sin refinar. Solución : Los gramos de sacarosa por 100 mL de solución se determinan: C = α α ' 1.9175 0.0066 t Donde: α = 12.648 ° α ' =− 3.922 ° 11 10 =− 4.3142 ° t = 25 °C Reemplazando: C = 12.648 ° −(− 4.3142 ° ) 1.9175 0.0066 ( 25 ) = 9.6788 g / 100 mL Determinamos el % de sacarosa en los 10g de azúcar sin refinar: %sacarosa = 9.6788 g 10 g 100
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%sacarosa = 96.788 4) Con los datos determinar ¿Cuál de las ecuaciones de Biot ha de aplicarse y calcúlese las constantes respectivas? P q ρ α 0.80 0.20 1.116 18.26° 0.50 0.50 1.290 13.74° 0.20 0.80 1.464 6.47° Graficar [ α ] vs q usando papel milimetrado u hoja de cálculo para hallar la ecuación y cálculo de constantes. El valor q = 0 = A , mediante la GRÁFICA. [ α ] = α l P ρ q = 100 P Solución : Determinamos [ α ] para cada uno de los casos ( l = 2 dm ): [ α ] 1 = 18.26 ° 2 0.80 1.116 = 10.23 ° [ α ] 2 = 13.74 ° 2 0.50 1.290 = 10.65 ° [ α ] 3 = 6.47 ° 2 0.20 1.464 = 11.05 ° Entonces: q [ α ] 0.20 10.23 0.50 10.65 0.80 11.05 Graficamos [ α ] vs q :
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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 9.9 10.1 10.3 10.5 10.7 10.9 11.1 f(x) = 1.37 x + 9.96 R² = 1 ? ] [? De la gráfica, la ecuación de la recta: [ α ] = 9.95 + 1.4 q Valores de las constantes: A = 9.95 B = 1.4 Si hacemos q = 0 [ α ] = 9.95 °
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  • Spring '18
  • KENDAll
  • Concentración, Disolvente, Ecuación, Puro, Universidad Nacional del Callao

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