Estudiar los parámetros que gobiernan el comportamiento de los gases. Analizar el efecto de la presión sobre el volumen de los gases a temperatura constante y establecer una relación entre la presión y el volumen (Ley de Boyle y Mariotte). Determinar el volumen molar de un gas a condiciones normales. Verificar la ley de Graham o difusión gaseosa contrastando las velocidades de dos sustancias gaseosas (NH 3) y el cloruro de hidrógeno (HCl).
PERA DE NIVEL
TUBO
TUBO PERA DE NIVEL
SOPORTE
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Calculamos el volumen muerto
TUBO
REGLA
Luego buscamos buscamos que el líquido de la pera y el tubo neumométrico neumométrico estén en el mismo nivel. TUBO REGLA
PERA DE NIVEL
Variamos la posición posición de la la pera. Estas mediciones mediciones serán serán el resultado resultado de hacer hacer variar un Δh los niveles del líquido del tubo y de la pera.
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Colocamos en una probeta agua casi llena.
Colocamos 10ml. de HCl 3M en la bureta.
Añadimos agua en la bureta haciendo que pase por la pared interna.
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Medimos la cinta de magnesio y colocamos un papel en la boca de la bureta.
Inmediatamente invertimos la bureta introduciéndola probeta y procedemos a observar la reacción.
Magnesio
Ácido clorhídrico
Ión magnesio
Ión cloro
Hidrógeno gaseoso
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Observamos que empiezan a aparecer burbujas, este es el hidrógeno liberado de la reacción en forma de gas.
Cuando el magnesio se ha disuelto igualamos niveles de agua de la bureta.
En cada tapón colocamos los algodones, en uno y en el otro HCl concentrado.
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Luego colocamos en el tubo al mismo tiempo.
Esperamos que se forme un anillo de color blanco.
Cloruro de
Cloruro de
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Marcamos el tubo de vidrio donde se formó el anillo y calculamos las distancias para ver la relación y comprobar la ley de Graham.
Condiciones ambientales T=19°C<>292K P=754mmHg P vapor O=16.5 mm Hg
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Experimento 2°
2 1 mol 8.46× 104mol
2
2mol
1mol
0.03mol
8.46× 104mol
Hallamos el volumen teórico: T lab = 292K Vol. (gas húmedo) =24.6 ml P1 =754.0mmHg Para calcular el volumen del H 2(gas seco) a C.N. aplicaremos la ley general de los gases ideales .
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De la ecuación:
2
2
Mg = 0.01 0.0125 25g/ g/cm cm
Longitud = 1.7 cm m Mg= 0.02125 g 1 mol Mg --------------> 1mol H2 0 .02125 /24.3mol Mg --------------> X X= 8.744104 mol H2(experimental)
V2 ----------> 8.744x104 mol H2 V (exp. a C.N.) ----------> 1 mol
V (exp. a C.N.) = 22.30 L
−
. . .
Peso de cinta de Mg Volumen de HCl 3M Temperatura de laboratorio
0.02125 g
10ml 292K
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1) Grafique la presión sobre el eje vertical y el volumen sobre el eje horizontal. Uno de los puntos obtenidos de la experiencia del laboratorio y trace t race la curva más apropiada, considerando la función P.V=cte (promedio obtenido de la experiencia)
765 760 755 750 745 740 .
VOLUMEN
PRESIÓN
37.5
737.3
37.1
744.65
36.75
752.005
36.4
759.358
37.85
729.94
38.3
722.5
38.7
715.29
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2)
765 760 755 750 745 740 735
En el mismo grafico anterior, con un color diferente graficar P versus 1/V. tenga cuidado que su nueva escala horizontal horizontal parta de 1/V = 0 en el eje Y. trace trace la línea recta más apropiada. VOLUMEN
PRESIÓN
0.0267
737.3
0.02695
744.65
0.0272
752.005
0.0275
759.358
0.0264
729.94
0.026
722.5
0.0258
715.29
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4)¿Qué factores pueden influir en que no se obtenga un valor constante para P.V en este experimento? El principal motivo que afecta al producto PV y causa su desviación en sus gráficas es que utilizamos a los gases ideales, de los cuales se puede comprobar que presenta diferencias a todo gas real, ya que la teoría cinética molecular en que están basadas las leyes sólo es una aproximación de la verdadera naturaleza de los gases. Siendo este uno de los motivos que dieron origen a la ecuación de Vander Waals para los gases reales. Otro de los factores fundamentales son las fallas experimentales que se pueden cometer en el momento de tomar las medidas correspondientes a las dimensiones del volumen utilizado.
8) ¿Cuál será el volumen del sistema gaseoso estudiado en el experimento 2 a 20° c y 800mmHg?
P.V -------------------- = CTE T
732.9 X 23.8 296
=
800.V.x 293
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1.-MENCIONE QUE OTROS COMPUESTOS USARÍA PARA PROBAR LA LEY DE GRAHAM . Bueno utilizaríamos utilizaríamos el perfume ya que con el se logra un mejor experimento experimento y además no se trabajan con gases tóxicos. 2.-INVESTIGUE DE QUE MANERA SE PODRÍA CONSTRUIR UN EQUIPO MEDIDOR DE LA PRESIÓN ATMOSFÉRICA O DE LA HUMEDAD AMBIENTAL(RELATIVA Y ABSOLUTA).
Es un aparato que sirve para medir la humedad. Esto se puede hacer sencillamente con un pelo de la cabeza, ya que cuando la humedad aumenta éste se alarga y viceversa. La variación es de un milímetro por cada 4 cm de longitud, luego para que se mote, será necesario usar un pelo bastante largo. Antiguamente se utilizaba el pelo de caballo que es más largo y fuerte. Otra forma es mojando este pelo, con cloruro de cobalto, de tal forma que cuando el aire es seco, se vuelve de color azul, y rosa cuando es húmedo.
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Material
Necesario
• Un vaso o bote de cristal, con la boca lo más grande mejor. de plástico grande. • Un globo •Una pajita. Cuanto más larga, más precisión obtendrás en la medida. • Una goma elástica. •Pegamento papel. •Una hoja de
2.Usa la parte superior del globo de
plástico.
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2.
Corta
el
globo
en
dos
partes,
utilizando
la
redonda.
3. Sitúa la parte redonda del globo tapando la boca del vaso. Asegúralo rodeándolo con la goma elástica. 4. Corta los dos extremos de la pajita de forma oblicua. Pega un extremo de la pajita en el centro de la tapa de globo por uno de sus extremos, utilizando el pegamento. 5. Fija la cartulina sobre la pared y sitúa el barómetro de tal forma que el extremo de la pajita esté delante de la cartulina. 6. Marca el nivel de la pajita en la cartulina, indicando el tiempo que hay existente. Explicación Las altas presiones o ausencia de nubes aplastarán el plástico haciendo que el extremo de la pajita, por efecto palanca, esté en la parte superior. A medida que la presión baja (llegada de una borrasca), el indicador irá bajando, indicando que llegan días nublados y lluviosos. Las variaciones de temperatura también influyen en nuestro barómetro. Por ello, no es conveniente dejar el barómetro en contacto directo con la luz del sol ya que esto provocará un aumento de temperatura, haciendo que el aire del interior del vaso se
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5. Tapa la botella y dale la vuelta, colocándola de una forma invertida en el plato. 6. Marca el nivel del agua en la cartulina, una vez que se ha estabilizado.
El peso de la columna de líquido se equilibra con la presión que ejerce la atmósfera sobre el líquido del plato...
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neón líquido, aunque no llega a temperaturas tan bajas como el helio líquido, también tiene aplicaciones en la criogenia, pues tiene una capacidad de refrigeración más de 40 veces superior a la del helio líquido y más de tres veces superior a la del hidrógeno líquido. El helio se utiliza como componente de los gases respirables para sustituir al nitrógeno, gracias a su baja solubilidad en fluidos, especialmente en lípidos. Los gases son absorbidos por la sangre y los tejidos corporales cuando hay presión, como en el submarinismo, lo que provoca un efecto anestésico conocido como "mal de profundidad". Los gases nobles se usan habitualmente para la iluminación debido a su falta de reactividad química. El argón, mezclado con nitrógeno, se utiliza como gas de
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EXPERIMENTO 3. En este experimento se pudo contrastar la ley de Graham, aunque con cierta desviación del valor teorico predicho, esto se da porque al realizar la colocación de los gases dentro del tubo no se realizan al mismo tiempo.
http://www.quiminet.com http://www.mitecnologico.com/Main/C http://www.mitecn ologico.com/Main/Conceptosde onceptosdeestequiometria estequiometria MANUAL DE LABORATORIO DE QUIMICA-Ing. Noemí Quintana A. http://www.taringa.com