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joja [24]
3 years ago
14

Imagine a beaker divided down the center by a rigid membrane that is freely permeable to water but impermeable to glucose. Side

1 contains a 10% glucose solution & side 2 contains the same volume of pure water. At equilibrium, what will be the situation?
1. water will continue to move from side 2 to side 1
2. the volume of liquid remain equal on both sides
3. water will continue to move from side 1 to side 2
4. the volume of liquid will be greater in side 2
5. the volume of liquid will be greater in side 1
Chemistry
1 answer:
damaskus [11]3 years ago
6 0

Answer:

D. The volume of liquid will be greater in side 1

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Una disolución contiene 50 gramos de KOH en 0.25 L de Determine la Molaridad de la disolución
snow_tiger [21]

Answer:

1. 3.56 M.

2. 0.99 M.

Explanation:

¡Hola!

1. En este caso, dado que la molaridad de una solution es calculada por medio de la siguiente ecuación:

M=\frac{n}{V}

Es posible calcular la molaridad de 50 gramos de hidróxido de potasio primero calculando las moles en dicha masa por medio de la masa molar:

n=50.0gKOH*\frac{1molKOH}{56.11gKOH}=0.891molKOH

Luego, dado el volumen de la solución, podemos calcular la molaridad:

M=\frac{0.891mol}{0.25L}=3.56M

2. En este segundo ejercicio, procedemos de la misma manera, pues primero calculamos las moles de nitrato de potasio:

n=75gKNO_3\frac{1molKNO_3}{101.1gKNO_3}=0.742mol

Luego, calculamos la molaridad justo como se hizo anteriormente:

M=\frac{0.742mol}{0.35L} \\\\M=0.99M

Best regards!

3 0
3 years ago
When 80.0 mL of a 0.812 M barium chloride solution is combined with 40 mL of a 1.52 M potassium sulfate solution, 10.8 g of bari
BabaBlast [244]

Answer:

76.1%

Explanation:

The reaction that takes place is:

  • BaCl₂ + K₂SO₄ → BaSO₄ + 2KCl

First we determine how many moles of each reactant were added:

  • BaCl₂ ⇒ 80 mL * 0.812 M = 64.96 mmol BaCl₂
  • K₂SO₄ ⇒ 40 mL * 1.52 M = 60.8 mmol K₂SO₄

Thus K₂SO₄ is the limiting reactant.

Using the <em>moles of the limiting reactant</em> we <u>calculate how many moles of BaSO₄ would have been produced if the % yield was 100%</u>:

  • 60.8 mmol K₂SO₄ * \frac{1mmolBaSO_4}{1mmolK_2SO_4} = 60.8 mmol BaSO₄

Then we <u>convert that theoretical amount into grams</u>, using the <em>molar mass of BaSO₄</em>:

  • 60.8 mmol BaSO₄ * 233.38 mg/mmol = 14189.504 mg BaSO₄
  • 14189.504 mg BaSO₄ / 1000 = 14.2 g BaSO₄

Finally we calculate the % yield:

  • % yield = 10.8 g / 14.2 g * 100 %
  • % yield = 76.1%
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