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Vanyuwa [196]
2 years ago
10

A balloon contains a 10% glucose solution. The balloon is permeable to water but not to glucose. A beaker contains a 5% glucose

solution. What will happen when the balloon is submerged in the beaker
Chemistry
1 answer:
deff fn [24]2 years ago
7 0

The state of the balloon:

When the balloon is submerged in the beaker, the amount of water in the beaker will get reduced.  

What is Osmosis:    

Based on the concentration of solutes on both sides of the membrane, water will flow through a permeable membrane in a specific direction.

<u><em>Hypertonic solution:</em></u>

It means that there are more solutes present in the surrounding environment than in the cell itself.

<u><em>Hypotonic solution:</em></u>

In a hypotonic solution, the concentration of solutes inside the cell is higher than that outside the cell.

  • When comparing two solutions, the one with the larger solute concentration is hypertonic, and the one with the lower solute concentration is hypotonic. Isotonic solutions have an identical solute concentration.

  • While the solution in the beaker is hypertonic, Meaning that will draw water molecules out of the cell. As water molecules move from a location of high water potential (dilute solution) to a region of reduced water potential (10% glucose solution), the water from the 5% glucose solution will flow into the 10% one (concentrated solution)
  • This is the reason why the amount of water decreases when the balloon is submerged in the beaker.

Learn more about the glucose solution and permeability here,

brainly.com/question/14748422

#SPJ4

 

 

 

 

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The Haber reaction for the manufacture of ammonia is: N2 + 3H2 → 2NH3 Without doing any experiments, which of the following can
Dimas [21]

Answer :  The correct statement is, \text{Rate of disappearance of }H_2=3\times (\text{Rate of disappearance of }N_2)

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

The expression for rate of reaction :

\text{Rate of disappearance of }N_2=-\frac{d[N_2]}{dt}

\text{Rate of disappearance of }H_2=-\frac{1}{3}\frac{d[H_2]}{dt}

\text{Rate of formation of }NH_3=+\frac{1}{2}\frac{d[NH_3]}{dt}

From this we conclude that,

\text{Rate of disappearance of }H_2=3\times (\text{Rate of disappearance of }N_2)

Hence, the correct statement is, \text{Rate of disappearance of }H_2=3\times (\text{Rate of disappearance of }N_2)

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Nuclear fission i think it is.
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