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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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Suppose a substance has a heat of fusion equal to 45 calg and a specific heat of 0.75
Varvara68 [4.7K]

Answer:

The substance will be in liquid state at a temperature of 97.3 °C

Note: The question is incomplete. The complete question is given below :

Suppose a substance has a heat of fusion equal to 45 cal/g and a specific heat of 0.75 cal/g°C in the liquid state. If 5.0 kcal of heat are applied to a 50 g sample of the substance at a temperature of 24°C, what will its new temperate be? What state will the sample be in? (melting point of the substance = 27°C; specific heat of the solid =0.48 cal/g°C; boiling point of the substance = 700°C)

Explanation:

1.a) Heat energy required to raise the temperature of the substance to its melting point, H = mcΔT

Mass of solid sample = 50 g; specific heat of solid = 0.75 cal/g; ΔT = 27 - 24 = 3 °C

H = 50 × 0.75 × 3 = 112.5 calories

b) Heat energy required to convert the solid to liquid at its melting point at 27°C, H = m×l, where l = 45 cal/g

H = 50 × 45 = 2250 cal

c) Total energy used so far = 112.5 cal + 2250 cal = 2362.5 calories.

Amount of energy left = 5000 - 2362.5 = 2637.5 cal

The remaining energy is used to heat the liquid

H = mcΔT

Where specific heat of the liquid, c = 0.75 cal/g/°C, H = 2637.5 cal, ΔT = temperature change

2637.5 = 50 × 0.75 x ΔT

ΔT = 2637.5 / ( 50*0.75)

ΔT = 70.3 °C

Final temperature of sample = (70.3 + 27) °C = 97.3 °C

The substance will be in liquid state at a temperature of 97.3 °C

5 0
3 years ago
For the process 2SO2(g) + O2(g) --&gt; 2SO3(g),
CaHeK987 [17]

Answer:

–187.9 J/K

Explanation:

The equation that relates the three quantities is:

\Delta G = \Delta H - T \Delta S

where

\Delta G is the Gibbs free energy

\Delta H is the change in enthalpy of the reaction

T is the absolute temperature

\Delta S is the change in entropy

In this reaction we have:

ΔS = –187.9 J/K

ΔH = –198.4 kJ = -198,400 J

T = 297.0 K

So the Gibbs free energy is

\Delta G=-198,400-(297.0)(-187.9)=254.2 kJ

However, here we are asked to say what is the entropy of the reaction, which is therefore

ΔS = –187.9 J/K

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