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7nadin3 [17]
3 years ago
10

Imagine a balloon is made from a membrane that's permeable to water but not sucrose molecules. How would the concentration of su

crose inside the balloon be affected if the balloon is filled with a 50% sucrose solution and then placed in a beaker of water?
Chemistry
1 answer:
Blababa [14]3 years ago
4 0

Answer:

The concentration of sucrose will decrease.

Explanation:

The balloon contains 50% sucrose solution.

Diffusion is the net movement of the substance from the region of the higher concentration to the region of the lower concentration.

Given that the membrane is only permeable to water not sucrose. So, movement of sucrose will not take place. <u>When the balloon is placed in a beaker of water, by the process of diffusion, the water moves from beaker to balloon and thus, the balloon will expand and the concentration of sucrose solution will decrease.</u>

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248 miles

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Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu
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Answer : The volume of CO_2 will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2

First we have to calculate the  mass of HCO_3^- in tablet.

\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g

Now we have to calculate the moles of HCO_3^-.

Molar mass of HCO_3^- = 1 + 12 + 3(16) = 61 g/mole

\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles

Now we have to calculate the moles of CO_2.

From the balanced chemical reaction, we conclude that

As, 1 mole of HCO_3^- react to give 1 mole of CO_2

So, 0.0202 mole of HCO_3^- react to give 0.0202 mole of CO_2

The moles of CO_2 = 0.0202 mole

Now we have to calculate the volume of CO_2 by using ideal gas equation.

PV=nRT

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = 37^oC=273+37=310K

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)

V=0.51411L=514.11ml

Therefore, the volume of CO_2 will be, 514.11 ml

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