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jolli1 [7]
3 years ago
8

Consider two solutions separated by a semipermeable membrane, as shown in the illustration to the right. The membrane allows the

passage of small molecules and ions, but not large molecules like polysaccharides or proteins. Solution A contains a 10% solution composed
of glucose and the protein albumin dissolved in water.
Solution B contains a 5% solution of NaCl in water. Indicate whether each substance in the system would flow into Solution A, Solution B, or neither.

1. Water
2. NaCl
3. glucose
4. Albumin
5. Glucose

Chemistry
1 answer:
Evgen [1.6K]3 years ago
6 0

Answer:

The correct movement would be -

1. Water  - into solution A.

2. NaCl  - into solution A.

3. glucose  - into Solution B.

4. Albumin  - neither.

Explanation:

All the substances are separated by the semipermeable membrane and the semipermeable membrane allows the only small molecule to pass through it. So the movement of the given substance would be -

1. Water  - into solution A.

Water molecules are small and can easily pass through the semipermeable membrane as it is given that the solution b has low solute concentration and solution A has high solute concentration. It is known that the movement of the solvent always takes place from low solute concentration to high so the movement of water will be into solution A.

2. NaCl  - into solution A.

The movement of small ionic molecule NaCl is always from high to low concentration as it is given that solution B has high concentration than solution A so movement will take place into solution A.

3. glucose  - into Solution B.

It is also a small molecule and moves from the high glucose region to the low glucose concentration region, in solution A the concentration of glucose is high than solution B so movement would be into solution B.

4. Albumin  - neither.

Albumin is a protein which is macromolecule and large in size to pass through the semipermeable membrane so, albumin move neither solution A nor solution B.

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en un recipiente se tiene 800 g de una solución al 35% en masa de ácido sulfuroso, de la cual se evapora 80ml de agua. ¿cuál es
Alinara [238K]

The mass percent of sulfurous acid in the new solution : 38.9%

<h3>Further explanation</h3>

<em>In a container you have 800 g of a 35% by mass solution of sulfurous acid, from which 80 ml of water evaporates. What is the mass percent of sulfurous acid in the new solution? data: density of water is 1g / ml.</em>

<em />

solution 1

composition :

  • 35% acid :

\tt 0.35\times 800~g=280~g

  • water :

\tt 800-280=520~g

solution 2(new solution)

composition :

  • water

\tt 520-(80~ml\times 1~g/ml)=440~g

  • Total mass of new solution after water evaporated

\tt 280(acid)+440(water)=720~g

  • %mass of acid in a new solution

\tt \dfrac{280}{720}\times 100\%=38.9\%

5 0
3 years ago
How many grams in 6.20 x 10^25 atoms of bromine (Br) ? image attached , will give brainliest
Deffense [45]

Answer:

8239.2g

Explanation:

Given parameters:

Number of atoms in Br  = 6.2 x 10²⁵atoms

Unknown:

Mass of Br = ?

Solution:

From mole concepts, we know that:

       1 mole of a substance contains 6.02 x 10²³ atoms/mol

 Molar mass of Br  = 80g/mol

6.2 x 10²⁵atoms  x \frac{1}{6.02 x 10^{23} } \frac{mol}{atoms} x  80 x \frac{g}{moles}  

          = 8239.2g

8 0
3 years ago
A student prepares a 1.8 M aqueous solution of 4-chlorobutanoic acid (C2H CICO,H. Calculate the fraction of 4-chlorobutanoic aci
Kruka [31]

Answer:

Percentage dissociated = 0.41%

Explanation:

The chemical equation for the reaction is:

C_3H_6ClCO_2H_{(aq)} \to C_3H_6ClCO_2^-_{(aq)}+ H^+_{(aq)}

The ICE table is then shown as:

                               C_3H_6ClCO_2H_{(aq)}  \ \ \ \ \to  \ \ \ \ C_3H_6ClCO_2^-_{(aq)} \ \ +  \ \ \ \ H^+_{(aq)}

Initial   (M)                     1.8                                       0                               0

Change  (M)                   - x                                     + x                           + x

Equilibrium   (M)            (1.8 -x)                                  x                              x

K_a  = \frac{[C_3H_6ClCO^-_2][H^+]}{[C_3H_6ClCO_2H]}

where ;

K_a = 3.02*10^{-5}

3.02*10^{-5} = \frac{(x)(x)}{(1.8-x)}

Since the value for K_a is infinitesimally small; then 1.8 - x ≅ 1.8

Then;

3.02*10^{-5} *(1.8) = {(x)(x)}

5.436*10^{-5}= {(x^2)

x = \sqrt{5.436*10^{-5}}

x = 0.0073729 \\ \\ x = 7.3729*10^{-3} \ M

Dissociated form of  4-chlorobutanoic acid = C_3H_6ClCO_2^- = x= 7.3729*10^{-3} \ M

Percentage dissociated = \frac{C_3H_6ClCO^-_2}{C_3H_6ClCO_2H} *100

Percentage dissociated = \frac{7.3729*10^{-3}}{1.8 }*100

Percentage dissociated = 0.4096

Percentage dissociated = 0.41%     (to two significant digits)

3 0
3 years ago
Read 2 more answers
Balance the following: ___ AlBr3+ ___ K---&gt; ___KBr+ ___ Al
Nataliya [291]
___AlBr3 + ___K -> ___KBr + ___ Al

1 AlBr3 + 3K -> 3KBr + 1 Al

hope this helps............
6 0
4 years ago
In the following reaction, if you wanted to produce more hydrochloric acid (HCl), what should you do? (2 points)
Paul [167]

Answer:

Add more H2O

Explanation:

Took the test

3 0
3 years ago
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