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kupik [55]
3 years ago
7

Consider two solutions made using water as the solvent: 1.0 m CaCl2 and 3.0 m sucrose. Describe how the osmotic pressures of the

se two solutions compare. Explain why this is true.
Chemistry
2 answers:
Setler [38]3 years ago
8 0

The osmotic pressures of the two solutions are the same.

Both solutions contain the same number of dissolved particles in the same quantity of the same solvent.

Osmotic pressure is a colligative property.

Because it is a colligative property, osmotic pressure is influenced only by the number of dissolved particles, not by the identities of the particles.

mash [69]3 years ago
7 0
Osmotic pressure is calculated by the product of the concentration in molarity, the  temperature, the vant Hoff factor (3 for CaCl2 and 1 for sucrose) and R, universal gas constant. At the same temperature, the osmotic pressures of both solutions are equal.

π = CRTi
For CaCl2,
π = (1)RT(3) = 3RT

For sucrose,
π = (3)RT(1) = 3RT
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The decomposition of HBr(g) into elemental species is found to have a rate constant of 4.2 ×10−3atm s−1. If 2.00 atm of HBr are
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Answer:

7,94 minutes

Explanation:

If the descomposition of HBr(gr) into elemental species have a rate constant, then this reaction belongs to a zero-order reaction kinetics, where the r<em>eaction rate does not depend on the concentration of the reactants. </em>

For the zero-order reactions, concentration-time equation can be written as follows:

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<u>To solve the problem, we just replace our data in the concentration-time equation, and we clear the value of t.</u>

Data:

K = 4.2 ×10−3atm/s,  

[A]o=[HBr]o= 2 atm,  

[A]=[HBr]=0 atm (all HBr(g) is gone)

<em>We clear the incognita :</em>

[A] = - Kt + [Ao]............. Kt =  [Ao] - [A]

                                        t  = ([Ao] - [A])/K

<em>We replace the numerical values:</em>

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In a aqueous solution of 4-chlorobutanoic acid , what is the percentage of -chlorobutanoic acid that is dissociated
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Answer:

Explanation:

Let assume that the missing aqueous solution of 4-chlorobutanoic acid = 0.76 M

Then, the dissociation of 4-chlorobutanoic acid can be expressed as:

\mathsf{C_3H_6ClCO_2H }          ⇄      \mathsf{C_3H_6ClCO_2^-}      +      \mathsf{H^+}

The ICE table can be computed as:

                   \mathsf{C_3H_6ClCO_2H }          ⇄      \mathsf{C_3H_6ClCO_2^-}      +      \mathsf{H^+}

Initial              0.76                                 -                           -

Change            -x                                  +x                         +x

Equilibrium   0.76 - x                              x                          x  

K_a = \dfrac{[\mathsf{C_3H_6ClCO_2^-}] [\mathsf{H^+}]}{\mathsf{[C_3H_6ClCO_2H ]}}

K_a = \dfrac{[x] [x]}{ [0.76-x]}

where:

K_a = 3.02*10^{-5}

3.02*10^{-5} = \dfrac{x^2}{ [0.76-x]}

however, the value of x is so negligible:

0.76 -x = 0.76

Then:

3.02*10^{-5}*0.76 = x^2

x=\sqrt{3.02*10^{-5}*0.76 }

x = 0.00479 M

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x = \mathsf{[C_3H_6ClCO_2^-] = [H^+]=} 0.00479 M

\mathsf{C_3H_6ClCO_2H }  = (0.76 - 0.00479) M

= 0.75521 M

Finally, the percentage of the acid dissociated is;

= ( 0.00479 / 0.76) × 100

= 0.630 M

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