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For a solution of H2SO4, determine the van\'t Hoff factor assuming 0% ionization and 100% ionization

Aleks [24]

The van't Hoff factor is the ratio between the amount of particles produced when a substance is dissolved and the amount of substance being dissolved. If ionization is 0 percent then it would mean that the substance do not dissociate into particles so the van't Hoff factor would be 1. However, when the percent ionization is 100%, then it fully dissociates into ions. For H2SO4 at 100% ionization, the vant hoff factor should be three, two hydrogen ions and one sulfate ion.

7 0

3 years ago

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A 0.2436-g sample of an unknown substance was dissolved in 20.0 ml of cyclohexane. The density of cyclohexane is 0.779 g/ml. The

Viktor [21]

The depression in freezing point is related to molality of solution as follows:

$\Delta T_{f}=k_{f}m$

Here, $k_{f}$ is freezing point depression constant, for cyclohexane it is equal to 20°C kg/mol.

The value of freezing-point depression is 2.5 °C, molality can be calculated as follows:

$m=\frac{\Delta T_{f}}{k_{f}}=\frac{2.5 ^{o}C}{20 ^{o}C kg/mol}=0.125 mol/kg$

Molality is defined as number of moles of solute in 1 kg of solvent.

Here, solvent is cyclohexane, its volume is given 20.0 mL and density is 0.779 g/mol thus, mass of cyclohexane can be calculated as follows:

$m=d\times V=0.779 g/mL\times 20 mL=15.58 g$

Converting this into kg,

$1 g=10^{-3} kg$

Thus, 15.58 g will be 0.01558 kg.

Now, number of moles of unknown solute is related to its mass and molar mass as follows:

$n=\frac{m}{M}$

Putting the values of mass of solute which is 0.2436 or 0.0002436 kg

$n=\frac{ 0.0002436 kg}{M}$

Now, expression for molality of solution is:

$m=\frac{n_{solute}}{m_{solvent}}$

Putting all the values,

$0.125 mol/kg=\frac{0.0002436 kg}{M\times 0.01558 kg}$

Or,

$0.125 mol/kg=\frac{0.015635}{M}$

On rearranging,

$M=\frac{0.015635}{0.125 mol/kg}=0.1250 kg/mol$

Or,

$M=0.125 kg/mol(\frac{1000 g}{1 kg})=125 g/mol$

Therefore, molar mass of unknown sample is 125 g/mol

5 0

4 years ago

Which statement below correctly describes the relationship between Q and K for both reactions? Are these reactions spontaneous a

nikklg [1K]

Answer:

Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.

Explanation:

Hello,

In this case, the undergoing chemical reactions with their proper Gibbs free energy of reaction are:

$A->B;\Delta _rG^o=-13 kJ/mol$

$C ->D ;\Delta _rG^o=3.5 kJ/mol$

The cellular concentrations are as follows: [A] = 0.050 mM, [B] = 4.0 mM, [C] = 0.060 mM and [D] = 0.010 mM.

For each case, the reaction quotient is:

$Q_1=\frac{4.0mM}{0.050mM}=80\\ Q_2=\frac{0.010mM}{0.060mM}=0.167$

A typical temperature at a cell is about 30°C, in such a way, the equilibrium constants are:

$K_1=exp(-\frac{-13000J/mol}{8.314J/mol*K*303.15K} )=173.8\\K_2=exp(-\frac{3500J/mol}{8.314J/mol*K*303.15K} )=0.249$

Therefore, Q < K for both reactions. Both are spontaneous at those concentrations of substrate and product.

Best regards.

6 0

3 years ago

____ occurs when the vapor pressure of a liquid is equal to the external atmospheric pressure.

vaieri [72.5K]

Answer:

huh be more clear

Explanation:

8 0

3 years ago

HELP PLEASE

Andreas93 [3]

Answer:

A.)

Explanation:

The first structure accurately represents ethyl ethanoate because there is an ethyl group (2 carbons) directly attached to the oxygen within the parent chain and there are 2 carbons (the eth- in ethanoate) attached to the double bonded oxygen.

The second structure is not ethyl ethanoate. There is an ethyl group directly attached to the oxygen within the parent chain. However, there are 3 carbons attached to the double bonded oxygen. This makes the structure ethyl propanoate.

3 0

2 years ago