Ask question

Ask question

All categories

3 years ago

10

The standard Gibbs energy of reaction, ÄG°, for the dissociation of phenol is 56.4 kJ mol-1 at 298 K. Calculate the pKa of pheno

l.
a.
9.88
c.
4.12
b.
22.8
d.
5.24

1 answer:

Neporo4naja [7]3 years ago

4 0

Answer:

a.

9.88

Explanation:

ΔG = - 2.303 RT log Ka

ΔG is change in free energy at temperature T , Ka is equilibrium constant

- 56.4 x 10³ = 2.303 x 8.31 x 298 logKa

- log Ka = 9.88 .

pKa = 9.88 .

You might be interested in

What is the electron configuration for Fe?

blsea [12.9K]

The answer is C) [Ar]4s2 3d6

5 0

3 years ago

Read 2 more answers

To help relieve heartburn, a person should take medicine that is

Sav [38]

Prescribed to you by your doctor.

I would use an over the counter antacids for occasional heart burn. If there are symptoms of acid reflux I would suggest a histamine blocker, or H2-Blockers, such as Ranitidine (Zantac) and Famotidine (Pepcid)

3 0

4 years ago

Read 2 more answers

Calculate for the following electrochemical cell at 25°C, Pt H2(g) (1.0 atm) H (0.010 M || Ag (0.020 M) Ag if E (H) - +0.000 V a

viva [34]

Answer : The correct option is, (b) +0.799 V

Solution :

The values of standard reduction electrode potential of the cell are:

$E^0_{[H^{+}/H_2]}=+0.00V$

$E^0_{[Ag^{+}/Ag]}=+0.799V$

From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $H_2\rightarrow 2H^{+}+2e^-$

Reaction at cathode (reduction) : $Ag^{+}+1e^-\rightarrow Ag$

The balanced cell reaction will be,

$H_2+2Ag^{+}\rightarrow 2H^{+}+2Ag$

Now we have to calculate the standard electrode potential of the cell.

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}$

$E^o_{cell}=(+0.799V)-(+0.00V)=+0.799V$

Therefore, the standard cell potential will be +0.799 V

4 0

3 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago

Explain why some pesticides bioaccumulate whereas others do not

KatRina [158]

Bioaccumulation refers to the accumulation of chemicals in a living organism. The compound or chemical accumulates at a rate faster than it is being metabolized or excreted by the organism. Chemicals bioaccumulate by binding to the proteins and fats in an organism while others bioaccumulate through the repeated consumption of contaminated organisms.

Pesticides containing chemicals that dissolve easily in fat but not in water tend to bioaccumulate. Pesticides that contain chemicals that can easily be metabolized by organisms do not bioaccumulate. In summary, the nature of the chemical used in pesticides and the capability of organisms to metabolize the said chemicals can dictate whether it will bioaccumulate or not.

7 0

3 years ago