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Sever21 [200]
3 years ago
8

A 0.100 M solution of bromoacetic acid (BrCH2COOH) is 13.2% ionized. Calculate [H+]? Calculate BrCH2COO^ - ? Calculate BrCH2COOH

. (looking for Molarity)
Chemistry
1 answer:
Natali [406]3 years ago
8 0

<u>Answer:</u> The concentration of hydrogen ion and bromoacetate ion is 0.0132 M and 0.0132 M resepectively and that of bromoacetic acid is 0.0868 M

<u>Explanation:</u>

We are given:

Molarity of bromoacetic acid = 0.100 M

Percent of ionization = 13.2 %

The chemical equation for the ionization of bromoacetic acid follows:

BrCH_2COOH\rightarrow BrCH_2COO^-+H^+

1 mole of bromoacetic acid produces 1 mole of bromoacetate ion and 1 mole of hydrogen ion

Molarity of hydrogen ion = 13.2 % of 0.100 = \frac{13.2}{100}\times 0.100=0.0132M

Molarity of bromoacetate ion = molarity of hydrogen ion = 0.0132 M

Molarity of bromoacetic acid = Molarity of solution - Molarity of ionized substance

Molarity of bromoacetic acid = 0.100 - 0.0132 = 0.0868 M

Hence, the concentration of hydrogen ion and bromoacetate ion is 0.0132 M and 0.0132 M resepectively and that of bromoacetic acid is 0.0868 M

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The protein lysozyme unfolds at a transition temperature of 75.5°C, and the standard enthalpy of transition is 509 kJ mol-1. Cal
spin [16.1K]

Answer:

0.4774 KJ/K.mol

Explanation:

We are told that the transition at 25.0°C occurs in three steps. Steps i, ii and iii.

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii

Now,

C_p,m(unfolded protein) = C_p,m(folded protein) + 6.28 kJ/K.mol

Now, for the first process, ΔS_i is given as;

ΔS_i = C_p,m × In(T2/T1)

We are given;

T1 = 25°C = 25 + 273.15K = 298.15 K

T2 = 75.5°C = 75.5 + 273.15 K=348.65 K

Thus;

ΔS_i = C_p,m × In(348.65/298.15)

Now, for the third process, ΔS_iii is given as;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(T1/T2)

Thus;

ΔS_iii = (C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)

Now, we don't know C_pm. So, we have to find a way to eliminate it. We will do it by rewriting In(298.15/348.65) in such a way that when ΔS_iii is added to ΔS_i, C_p,m will cancel out. Thus;

In(298.15/348.65) can also be written as;

In(348.65/298.15)^(-1) or

- In(348.65/298.15)

Thus;

ΔS_iii = - [(C_p,m + 6.28 kJ/K.mol) × In(298.15/348.65)]

Now, let's add ΔS_iii to ΔS_i to get;

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] + [(-C_p,m - 6.28 kJ/K.mol) × In(348.65/298.15)]

ΔS_i + ΔS_iii = [C_p,m × In(348.65/298.15)] - [C_p,m × In(348.65/298.15)] - [6.28In(348.65/298.15)]

First 2 terms will cancel out to give;

ΔS_i + ΔS_iii = -6.28In(348.65/298.15)

ΔS_i + ΔS_iii = -0.9826 KJ/K.mol

Now,for process ii;

ΔS_ii = standard enthalpy of transition/Transition Temperature

Thus;

ΔS_ii = (509 KJ/K.mol)/348.65

ΔS_ii = 1.46 KJ/K.mol

Thus;

the entropy of unfolding of lysozyme = ΔS_i + ΔS_ii + ΔS_iii = -0.9826 + 1.46 = 0.4774 KJ/K.mol

5 0
4 years ago
Which of the followings are true 1,3-BPG? A. It regulates Hb B. It contains a high-energy bond C. It contains two ester bonds D.
weeeeeb [17]

Answer:

D. It contains a phosphate with higher phosphoryl transfer potential than ATP

Explanation:

1,3-Bisphosphoglycerate contains a phosphate group that has high phosphoryl transfer potential than ATP (they can transfer the phosphoryl group to ATP). Other high phosphoryl transfer potential groups include :Creatine kinase and phosphoenolpyruvate.

7 0
3 years ago
What is the boiling point of a solution of 76 g of water dissolved in 500 mL of acetic acid, CH3COOH?
tatuchka [14]

Answer:

127.3° C, (This is not a choice)

Explanation:

This is about the colligative property of boiling point.

ΔT = Kb . m . i

Where:

ΔT = T° boling of solution - T° boiling of pure solvent

Kb = Boiling constant

m = molal (mol/kg)

i = Van't Hoff factor (number of particles dissolved in solution)

Water is not a ionic compound, but we assume that i = 2

H₂O →  H⁺  +  OH⁻

T° boling of solution - 118.1°C =  0.52°C . m . 2

Mass of solvent =  Solvent volume / Solvent density

Mass of solvent = 500 mL / 1.049g/mL → 476.6 g

Mol of water are mass / molar mass

76 g / 18g/m = 4.22 moles

These moles are in 476.6 g

Mol / kg = molal → 4.22 m / 0.4766 kg = 8.85 m

T° boling of solution =  0.52°C . 8.85 m . 2 + 118.1°C =  127.3°C

6 0
3 years ago
Read 2 more answers
The number of electrons in the outermost principal energy level of a chlorine atom is
Hitman42 [59]

Answer:

seven electrons

Explanation:

Chlorine is present in group seventeen of periodic table. It is halogen element. All halogens have seven electrons in outer most valance shell.

The require only one electron to gain the stable electronic configuration or to complete the octet.

Electronic configuration of chlorine:

Cl₁₇ = 1s² 2s² 2p⁶ 3s² 3p⁵

Abbreviated electronic configuration:

Cl₁₇ = [Ne] 3s² 3p⁵

Properties of chlorine:

1. it is greenish-yellow irritating gas.

2. its melting point is 172.2 K

3. its boiling point is 238.6 K

4. it is disinfectant and can kill the bacteria.

5. it is also used in manufacturing of paper, paints and textile industries.

5 0
3 years ago
11. Fill in the blanks below for a general chemical reaction (products, reactants).
Bad White [126]
Reactants on the left and products on the right
5 0
3 years ago
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