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Elza [17]
3 years ago
6

For the following reaction, the products are favored at equilibrium. Classify each of the reactants and products based on their

strength as Bronsted Lowry acids or bases.
(C2H5)2NH + C6H5OH -> (C2H5)2NH2+ + C6H5O-

(C2H5)2NH

C2H5)2NH2

C6H5O-

C6H5OH

Match with the following.

a. Stronger Bronsted Lowry acid
b. Weaker Bronsted Lowry acid
c. Stronger Bronsted Lowry base
d. Weaker Bronsted Lowry base
Chemistry
1 answer:
andrey2020 [161]3 years ago
4 0

Answer:

(C2H5)2NH    -------------> c. Stronger Bronsted Lowry base

C2H5)2NH2+  ------------>  b. Weaker Bronsted Lowry acid

C6H5O-          ------------->  d. Weaker Bronsted Lowry base

C6H5OH       --------------->   a. Stronger Bronsted Lowry acid

Explanation:

Bronsted-Lowry defined an acid as a molecule or ion which donates a proton, while a base is a molecule or ion which accepts the proton. Bronsted-Lowry Theory of definition of an acid and base could also be extended to include acid-base titrations in non aqueous solution. In this theory, the reaction of an acid with a base constitutes the transfer of a proton from an acid to the base.

From the question given, we are to Classify each of the reactants and products based on their strength as Bronsted Lowry acids or bases:

(C2H5)2NH + C6H5OH    ⇄  (C2H5)2NH2+ + C6H5O-

(C2H5)2NH    -------------> c. Stronger Bronsted Lowry base

C2H5)2NH2+  ------------>  b. Weaker Bronsted Lowry acid

C6H5O-          ------------->  d. Weaker Bronsted Lowry base

C6H5OH       --------------->   a. Stronger Bronsted Lowry acid

In the forward reaction, Phenol (C6H5OH) donates a proton to diethylamine (C2H5)2NH which accepts it. The phenol behaves as an acid and the diethylamine behaves as a base.

In the backward reaction, diethylammonium(C2H5)2NH2+) ion donates a proton to the phenolate (C6H5O-), Therefore, diethylammonium behaving as an acid (proton donor) and the phenolate behaves as a base(proton acceptor)

The equilibrium of this reaction lies to the far right (which is indicated by the longer arrow of the forward reaction) because Phenol is a stronger Bronsted Lowry acid than the diethylammonium. (i.e Weaker Bronsted Lowry acid).

Similarly, the diethylamine is a stronger Bronsted Lowry base while the phenolate which is a weaker Bronsted Lowry base.

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<u>Answer:</u> The percentage abundance of _{77}^{191}\textrm{Ir} and _{77}^{193}\textrm{Ir} isotopes are 37.10% and 62.90% respectively.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the fractional abundance of _{77}^{191}\textrm{Ir} isotope be 'x'. So, fractional abundance of _{77}^{193}\textrm{Ir} isotope will be '1 - x'

  • <u>For _{77}^{191}\textrm{Ir} isotope:</u>

Mass of _{77}^{191}\textrm{Ir} isotope = 190.9606 amu

Fractional abundance of _{77}^{191}\textrm{Ir} isotope = x

  • <u>For _{77}^{193}\textrm{Ir} isotope:</u>

Mass of _{77}^{193}\textrm{Ir} isotope = 192.9629 amu

Fractional abundance of _{77}^{193}\textrm{Ir} isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710

Percentage abundance of _{77}^{191}\textrm{Ir} isotope = 0.3710\times 100=37.10\%

Percentage abundance of _{77}^{193}\textrm{Ir} isotope = (1-0.3710)=0.6290\times 100=62.90\%

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3 years ago
Determine the concentrations of MgCl2, Mg2+, and Cl− in a solution prepared by dissolving 2.52 × 10−4 g MgCl2 in 2.00 L of water
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Answer:

Explanation:

Given parameters:

Mass of MgCl₂ = 2.52 x 10⁻⁴g

Volume of water = 2L

Unkown:

Concentration of MgCl₂ =?

Concentration of Mg²⁺ = ?

Concentration of Cl⁻ =?

Solution:

  Concentration is defined as the number of moles of a solute contained in a solution.

   Concentration = \frac{number of moles }{volume}

 To find the number of moles"

           number of moles = \frac{mass}{molar mass}

   Molar mass of MgCl₂ = 24.3 + (2 x 35.5) = 95.3g/mol

       number of moles = \frac{0.000}{molar mass} = 0.000252moles

 Concentration of MgCl₂ = \frac{0.000252}{95.3} = 2.64 x 10⁻⁶moldm⁻³

 

from the formula of the compound;

    1 mole of MgCl₂ contains 1 mole of Mg²⁺

Therefore, 2.64 x 10⁻⁶moldm⁻³ of MgCl₂ will contains 2.64 x 10⁻⁶moldm⁻³ of Mg

Also;

   1 mole of MgCl₂ contains 2 mole of Cl⁻

  2.64 x 10⁻⁶moldm⁻³ contains 2 x 2.64 x 10⁻⁶moldm⁻³; 5.29 x 10⁻⁶moldm⁻³

 

Expressing in ppm;

               1ppm = 1mg/L

   2.64 x 10⁻⁶moldm⁻³ to mg/L for Mg²⁺

2.64 x 10⁻⁶moldm⁻³  = 2.64 x 10⁻⁶moldm⁻³  x molar mass(g/mol)

                                     = 2.64 x 10⁻⁶moldm⁻³ x 24.3 = 6.43 x 10⁻⁵g/L

    g/L to mg/L; 6.43 x 10⁻⁵g/L x 1000 =   6.43 x 10⁻²mg/L = 6.43 x 10⁻²ppm

5.29 x 10⁻⁶moldm⁻³ to mg/L for Cl⁻;

   5.29 x 10⁻⁶moldm⁻³ = 5.29 x 10⁻⁶moldm⁻³ x 35.5 = 1.88 x 10⁻⁴g/L

   g/L to mg/L; 1.88 x 10⁻⁴g/L x 1000 = 1.88 x 10⁻¹mg/L = 1.88 x 10⁻¹ppm

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Answer:

The answer is

<h2>0.087 g/mL</h2>

Explanation:

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The density is

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We have the final answer as

<h3>0.087 g/mL</h3>

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