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GrogVix [38]
3 years ago
14

In the process of attempting to characterize a substance, a chemist makes the following observations: (a) The substance is a sil

very white, lustrous metal. (b) The substance melts at 649 ∘C, (c) The substance boils at 1105 ∘C. (d) The density of the substance at 20 ∘C is 1.738 g/cm3. (e) The substance burns in air, producing an intense white light. (f) The substance reacts with chlorine to give a brittle white solid. (g) The substance can be pounded into thin sheets or drawn into wires. (h) The substance is a good conductor of electricity.
Chemistry
1 answer:
Amiraneli [1.4K]3 years ago
7 0

The question is incomplete, this is the complete question;

In the process of attempting to characterize a substance, a chemist makes the following observations. Which are physical properties and which are chemical properties?

(a) The substance is a silvery white, lustrous metal.

(b) The substance melts at 649 ∘C

(c) The substance boils at 1105 ∘C.

(d) The density of the substance at 20 ∘C is 1.738 g/cm³.

(e) The substance burns in air, producing an intense white light.

(f) The substance reacts with chlorine to give a brittle white solid.

(g) The substance can be pounded into thin sheets or drawn into wires.

(h) The substance is a good conductor of electricity.

Answer:

(a) Physical

(b) Physical

(c) Physical

(d) Physical

(e) Chemical

(f) Chemical

(g) Physical

(h) Physical

Explanation:

(a) The substance is a silvery white, lustrous metal - is appearance which is physical.

(b) The substance melts at 649 ∘C metling point is a physical measurement

(c) The substance boils at 1105 ∘C boiling point is a physical measurement

(d) The density of the substance at 20 ∘C is 1.738 g/cm3 how heavy a substance is,is physical.

(e) The substance burns in air, producing an intense white light. Burning is a chemical change. It is a reaction of a substance with oxygen

(f) The substance reacts with chlorine to give a brittle white solid. This is a chemical reaction as the question says.

(g) The substance can be pounded into thin sheets or drawn into wires. Ductility is a physical change, the substance remains chemically unchanged

(h) The substance is a good conductor of electricity. Any conductivity is physical, the substance remains chemically unchanged

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Calculate the pH of each of the following solutions: (a) 0.1000M Propanoic acid (HC3H5O2, Ka= 1.3x10-5 ) (b) 0.1000M sodium prop
jek_recluse [69]

(a) The pH of 0.1000 M propanoic acid (HC3H5O2) is 2.9.

(b) The pH of 0.1000 M sodium propanoate (NaC3H5O2) is 8.9.

(c) The pH of 0.1000 M propanoic acid (HC3H5O2) and 0.1000 M sodium propanoate (NaC3H5O2) is 4.9.

<h3>Further explanation:</h3>

(a)

Given information:

The value of acid ionization constant for propanoic acid is  1.3 x 10^{-5} .

The initial concentration of propanoic acid is  .

To calculate:

The pH of 0.1000 M propanoic acid solution.

Solution:

Propanoic acid  is a weak acid. It ionizes partially in water as follows:

 

The expression for acid dissociation constant is,

                                                            …… (1)

Here,

 is ionization constant of propanoic acid.

is the equilibrium concentration of propanoate ion.

is the equilibrium concentration of hydronium ion.

 is the equilibrium concentration of propanoic acid.

ICE table (1):

 

Refer ICE table (1),

 

Substitute the values form the ICE table (1) in equation (1).

 

The approximation x is very small is valid. Therefore, the value of x can be neglected. Above equation can be modified as,

 

Rearrange above equation for x.

                                                                                                           …… (2)

Substitute   for   in equation (2) to calculate the value of x.

 

Therefore, from the ICE table (1) the concentration of hydronium ion is,

 .

The negative logarithm of hydronium ion concentration is defined as the pH of the solution. Mathematically,

                                                                                                               …… (3)

Substitute    for    in equation (3) to calculate the pH of the solution.

 

(b)

Given information:

The value of acid ionization constant for propanoic acid is  .

The initial concentration of sodium propanoate is  .

To calculate:

The pH of 0.1000 M sodium propanoate solution.

Solution:

Sodium propanoate  is conjugate base of weak propanoic acid. It undergoes hydrolysis in water to yield hydroxide ion in the solution as follows:

                                                        …… (4)

Propanoic acid  is a weak acid. It ionizes partially in water as follows:

                                                       …… (5)

Dissociation reaction for water is written as follows:

                                                                                       …… (6)

From equation (4), (5), and (6) the relationship between   and   is,

                                                                                                                              …… (7)

Substitute   for   and   for   in equation (7).

 

ICE table (2):

 

The expression for base dissociation constant is,

                                                                                                     …… (8)

Here,

is base ionization constant.

is the equilibrium concentration of propanoate ion.

is the equilibrium concentration of hydroxide ion.

 is the equilibrium concentration of propanoic acid.

From the ICE table (2),

 

Substitute the values form the ICE table (2) in equation (8).

 

The approximation y is very small is valid. Therefore, the value of y can be neglected. Above equation can be modified as,

 

Rearrange above equation for y.

                                                                                                           …… (9)

Substitute   for   in equation (9) to calculate the value of y.

 

Therefore, from the ICE table (2) the concentration of hydroxide ion is,

 

The negative logarithm of hydroxide ion concentration is defined as pOH of the solution. Mathematically,

                                                                                                           …… (10)

Substitute    for    in equation (10) to calculate pOH of the solution.

 

The relation between pH and pOH is as follows:

                                                                                                                   …… (11)

Substitute 5.057 for pOH in equation (11) to calculate the pH of the solution.

 

(c)

Given information:

The value of acid ionization constant for propanoic acid is  .

The initial concentration of sodium propanoate is  .

The initial concentration of sodium propanoate is  .

To calculate:

The pH of 0.1000 M sodium propanoate and 0.1000 M propanoic acid solution.

Solution:

Propanoic acid is a weak acid, and sodium propanoate is salt of the conjugate base of propanoic acid. Thus, propanoic acid and sodium propanoate will form a buffer system.

The pH of the buffer solution can be determined with the help of the Henderson-Hasselbalch equation. Mathematically,

 

For propanoic acid and sodium propanoate buffer system, the Henderson-Hasselbalch equation can be modified as,

                                                                                               …… (12)

The negative logarithm of acid ionization constant is equal to  .

                                                                                                                …… (13)

Substitute   for  in equation (13).

 

Substitute    for  ,   for   and 4.9 for    in equation (12).

 

Learn more:

1. About Henderson-Hasselbalch equation brainly.com/question/12999557

2. Learn more about how to calculate moles of the base in given volume brainly.com/question/4283309

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ionic equilibria

Keywords: ionic equilibrium, propanoic acid, sodium propanoate, ionization constant, weak acid, conjugate base, equilibrium concentration, hydronium ion, hydroxide ion, pH, pOH, ICE table, negative logarithm, buffer solution, Henderson-Hasselbalch equation, 0.1000 M, 4.9, 8.9, 2.9.

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