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Elanso [62]
3 years ago
7

A. How many moles of copper equal 8.00 × 109 copper atoms?

Chemistry
1 answer:
BARSIC [14]3 years ago
7 0

Answer:

Explanation:

a ) one mole = 6.02 x 10²³ atoms

no of moles in given no of atoms

= 8 x 10⁹ / 6.02 x 10²³

= 1.329 x 10⁻¹⁴ moles .

b ) one mole of calcium = 40 gram

102 .5 g calcium

= 102 .5 / 40 moles

= 2.5625 moles

c )

no of moles in 5.04 g lead = 5.04 / 207

= 2.4347 x 10⁻² moles

= 2.4347 x 10⁻²x 6.02 x 10²³ no of atoms of lead

= 14.6568 x 10²¹ no of atoms .

d)  

one mole = 6.02 x 10²³ atoms

2.85 mole = 17.157 x 10²³ atoms .

e )

moles of fluorine gas = 1.08 x 10³ / 6.02 x 10²³

= .1794 x 10⁻²⁰ moles

mass in grams =  .1794 x 10⁻²⁰ x 38

= 6.8172 x 10⁻²⁰ grams

f )

no of moles in .584 g of benzene = .584 / 78

= 7.487 x 10⁻³ moles

no of molecules = 6.02 x 10²³ x  7.487 x 10⁻³

= 45.07 x 10²⁰ molecules .

g )

moles of atoms = 5.09 x 10⁹ / 6.02 x 10²³

= .8455 x 10⁻¹⁴ moles

mass in gram = .8455 x 10⁻¹⁴ x 1

=  .8455 x 10⁻¹⁴ g

h )

.45 moles of Ca₃PO₄ = .45 x 6.02 x 10²³ molecules

= 2.709 x 10²³ molecules of Ca₃PO₄

no of atoms of Ca = 3 x 2.709 x 10²³

= 8.127 x 10²³ atoms of Ca .

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calculate the ph for each case in the titration of 50.0 ml of 0.130 m hclo(aq) with 0.130 m koh(aq). use the ionization constant
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The ph before the addition of any Koh is<u> 10.105.</u>

Concentration is the abundance of a constituent divided by way of the overall volume of an aggregate. several sorts of mathematical descriptions may be outstanding: mass concentration, molar concentration, variety concentration, and extent awareness.

After the addition of 50 ml KOH,

moles of KOH = 50 * 0.13 =<u> 6.5 mmol </u>

<u>moles </u><u>of HClO = 50 * 0.13 = 6.5 mmol </u>

occurred hydrolysis solution,

pH = 0.5(14 + pKa + log [base conjugate])

pH = 0.5(14 + (- log (4 * 10^-8)) + log (6.5/(50 + 50)))

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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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