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ahrayia [7]
2 years ago
8

How many moles are there in 6.33x10^23 molecules of NH3?

Chemistry
1 answer:
nadezda [96]2 years ago
6 0
About 1.051 moles (depends on how u round it)
You might be interested in
What temperature in degrees c is the freezing point of water?
guajiro [1.7K]
The freezing point of water is 0° C. The Celsius (centigrade) scale is based off of water, with the freezing point at 0° and the boiling point at 100°. (Google will probably tell you this in 10 seconds, then you wouldn't have had to wait 7 hours!)
5 0
3 years ago
Given the following reaction: 2D(g) + 3E(g) + F(g) \longrightarrow⟶ 2G(g) + H(g) When the concentration of D is decreasing by 0.
Zepler [3.9K]

Answer:

Rate of reaction = -d[D] / 2dt  = -d[E]/ 3dt = -d[F]/dt  = d[G]/2dt = d[H]/dt

The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1

E decreseas 3/2 as fast as G increases = 0.30 M/s

Explanation:

Rate of reaction = -d[D] / 2dt  = -d[E]/ 3dt = -d[F]/dt  = d[G]/2dt = d[H]/dt

When the concentration of D is decreasing by 0.10 M/s, how fast is the concentration of H increasing:

Given data = d[D]/dt = 0.10 M/s

-d[D] / 2dt  = d[H]/dt

d[H]/dt = 0.05 M/s

The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1

When the concentration of G is increasing by 0.20 M/s, how fast is the concentration of E decreasing:

d[G] / 2dt  = -d[H]/3dt

E decreseas 3/2 as fast as G increases = 0.30 M/s

5 0
3 years ago
Which element is classified as a
Yuki888 [10]

Answer: Boron

Explanation:

5 0
2 years ago
A pipe leaks water at a rate of 1.24 mL/s. What is the rate of the water leak in L/hr?
Umnica [9.8K]

Answer:

b)4.46 L/hr

Explanation:

To solve this question we need to convert the mL to liters (Using the conversion of 1000mL = 1L) and convert the time from seconds to hours (3600s = 1hr)

<em>mL to L:</em>

1.24mL/s * (1L / 1000mL) = 0.00124L/s

<em>seconds to hours:</em>

0.00124L/s * (3600s / 1hr) = 4.46L/hr

Right answer is:

<h3>b)4.46 L/hr </h3>

3 0
3 years ago
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.

3 0
3 years ago
Read 2 more answers
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