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3 years ago

What volume will 2.50 mol of hydrogen (H2) occupy at -20.0 °C and 1.5 atm?

Chemistry

1 answer:

svet-max [94.6K]3 years ago

5 0

Answer:

v = 34.62 L

Explanation:

Given data:

Moles of H₂ = 2.50 mol

Temperature of gas = -20.0°C

Pressure of gas = 1.5 atm

Volume of gas = ?

Solution:

Formula:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

Now we will convert the temperature.

-20 + 273 = 253 K

Now we will put the values in formula.

1.5 atm × v = 2.50 mol ×0.0821 atm.L/ mol.K × 253 K

1.5 atm × v = 51.93 atm.L

v = 51.93 atm.L/1.5 atm

v = 34.62 L

The volume of hydrogen is 34.62 L.

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Determine how many carbon dioxide (CO2) molecules are produced if 8.45 x 1023 molecules of water (H2O) are produced 2 C2H6 (g) +

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3 years ago

Column A 1. Reproduction: Reproduction 2. offspring: offspring 3. gamete: gamete 4. Budding: Budding 5. Binary fission: Binary f

7nadin3 [17]

Answer:

1. Reproduction: One or two parents, organisms make new offspring

2. Offspring: A new living organism

3. Gamete: female and male create mini-me by cell division

4. Budding: Hydra

5. Binary fission: Bacteria

6. Zygote: The original cell after egg cell and sperm cell join

7. Fragmentation/Regeneration: Sea star / Planarian

8. Mitosis: Healling and growth occurs in budding

9. Meiosis: Egg or sperm cell/ specialized cells

Explanation:

1. Reproduction is the process by which organisms make new offspring either sexually (involving two parents) or asexually (involving one parent only).

2. An Offspring is a new living organism

3. A gamete is a mature sexual reproductive cell, as a sperm or egg, that unites with another cell to form a new organism. It formed from female and male parents by cell division known as meiosis

4. Budding is a form of asexual reproduction in which a new offspring grows out from a part of the parent. It occurs in Hydra.

5. Biinary fission is a form of asexual reproduction which involves a separation of the parent into two new offspring. It commonly occurs in Bacteria

6. Zygote is the original cell formed after egg cell and sperm cell fuse.

7. Fragmentation/Regeneration: this is a form of reproduction in which a parent splits into fragments which are then able to develop into new organisms. It occurs in Sea star and Planarians

8. Mitosis is a form of cell division which a cell divides into two identical daughter cells with the same genetic components as the parent cell. It functions in healing and growth as well as in budding

9. Meiosis is a type of cell division that results in the production of sex cells or gametes. The number of chromosomes in the parent cell is halved and four gamete cells are produced. Egg or sperm cell/ specialized cells are produced by meiosis.

5 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.

<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 .

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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