How many protons, neutrons, and electrons make up an ion of

n the laboratory, a general chemistry student measured the pH of a 0.342 M aqueous solution of ethylamine, C2H5NH2 to be 12.067.

Tasya [4]

Answer:

The value of $K_b$ of the an ethylamine is $4.121\times 10^{-4}$ .

Explanation:

The pH of the solution = 12.067

The pOH of the solution = 14 - pH =14-12.607 =1.933

$pOH=-\log[OH^-]$

$1.933=-\log[OH^-]$

$[OH^-]=0.0117 M$

$C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-$

Initially

0.342 M 0 0

At equilibrium

(0.342-x) x x

The value of x = $[OH^-]=0.0117 M$

The expression of $K_b$ is given as:

$K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}$

$K_b=\frac{x^2}{(0.342-x)}$

$K_b=\frac{(0.0117 )^2}{(0.342-0.0117)}=4.121\times 10^{-4}$

The value of $K_b$ of the an ethylamine is $4.121\times 10^{-4}$ .

7 0

3 years ago

Cytokinesis happens differently for plant and animal cells. Both separate cytoplasm between two new daughter cells. However, whi

victus00 [196]

Answer:

animal cell

Explanation:

Cytokinesis -

It refers to the process of cell division, which takes place during mitosis, is known as cytokinesis.

The process of cytokinesis is different for animal cell and plant cell.

Where, in case of plant cell, a separation, known as cell plate is formed along the center of the parent cell, which is responsible for the separation of the cells.

Whereas, in case of animal cell, the plasma membrane contracts itself along the center, until the two daughter cells are separated from each other.

Hence, from the given information of the question,

The correct answer is animal cell.

7 0

3 years ago

The molecular weight of acetic acid is

kolezko [41]

Answer:

60.052 grams per mole

8 0

3 years ago

A 200 g of sugar was dissolved in 100 ml of water describe the solution

yarga [219]

Answer:

b

Explanation:

mark me as brainliest have a great day

7 0

2 years ago

What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91

Vikentia [17]

Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

$1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L$

Temperature = $91^{o}C = (91 + 273) K = 364 K$

As molar mass of $O_2$ is 32 g/mol. Hence, the number of moles of $O_2$ are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol$

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

$PV = nRT\\P \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm$

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

4 0

2 years ago