Ask question

Ask question

All categories

2 years ago

11

I need help with this. Read instructions and help plzz

1 answer:

Semenov [28]2 years ago

6 0

Answer:

e

Explanation:

You might be interested in

Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolv

Triss [41]

Answer:

1. $3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

2. $5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

Explanation:

1.

The $pH$ value of Aspirin solution = 2.62

$pH=-\log[H^+]$

$[H^+]=10^{-2.62}=0.00240 M$

Moles of s asprin = $\frac{2.00 g}{180 g/mol}=0.01111 mol$

Volume of the solution = 0.600 L

The initial concentration of Aspirin = c = $\frac{0.01111 mol}{0.600 L}=0.0185 M$

$HAs\rightleftharpoons As^-+H^+$

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

$K_a=\frac{[As^-][H^+]}{[HAs]}$ :

$K_a=\frac{x\times x }{(c-x)}$

$=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}$

$K_a=3.57\times 10^{-4}$

$3.57\times 10^{-4}$ was the $K_a$ value calculated by the student.

2.

The $pH$ value of ethylamine = 11.87

$pH+pOH=14$

$pOH=14-11.87=2.13$

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

$[OH^-]=10^{-2.13}=0.00741 M$

The initial concentration of ethylamine = c = 0.100 M

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

initially

c 0 0

At equilibrium

(c-x) x x

The expression of dissociation constant :

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

$K_b=\frac{x\times x}{(c-x)}$

$=\frac{0.00741\times 0.00741}{(0.100-0.00741)}$

$K_b=5.93\times 10^{-4}$

$5.93\times 10^{-4}$ was the $K_b$ of ethylamine value calculated by the student.

3 0

3 years ago

How many grams in 11.9 moles of sulfur? Avogadro’s number: 6.02x1023 atoms = 1 mole Molar mass of sulfur: 32.06 g sulfur = 1 mol

sergiy2304 [10]

Answer:

Mass = 0.37 g

Explanation:

Given data:

Number of moles of sulfur = 11.9 mol

Mass of sulfur in 11.9 mol = ?

Molar mass of sulfur = 32.06 g

Solution:

Number of moles = mass/molar mass

by putting values,

11.9 mol = mass/ 32.06 g/mol

Mass = 11.9 mol × 32.06 g/mol

Mass = 0.37 g

4 0

2 years ago

A buffer solution is composed of 1.00 mol of acid and 2.25 mol of the conjugate base. If the p K a of the acid is 4.90 , what is

Gemiola [76]

<u>Answer:</u> The pH of the buffer is 5.25

<u>Explanation:</u>

Let the volume of buffer solution be V

We know that:

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution}}$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{conjugate base}]}{[acid]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of weak acid = 4.90

$[\text{conjugate base}]=\frac{2.25}{V}$

$[acid]=\frac{1.00}{V}$

pH = ?

Putting values in above equation, we get:

$pH=4.90+\log(\frac{2.25/V}{1.00/V})\\\\pH=5.25$

Hence, the pH of the buffer is 5.25

4 0

3 years ago

What is number 1 for b, c, and d?

SVETLANKA909090 [29]

B because....................

4 0

3 years ago

Calculate the density for a rectangular block using the following measurements: Length = 10 cm, Width = 1.1 cm, Height = 15 cm,

Ksivusya [100]

Answer:

0.40 g/cm3

Explanation:

density = mass / volume.

mass = 65.2 grams

volume = 10*1.1*15=165 cm3

so density = 65.2/165=0.40 g/cm3

3 0

3 years ago