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2 years ago

A solution of benzoic acid (C6H5COOH) that is 0.20 M has a pH of 2.45. What is Ka for benzoic acid?

1 answer:

8 0

We first calculate for the H+ ion concentration from the pH of the solution of benzoic acid. We do as follows:

pH = -log[H+]
2.45 = -log [H+]
[H+] = 3.55x10^-3

We then use the initial concentration and the h+ ion concentration as follows for Ka calculation:
C6H5COOH = C6H5COO- + H+
I 0.20 0 0
C -x x x
E 0.20-x x x

x = [H+] = 3.55x10^-3
[C6H5COO-] = 3.55x10^-3
[C6H5COOH] = 0.1965

Ka = 3.55x10^-3 (3.55x10^-3) / 0.1965 = 6.42x10^-5

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The nucleus of a neutral potassium atom is "surrounded" by

ddd [48]

The answer is electron.

The nucleus of a neutral potassium atom is "surrounded" by electron.

The neutral potassium atom contains equal number of protons and neutrons, and there are 19 electrons and 19 protons while 20 neutrons.

20 protons and 20 neutrons are there in the nucleus while 19 electrons surrounds the nucleus in different orbits .

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2 years ago

Read 2 more answers

The number of p atoms in 1.0 g of ba3(po4)2 is:

marta [7]

First, we determine the number of moles of barium phosphate. This is done using:

Moles = mass / Mr
Moles = 1 / 602
Moles = 0.002

Now, we see from the formula of the compound that each mole of barium phosphate has 2 moles of phosphorus (P). Therefore, the moles of phosphorus are:
0.002 * 2 = 0.004

The number of particles in one mole of substance is 6.02 x 10²³. The number of phosphorus atoms will be:
0.004 * 6.02 x 10²³

2.41 x 10²¹ atoms of phosphorus are present

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2 years ago

A 15.0-L rigid container was charged with 0.500 atm of kryp‑ ton gas and 1.50 atm of chlorine gas at 350.8C. The krypton and chl

Alecsey [184]

Answer: 32.94 g

Explanation: It's stoichiometry problem so balanced equation is required. The balanced equation is given below:

$Kr+2Cl_2\rightarrow KrCl_4$

From the balanced equation, krypton and chlorine react in 1:2 mol ratio. We will calculate the moles of each reactant gas using ideal gas law equation(PV = nRT) and then using mol ratio the limiting reactant is figured out that helps to calculate the amount of the product formed.

for Krypton, P = 0.500 atm and for chlorine, P = 1.50 atm

V = 15.0 L

T = 350.8 + 273 = 623.8 K

For krypton, $n=\frac{0.500*15.0}{0.0821*623.8}$

n = 0.146 moles

for chlorine, $n=\frac{1.50*15.0}{0.0821*623.8}$

n = 0.439

From the mole ratio, 1 mol of krypton reacts with 2 moles of chlorine. So 0.146 moles of krypton will react with 2 x 0.146 = 0.292 moles of chlorine.

Since 0.439 moles of chlorine are available, it is present in excess and hence the limiting reactant is krypton.

So, the amount of product formed is calculated from moles of krypton.

Molar mass of krypton tetrachloride is 225.61 gram per mol.

There is 1:1 mol ratio between krypton and krypton tetrachloride.

$0.146molKr(\frac{1molKrCl_4}{molKr})(\frac{225.61gKrCl_4}{1molKrCl_4})$

= 32.94 g of $KrCl_4$

So, 32.94 g of the product will form.

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2 years ago

Copper(II) sulfide, CuS, is used in the development of aniline black dye in textile printing. What is the maximum mass of CuS wh

Naya [18.7K]

Answer:

1.82 g is the maximum mass of CuS.

Explanation:

Considering:

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

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $CuCl_2$ :

Molarity = 0.500 M

Volume = 38.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 38.0×10⁻³ L

Thus, moles of $CuCl_2$ :

$Moles=0.500 \times {38.0\times 10^{-3}}\ moles$

Moles of $CuCl_2$ = 0.019 moles

For $(NH_4)_2S$ :

Molarity = 0.600 M

Volume = 42.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 42.0×10⁻³ L

Thus, moles of $(NH_4)_2S$ :

$Moles=0.600 \times {42.0\times 10^{-3}}\ moles$

Moles of $(NH_4)_2S$ = 0.0252 moles

According to the given reaction:

$CuCl_2_{(aq)}+(NH_4)_2S_{(aq)}\rightarrow CuS_{(s)}+2NH_4Cl_{(aq)}$

1 mole of $CuCl_2$ reacts with 1 mole of $(NH_4)_2S$

So,

0.019 mole of $CuCl_2$ reacts with 0.019 mole of $(NH_4)_2S$

Moles of $(NH_4)_2S$ = 0.019 mole

Available moles of $(NH_4)_2S$ = 0.0252 mole

Limiting reagent is the one which is present in small amount. Thus, $CuCl_2$ is limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of $CuCl_2$ gives 1 mole of $CuS$

0.019 mole of $CuCl_2$ gives 0.019 mole of $CuS$

Moles of $CuS$ formed = 0.019 moles

Molar mass of $CuS$ = 95.611 g/mol

Mass of $CuS$ = Moles × Molar mass = 0.019 × 95.611 g = 1.82 g

1.82 g is the maximum mass of CuS.

5 0

2 years ago

Watch the video to determine which of the following relationships correctly depict the relationship between pressure and volume

AnnZ [28]

Answer : The correct options are,

(B) $V\propto \frac{1}{P}$

(C) $P\propto \frac{1}{V}$

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

or,

$V\propto \frac{1}{P}$

The relation between the pressure and volume of two gases are:

$P_1V_1=P_2V_2$