All categories

3 years ago

The concentration of hydrogen peroxide solution can be determined by

Chemistry

1 answer:

max2010maxim [7]3 years ago

8 0

The question is incomplete, the complete reaction equation is;

The concentration of a hydrogen peroxide solution can be determined by titration

with acidified potassium manganate(VII) solution. In this reaction the hydrogen

peroxide is oxidised to oxygen gas.

A 5.00 cm3 sample of the hydrogen peroxide solution was added to a volumetric flask

and made up to 250 cm3 of aqueous solution. A 25.0 cm3 sample of this diluted

solution was acidified and reacted completely with 24.35 cm3 of 0.0187 mol dm–3

potassium manganate(VII) solution.

Write an equation for the reaction between acidified potassium manganate(VII)

solution and hydrogen peroxide.

Use this equation and the results given to calculate a value for the concentration,

in mol dm–3, of the original hydrogen peroxide solution.

(If you have been unable to write an equation for this reaction you may assume that

3 mol of KMnO4 react with 7 mol of H2O2. This is not the correct reacting ratio.)

Answer:

2.275 M

Explanation:

The equation of the reaction is;

2 MnO4^-(aq) + 16 H^+(aq) + 5H2O2(aq) -------> 2Mn^+(aq) + 10H^+ (aq) + 8H2O(l)

Let;

CA= concentration of MnO4^- = 0.0187 mol dm–3

CB = concentration of H2O2 = ?

VA = volume of MnO4^- = 24.35 cm3

VB = volume of H2O2 = 25.0 cm3

NA = number of moles of MnO4^- = 2

NB = number of moles of H2O2 = 5

From;

CAVA/CBVB = NA/NB

CAVANB = CBVBNA

CB = CAVANB/VBNA

CB = 0.0187 * 24.35 * 5/25.0 * 2

CB = 0.0455 M

Since

C1V1 = C2V2

C1 = initial concentration of H2O2 solution = ?

V1 = initial volume of H2O2 solution = 5.0 cm3

C2 = final concentration of H2O2 solution= 0.0455 M

V2 = final volume of H2O2 solution = 250 cm3

C1 = C2V2/V1

C1 = 0.0455 * 250/5

C1 = 2.275 M

You might be interested in

Compare and contrast the burning of wood and the metabolism of glucose in your cells. How are they similar, and how are they dif

Alisiya [41]

Burning of wood is a combustion reaction and the metabolism of glucose in your cells is cellular respiratory reaction.

Cellular respiration releases stored energy in glucose molecules and converts it into a form of energy that can be used by cells.

Cellular respiration is the process by which organisms combine oxygen with foodstuff molecules, diverting the chemical energy in these substances into life-sustaining activities and discarding, as waste products, carbon dioxide and water.

Wood as well as many common items that combust are organic (i.e., they are made up of carbon, hydrogen and oxygen). When organic molecules combust the reaction products are carbon dioxide and water (as well as heat)

Similarities:

1. Combustion reaction and metabolism of glucose both require oxygen.

2. Combustion and metabolism of glucose both product carbon dioxide and water

3. Both produces by-products: After cellular respiration and combustion have gotten what they needed from the wood, there will be byproducts from the conversion. In the case of combustion, they are noxious gases like carbon monoxide. In the case of respiration, the sugar molecule is broken into two molecules of pyruvic acid.

4. Catalyst: While breaking apart the bonds to release the stored energy either combustion or sugars for respiration the bonds will not broken by themselves. In each case, a catalyst is required to start the reaction that will break the bonds apart. In the case of combustion, the catalyst is a spark. Wood are flammable, so the spark will ignite the burning, breaking apart the bonds and releasing the energy. For respiration, enzymes are used to break the sugar molecule apart.

Differences

1. Glucose metabolism produces a chemical energy, while combustion produces light and heat energy.

2. Glucose metabolism is an endothermic reaction while combustion is an exothermic reaction (produces heat)

8 0

3 years ago

A compound is found to have 55.7% hafnium+and+44.3%+chlorine.+what+is+the+empirical+formula?

uranmaximum [27]

The empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

<h3>How to calculate empirical formula?</h3>

The empirical formula of a compound is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.

The empirical formula of the given compound can be calculated as follows:

Hafnium = 55.7% = 55.7g
Chlorine = 44.3% = 44.3g

First, we convert mass values to moles by dividing by the molar mass of each element

Hafnium = 55.7g ÷ 178.49g/mol = 0.312mol
Chlorine = 44.3g ÷ 35.5g/mol = 1.25mol

Next, we divide each mole value by the smallest

Hafnium = 0.312 ÷ 0.312 = 1
Chlorine = 1.25 ÷ 0.312 = 4

Therefore, the empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

Learn more about empirical formula at: brainly.com/question/14044066

#SPJ1

7 0

1 year ago

a doctor's order is 0.125 g of ampicillin. the liquid suspension on hand contains 250 mg/5.0 ml. how many milliliters of the sus

disa [49]

0.125 g=(0.125 g)(1000 mg/1g)=125 mg.

Then, we need 125 mg of ampicillin.
5 ml of liquid suspension contains 250 mg of ampicilling , therefore:

5 ml----------------250 mg of ampicilling
x--------------------125 mg of ampicilling

x=(5 ml * 125 mg of ampicilling) / 250 mg of ampicilling=2.5 ml

Answer: we require 2.5 ml

5 0

3 years ago

Read 2 more answers

Limestone (CaCO3) is decomposed by heating to quicklime (CaO) and carbon dioxide. Calculate how many grams of quicklime can be p

Harman [31]

Answer: 2800 g

Explanation:

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 5 kg = 5000 g

$\text{Number of moles}=\frac{5000g}{100g/mol}=50moles$

1 mole of $CaCO_3$ produces = 1 mole of $CaO$

50 moles of $CaCO_3$ produces = $\frac{1}{1}\times 50=50moles$ of $CaO$

Mass of $CaO=moles\times {\text{Molar mass}}=50moles\times 56g/mole=2800g$

2800 g of $CaO$ is produced from 5.0 kg of limestone.

3 0

3 years ago

How would you prepare 1.00 L of a 0.400M solution of copper(II)sulfate, CuSO4?

Ber [7]

We need to dilute 0.400 mol of copper (II) sulfate, how do we know, how many weigh of $CuSO_4$ we have to dilute??

It's simple.

$\eta=\frac{m}{MM}$

Using a periodic table we can find the molar mass of $Cu,~~S~~and~~O$

$MM_{CuSO_4}=153.9~g/mol$

Then

$m=\eta*MM$

now we can replace it

$m=0.400*159.6$

$\boxed{\boxed{m=63.84~g}}$

Then we have to dilute 63.84 grams of copper (II) sulfate in 1 L of water to obtain a solution with 0.400M

3 0

3 years ago

Read 2 more answers