chemistry tasks

Chemistry tasks

Solution of chemistry tasks

What you need to know to solve chemistry tasks

Solution of chemistry tasks 1: How to determine the concentration of the solution

Solution of chemistry tasks 2: How to determine the number of substances for the reaction

Solution of chemistry tasks 3: How to calculate the mass fraction of elements in a complex substance

Solution of chemistry tasks 4: How to determine the shelf life of food products

Solution of chemistry tasks
In practice, we sometimes encounter problems associated with determining a quantitative assessment of the substances entering into a chemical reaction, the estimate of the number obtained in this reaction products, substances of interest to us. The hosts certainly often determine the percentage of a substance in solution and how much one or the other substances you have to take. How to calculate the amount of a substance that would be neutralized or neutralized another substance? How much gas will be allocated, if to the citric acid solution add soda? How to cook, for example, a 5% solution of potassium permanganate, and many other tasks encountered in everyday practice.
The purpose of it article Solution of chemistry tasks - to show how you can easily solve these tasks without resorting to complex chemical calculations, and applying only a general knowledge of mathematics and a bit of general knowledge of chemistry.

What you need to know to solve chemistry tasks

By the way, what You need to know to solve simple chemistry tasks:
M - molar mass (molecular weight substances) - this value for simple substances taken from the periodic table (the number written in the bottom right-hand corner of each element, for example, carbon M(C)=12,01115 g/mol, while the fractional part is usually cast). If it is a gas (e.g. hydrogen), then M(H2) =1 x 2 =2 g/mol), and so for all gases elements.
Basically we are dealing with complex substances, the molar mass of which is equal to the sum of the molar masses of its constituent simple elements, for example, carbon dioxide (CO2): M(CO2) = 12+16x2 = 44 g/mol.
W - concentration is how much of a substance by mass are contained in 100 g of the solution, such as 5% solution contains 5 g of dry matter and 95 g of solvent.

Well, let's look at the most widespread task, how to determine a percentage of a substance in solution and how much you need to take this substance and the solution.

Chemistry task 1:
How much to take water to 20% acetic acid (CH3 -COOH), to prepare a 5% solution of this acid?

Solution: what % of a substance in solution? Is as pure substances contained in 100 ml of solution. The solution consists of a pure substance and a solvent therefore, 100 ml of the solution will consist of 5 grams of pure substances (vinegar) and 95 ml of solvent (water) it is important to remember!
Further problem can be solved by using 1 formula or to make a proportion:
- if you use the formula,

% = (M substance) / (M solution) х 100.

In the task we have 20% solution of acetic acid. Find the mass of the new solution obtained by adding water to the 20% solution. From the formula we obtain: (M substances) / % x 100, i.e., 20 ml / 5 x 100 = 400 ml. We received 400 ml of a solution containing 20g of pure substances (vinegar), then 400 ml - 20 = 380 ml of water in a new solution. But don't forget, 20% solution contained water, and it was 80 g of water per 100 ml (95 + 5 =100). Therefore, in order to dilute our 20% acetic acid containing 80 ml of water, we need for every 100 ml of 20% of the initial solution add another 380 - 80 =300 ml of pure water.

- if you do not use the formula, you will make a proportion:
a 20% solution contains 20 g of vinegar and 80 ml of water
5 % solution contains 5 g of vinegar and 95 ml of water, 5 / 95 = 20 / X, here X is the amount of water contained in a 5% solution prepared from 20% solution. We find that X= 380 ml. From this amount of the water subtract water already contained in the 20% solution (before dilution), and get a 380 - 80 =300 ml of water, it need to be added to every 100 ml of 20% solution. That's all!

Chemistry task 2
The task of the determination of the amount of reaction products or reactants.
Suppose this condition. We need to neutralize the acidic environment. You accidentally poured into 200 ml of hydrochloric acid (HCl), and you have washing soda (Na2CO3) (or its second name - soda ash). Now: how much to take take washing soda to neutralize 200 ml of hydrochloric acid!?

write the chemical reaction of hydrochloric acid with washing soda:
2HCl + Na2CO3 => 2NaCl + H2O + CO2.
From elementary chemistry course we know that reaction of acid with the salt is formed another (weaker) and the other salt. In our case form carbonic acid, which immediately breaks down into water and carbon dioxide, and salt solution. Now on the periodic table you need to determine the molecular weight substances hydrochloric acid (2 x HCl) and washing soda (Na2CO3). The molecular weight of substances is considered as the sum of the masses of its constituent substances taking into account the number of atoms in the molecule and the number of molecules!
For example, the molecular weight of 2 molecules of HCl: Mr (HCl) = 2 x (1+ 35,5)= 73 g/mol;
molecular weight of 1 molecule of Na2CO3: Mr (Na2CO3) = 2 x 23 + 12 + 16 x 3)= 106 g/mol
Now make a proportion:
73 ml (HCl) react with 106 g of Na2CO3
200 ml (HCl) react with X g of Na2CO3, where we find that X = 200 x 106 / 73 = 290 g.
So, to neutralize 73 ml spilled acid (concentrated) you need to mix it with 290 g of washing soda.

Such chemical task can be a little complicated, given the concentration of acid. In this case, we need to calculate how much of a pure substance is in solution (as in the previous task 1), and then to make a proportion with the amount of hydrochloric acid. But in any case, even if the acid have a lower concentration, then our solution will fulfill the solution because, in this case, the soda will be taken in excess!

Chemistry task 3
So, now we want to calculate how much % of each simple substance is in a complex substance - CaSO4 - plaster.
Take periodic table and find a common relative molecular mass CaSO4: (remember that the total molecular complex substances equal to the sum of molecular masses of its constituent simple substances - choose from a table)
M (CaSO4) = 40 + 32 +16x4 = 136 г/моль
Now calculate the share of each item individually:
40:136 =0,29 (29%)
32:136 = 0,24 (24%)
The percentage of oxygen we find as the rest of the task: (100%-29%-24% = 47%). That's all.

Chemistry task 4

But, have You ever read the packaging of the products expiration date products! For sure! There is the date of manufacture and shelf life at a certain temperature. But it is not always possible to store the products at the recommended temperature. Now! You can define shelf life, if you know the conditions recommended storage. So, task in chemistry:
There is a cake with cream, shelf life of 4 days at a temperature of +50C. So, how long this cake can be stored at a temperature of +250C?
Solution: first of all, when solving such taskschemistry task about the rate of chemical reactions) there is a rule:
when the temperature is increased for every 10 0C the speed of a chemical reaction is increased by 2...4 times
how should the formula
Vt2 = Vt2*Y (t2-t1) / 10 , where
Vt1 is the reaction rate (recommended)
Vt2 is the reaction rate for the new environment
Y - coefficient equal to 2 or 3, or 4 (is usually taken 3)
t1 - storage temperature (recommended, listed on the package)
t2 - temperature storage in the new environment
So, in our case: t1 = +50C
; t2 = +250C; coefficient Y = 3
then Vt2 / Vt1 = Y(25 - 5) / 10C = 31,5C = 32 = 9.
So, if you store the cake at a temperature of +250C, instead of +50C, then it can be stored in time is 9 times smaller, namely: 4 days = 96 hours, then 96 / 9 = 1.5 hours.

Properties of acids
Getting acids

Complex substances. Water hardness

Alloys of pure metals

Simple substance

Color of gold. Alloys of gold. Stamp of gold


Chemical reaction. Types of chemical reactions

Crystal growing. How to grow crystal

Solid grease.liquid grease.Grease properties

Resin. Phenol-formaldehyde resin

Saturated hydrocarbons.
Homologous series

Organic glass. Ceramics

Protection against corrosion, electrochemical protection from corrosion

Metal alloys, solid solutions

Metal zinc, pure lead

Pure aluminum, pure tin

Natrium. Titanium

Decoration by your hands. Polymer clay

Potassium permanganate. Peroxide

Manganese. Metal uranium

Nickel. Metal germanium

Metal chrome. Vanadium

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