Advice 1: How to solve chemistry problems 8th grade

In most cases, current or former students there are at least some theoretical understanding of chemical processes. But here is the solution of tasks on chemistry is quite a difficult situation if there are certain skills. But chemical task is to help in the kitchen when breeding, for example, acetic essence or just a friendly hint own son or sister. Remember how to solve chemistry problems? Usually, in the 8th grade, the first problem using the equations of chemical reactions are of the type "Calculate the mass of one of the reaction products by a known mass of one of the reacting substances. The problem is solved with the help of chemical formulas, because often jobs use this method.
How to solve chemistry problems 8th grade
Task. Calculate the mass of aluminium sulfide if the reaction of sulfuric acid entered the 2.7 g of aluminium.

Recorded a short condition


m(Al) =2, 7 g



m(Al2 (SO4) 3)-?
How to solve chemistry problems 8th grade

Before to solve problems in chemistry, a chemical equation. In the interaction of metals with dilute acid forming salt and released gaseous substance is hydrogen. We arrange the coefficients.

2Al + 3H2SO4 = Al2 (SO4) 3 + 3H2

When you always need to pay attention only to substances for which known and you need to find, options. All the rest are not taken into account. In this case it will be: Al and Al2 (SO4) 3
How to solve chemistry problems 8th grade

Find the relative molecular mass of these substances in the table of D. I. Mendeleev

Mr(Al) =27

Mr(Al2 (SO4) 3) =27•2(32•3+16•4•3) =342

Translate these values into molar mass (M) multiplied by 1 g/mol

M(Al) =27G/mol

M(Al2 (SO4) 3) =342g/mol
How to solve chemistry problems 8th grade

Recorded a basic formula, which connects the amount of substance (n), mass (m) and molar mass (M).


Carry out calculations according to the formula

n(Al) =2.7 g/27G/mol=0.1 mol
How to solve chemistry problems 8th grade

Make up two ratio. The first relation based on equation based on the odds faced by the formulas of the substances whose parameters are given or need to find.

First ratio: 2 mol Al you have 1 mole of Al2 (SO4) 3

The second ratio of 0.1 mol Al we have X mole Al2 (SO4) 3

(compiled on the basis of the obtained results)

Solve the proportion, given that X is the amount of substance

Al2 (SO4) 3 and has the unit mol


n(Al2 (SO4) 3)=0.1 mol(Al)•1 mol(Al2 (SO4) 3):2моль Al=0.05 mol
How to solve chemistry problems 8th grade

Now there is the amount of substance and molar mass of Al2(SO4)3, consequently, you can find lots of which derive from the basic formula


m(Al2 (SO4) 3)=0.05 mol•342g/mol=17,1 g


Answer: m(Al2 (SO4) 3)=17,1 g
How to solve chemistry problems 8th grade
At first glance, it seems that to solve chemistry problems is very difficult, but it is not. And to verify the degree of assimilation, first try to solve the same problem, but only yourself. Then substitute other values using the same equation. And the last step is the solution of the new equation. And if you were able to handle what – you're to be congratulated!
Useful advice
Wonderful helper during solving tasks is a manual, time-tested "problems in chemistry for entering Universities" G. P. Khomchenko. And don't be afraid to use it – it proposes the solution of problems from the very beginning!

Advice 2: How to solve problems on proportions

Leaves no doubt that the proportions of the thing desired. The proportions in our lives everywhere. Calculate salary for the year, knowing the monthly income. How much to buy of goods for money, if you know the price. It's all proportions.
How to solve problems on proportions
When solving problems on proportions is always possible to use the same principle. That they are comfortable. When dealing with proportion, always proceed as follows:Define the unknown and label it with the letter H.
Record the condition of tasks as a table.
Determine the type of dependence. They can be direct or reverse. How to determine the type? If the proportion is subject to the rule "the more, the better", so a direct relationship. If on the contrary, "the more, the less", it means an inverse relationship.
Put your hands with the edges of the table in accordance with the type of dependency. Remember: the arrow pointing upwards.
Using the table, write a proportion.
Solve the proportion.
Now let us examine two examples of different types of dependence.Task 1. 8 yards of cloth cost 30 R. How much are the 16 yards of this cloth?
1) the Unknown - the cost of 16 yards of cloth. Let's denote it as x.
2) let's Make the table:8 yards 30 p.
16 yards x R. 3) to Define the type of dependency. Think: the more cloth you buy, the more you will pay. Therefore, a direct relationship.4) Put the arrow in the table:^ 8 yards 30 p. ^
| 16 yards R. x |5) we form the ratio:8/16=30/xx=60 p answer: the cost of 16 yards of cloth is 60 p.
Task 2. The motorist noticed that at a speed of 60 km/h he drove a bridge across the river for 40 s. On the way back he passed the bridge in 30 s. Determine the speed of the car on the way back.1) the Unknown - the speed of the car on the way back.2) let's Make the table:60km/h 40
x km/h 30 C3) Define the type of dependency. The greater the speed, the faster a motorist will pass a bridge. Hence the inverse relationship.4) will form the proportion. In the case of inverse relationship here a little trick: one of the table columns you need to flip. In our case, we get the following proportion:60/x=30/40x=80 km/cotvet: back on the bridge a motorist drove at speeds of 80 km/h.
Useful advice
A direct correlation obeys the rule of "the more, the better".

Reverse obeys the rule of "bigger is less."

In the case of reverse dependencies when preparing the proportions of one of the columns of the table should turn.

Advice 3: How to solve problems with parameters

To solve the problem with the - a option means to find out what is the variable if any, or a specified parameter value. Or the task may be to search for those parameter values under which the variable meets certain conditions.
How to solve problems with parameters
If the given equation or inequality can be simplified, be sure to use it. Apply standard methods for solving equations, as if the argument were a normal number. As a result, you will be able to Express the variable via a parameter, for example, x=R/2. If the solution of equation you met with no restrictions on the parameter value (it is not necessary under the root sign, under the sign of the logarithm in the denominator), record the answer, putting that he found for all valid values of the parameter R.
For solving problems with standard charts (e.g., line, parabola, hyperbola) use the graphical method. Divide the area of parameter values for such intervals where the value of the variable (or variables) will be different, and for each interval, draw a line graph. Pay special attention to the extreme points of the lines to precisely determine their belonging to the schedule, substitute this value into the function and solve the equation with him. If the equation at this point has no solution (for example, is obtained by dividing by zero), eliminate it from the graph, noting an empty circle.
To solve the problem with respect to the parameter, pick the variable and the argument for equal members of equations or inequalities and simplify the expression. Then return to the original sense of the members and consider the solution of the problem for all possible parameter values. For this set of parameter values you need to divide into intervals.
When you search for the boundaries of intervals, pay attention to those expressions involving a parameter. For example, you have the expression (a-5), among the borders of intervals have to be number 5, since this value reverses the value in the parentheses to 0. Great importance is the expression under the sign of the parameter of division, root, module , etc.
When you find the limits of intervals, consider its function for each of them. To simplify this, just substitute into the function one of the numbers from this interval and solve the resulting problem. Often, just by substituting different values, it is possible to find the right way of solving the problem.

Advice 4: How to learn to solve chemistry problems

The curriculum is quite intense, absorbed theoretical knowledge, but practical skills there is no solution. What to do and how to learn to solve problems in chemistry? What is primarily required from the student?
The task of chemistry
Solution of tasks on chemistry is specific, and you need to find a starting point that will help to learn to understand this difficult matter.

What you need to know for solving problems in chemistry

To correctly solve chemistry problems, you first need to know what is the valence of the elements. Depends on the formulation of the substance, equation of a chemical reaction without taking into account the valence not be, and not to equalize. The periodic table is used in almost every job, you need to learn how to use it correctly to obtain the necessary information about the chemical elements, their mass, the electronic levels. Often tasks require us to calculate the mass or volume of the resulting product, this is the basis.
If the valence to determine incorrectly that all the calculations are incorrect.

And then the other, more complicated tasks will be solved easier. But first of all, formulas of substances, well written reaction equations, showing that in the end will succeed and in what form. It could be fluid, freely evolved gas, solid, drop down into the sediment or dissolved in water or any other liquid.

Where to start when solving problems in chemistry

To solve the problem briefly recorded her condition. After this is a balanced equation for the reaction. For example, consider a specific data need to determine the mass of the received substance, sulphide of aluminium, the reaction of aluminum metal with sulfuric acid, if taken aluminum is 2.7 grams. We should pay attention only to substances that are known, then those that want to find.

To start you need to solve with the translation of the mass in grams per mole. To make the reaction formula, substitute into it the values of mass and calculate the ratio. After solved a simple problem, you can try to learn on their own similar, but with other elements, that is, to get hand. The formulas are the same, only the elements change. All the solution of tasks on chemistry is reduced to writing the correct formula of the substance, to the proper preparation of the reaction equations.
All the problems are solved on the same principle, the main thing is to place coefficients in the equation.

For exercise, you can use the Internet, there's a huge number of different jobs, and then you can see the solution algorithm, which continue to apply independently. The advantage is that you can always see the correct answer, and if your result does not coincide, to understand, to find the error. More for learning to use reference books and collections of problems.
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