Instruction

1

In the left side of the equation write down the substances entering into a chemical reaction. They are called "original substances". On the right side, respectively, of the resulting substances ( "reaction products").

2

When writing formulas for molecules use chemical symbols of common atoms. The index of each atom is determined by the formula of the compound and valence.

3

Remember that unlike mathematical equations, the equations of chemical reactions in any case it is impossible to swap the right and left parts! Because it completely changes the meaning of the record. In addition, this reaction is often simply impossible.

4

The number of atoms of all elements in the left and right side of the reaction must be the same. If necessary, the "trim" number produced by the selection of coefficients.

5

When writing equations chemical reactions first make sure that it is at all possible. That is, that its occurrence does not contradict to the known physico-chemical rules and properties of substances. For example, the reaction:

NaI + AgNO3 = NaNO3 + AgI

NaI + AgNO3 = NaNO3 + AgI

6

She proceeds quickly to the end, in the course of the reaction forms insoluble light yellow precipitate of silver iodide. And the reverse reaction:

AgI + NaNO3 = AgNO3 + NaI - impossible, though, and recorded the correct symbols, and the number of atoms of all elements in the left and right sides the same way.

AgI + NaNO3 = AgNO3 + NaI - impossible, though, and recorded the correct symbols, and the number of atoms of all elements in the left and right sides the same way.

7

Write down the equation in the "full" form, that is, using their molecular formula. For example, the reaction form a precipitate of barium sulfate:

BaCl2 + Na2SO4 = 2NaCl + BaSO4

BaCl2 + Na2SO4 = 2NaCl + BaSO4

8

But can the same reaction be written in ionic form:

Ba 2+ + 2Cl- + 2Na+ + SO4 2- = 2Na+ + 2Cl- + BaSO4

Ba 2+ + 2Cl- + 2Na+ + SO4 2- = 2Na+ + 2Cl- + BaSO4

9

You see that in the left and right parts of the equations contain exactly the same ions of chlorine and sodium. Discard them and get final abbreviated equation of the reaction in ionic form:

Ba 2+ + SO4 2- = BaSO4

Ba 2+ + SO4 2- = BaSO4

10

Similarly, you can write in ionic form the equation for another reaction. Remember that each molecule soluble (dissociated) substances are written as ions, the same ions in the left and right side of the equation are eliminated.

# Advice 2: How to write the equation of the tangent

The tangent to the curve is a straight line, which is adjacent to this curve at a given point, that is, passes through it so that in a small area around this point is possible without much loss of accuracy, replace curve segment tangent. If this curve is the graph of a function, tangent to it is possible to build a special equation.

Instruction

1

Suppose you have a graph of some function. Through two points lying on the graph, you can draw a straight line. Such a line crossing the graph of the given function at two points is called a secant.

If, leaving the first location on the ground, gradually moving in the direction of the second point, then the secant line will gradually turn around, trying to any particular position. In the end, when the two points will merge into one, clipping will fit tightly to your sense of in this one point. In other words, the secant will become a tangent.

If, leaving the first location on the ground, gradually moving in the direction of the second point, then the secant line will gradually turn around, trying to any particular position. In the end, when the two points will merge into one, clipping will fit tightly to your sense of in this one point. In other words, the secant will become a tangent.

2

Any sloping (i.e. not vertical) straight line on the coordinate plane is the graph of the equation y = kx + b. The secant passing through the points (x1, y1) and (x2, y2), must, therefore, satisfy the conditions:

kx1 + b = y1, kx2 + b = y2.

Solving this system of two linear equations, we get: kx2 - kx1 = y2 - y1. Thus, k = (y2 - y1)/(x2 - x1).

kx1 + b = y1, kx2 + b = y2.

Solving this system of two linear equations, we get: kx2 - kx1 = y2 - y1. Thus, k = (y2 - y1)/(x2 - x1).

3

When the distance between x1 and x2 tends to zero, the differences become differentials. Thus, the equation of the tangent passing through the point (x0, y0) the factor k is equal to ∂y0/∂x0 = f'(x0), that is, the value of the derivative of f(x) at the point x0.

4

To find the coefficient b, substitute the computed value of k in the equation f'(x0)*x0 + b = f(x0). Solving this equation for b, we get b = f(x0) - f'(x0)*x0.

5

The final version of the equation of the tangent to the graph of the given function at the point x0 looks like this:

y = f'(x0)*(x - x0) + f(x0).

y = f'(x0)*(x - x0) + f(x0).

6

As an example, consider the equation of the tangent to the function f(x) = x^2 at the point x0 = 3. Derivative of x^2 equal to 2x. Therefore, the equation of the tangent becomes:

y = 6*(x - 3) + 9 = 6x - 9.

The correctness of this equation easy to check. Graph the line y = 6x - 9 passes through the same point (3;9) as the original parabola. By building two graphs, you can see that this video really fits to the parabola at this point.

y = 6*(x - 3) + 9 = 6x - 9.

The correctness of this equation easy to check. Graph the line y = 6x - 9 passes through the same point (3;9) as the original parabola. By building two graphs, you can see that this video really fits to the parabola at this point.

7

Thus, the graph of the function has a tangent at the point x0 only when the function has a derivative at this point. If at the point x0, the function has a discontinuity of the second kind, then the tangent becomes vertical asymptote. However, the mere existence of the derivative at the point x0 does not guarantee the indispensable existence of the tangent at that point. For example, the function f(x) = |x| at the point x0 = 0 is continuous and differentiable, but draw a tangent to it at this point is impossible. The standard formula in this case yields the equation y = 0, but this line is not tangent to the graph module.