How did high school chemistry c 1000 w m come about

c = n v (n is the amount of the substance in the solution, v is the volume) n = m m (m is the mass of the solution, m is the relative molecular mass of the substance.

v = m p multiplied by 1000 (v is the volume of the solution approximately equal to the volume of water, p is the density of the solution is approximately equal to the density of water, because the volume is liters, so the density should be 1000 substituted to get the above formula.

1. The physical quantities in the formula.

Meaning: c is the concentration of the amount of the substance.

mol/l）

m is the molar mass of the solute.

g/mol)

is the density of the solution (g cm3 or g ml).

w is the mass fraction of the solute in solution.

2. The relationship between the quantities.

The amount of solute contained in 1L solution is concentration C, that is, the amount of solute contained in 1000ml is the concentration of the substance C

Solution volume: 1000 (ml).

Solution mass: 1000*g).

Soluble mass: 1000* w (g).

Amount of solute substance: 1000* w m (mol).

Therefore, the amount of substance contained in 1L solution is: (1000* W m) mol, which is the amount concentration of the substance c

The unit is: mol l

The amount of solute substance--n, the volume of the solution --v, the density of the solution-- the mass fraction of the solution--w, the molar mass of the solute m, the mass of the solute m= vw=nm, gives n= vw m, whereas the amount concentration of the solute is c=n v= vw m v= w m, but in chemistry, the unit of volume.

Generally L (liter), the unit of density is g cm3, then the unit of C = W M introduced above is mol cm3, which is converted to 1000 mol L, (because 1L = 1000 cm3), therefore: C = 1000 * W M, is the formula for the unit of the standard unit mol L.

c = n v(l) = ((m total * ) m) (m total ) = m, and finally unify the units into commonly used units to obtain c = 1000 m

BelowWrongn=m m (m is the mass of the solution, m is the relative molecular mass of the substance).

Correct: M is the mass of the solute, so there is a mass percentage W added. n = m (mass of solution) w m, while m (mass of solution) = (density of solution) v, n = vw m, because molar concentration c = n v (how many molar solutes per liter of solution).

So c= vw m v= w m

Because the density is g ml and c is mol l, it is multiplied by 1000

c = n v (n is the amount of the substance of the solution, v is the volume) n = m m (m is the mass of the solution, m is the relative molecular mass of the substance) v = m p multiplied by 1000 (v is the volume of the solution is approximately equal to the volume of water, p is the density of the solution is approximately equal to the density of water, because the volume is liters, so the density should be 1000 to get the above formula.

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Hello, c = n v (n is the amount of the substance of the solution, v is the volume) n = m m (m is the mass of the solution, m is the relative molecular mass of the substance) v = m p multiplied by 1000 (v is the volume of the solution is approximately equal to the volume of water, p is the density of the solution is approximately equal to the density of water, because the volume is liters, so the density should be 1000 to get the above formula.

The principle is very simple, you only need to figure out the effect of the relationship between quantities on the reaction result.

The solution system is stored in Fe3+, Ba2+, NH3+, OH-, SO42-, of course, you will know that there is also H+ but it does not affect you. It is clear that some of these ions do not coexist and tend to produce less ionizing substances.

There is no doubt that there are BaSo4, Fe(OH)3 and can be generated, as long as there are two left, it should be noted that they are both made by OH- and a kind of ionic dust, and OH- is added gradually, so there is a sequential relationship. Obviously, Fe(OH)3 is less susceptible to ionization than NH3H2O, and Mr. becomes the former. c is established.

NH3+ can be reacted when Fe3+ is completed, but the D error is that the ratio of NH3+ to Fe3+ is wrong. In the case indicated by d, Fe3+ has all participated in the reaction, and according to the equation, NH3+ is three times that of Fe3+, and the extra NH3+ is **? So d is wrong.

As for B, in fact, just looking at the equation, he is purely suspicious and ...... to sayHowever, when Fe3+ is precipitated, it is equivalent to adding alkali to the NH3+ solution, which will definitely generate NH3, and part of it will definitely run away. Therefore, although there is nothing wrong with the amount of b and the addition of oh- can be written as NH3H2O, it is actually impossible in the real reaction process.

c = n vn is the amount of substance in the solution, v is the volume).

n = m mm is the mass of the solution, and the number m is the relative molecular mass of the substance) v = m p multiplied by 1000

v is the volume of the solution approximately equal to the volume of water, p is the density of the solution is approximately equal to the density of water, and the density should be 1000 because the volume is liters

Substituting it will get the formula of Shangyan Bizi's rough face.

The solid is only caCO3, because the first section of the lime water is excessive, and it burns completely.

All the C elements in the organic matter are converted into CO2=>CaCO3

The content of element C in 10g of CaCO3 is the mass of element C m(c) contained in organic matter