Chemical Concentration Calculation Problem High Fraction o Fast Fast Fast

The addition of positive ions is mol l

Subtract the negative ion mol l

The remaining l is due to the negative bivalent of SO42 (in other words, one can be used as two) so the answer is d

Balance of anions and cations in the solution to do.

Suppose there is a 1 L solution.

Cations, then anions, too.

The amount of anion left is.

And SO42- is a 2-valent ion, then the amount of matter is.

The volume is 1 L, and the concentration is selected D

Let NaCl, KCl, Na2SO4 and K2SO4 be A, B, C, D MOL L, respectively

Na is mol l to get a+2c=

Cl is mol l to get a+b=

k is mol l to get b+2d=

SO42 is c+d=

So choose D

Conservation of positive and negative charges in solution.

The positive charge na is mol l....k is mol l negative charge cl is mol l....SO4 is X in the same solution and has equal volumes.

Get x = mol l .Pick D

It is easiest to calculate according to the positive and negative electrons.

The total number of positives = the total number of negatives.

Positive: Negative: So n=

The solution is electrically neutral, and the positive charge is equal to the negative charge.

1. Equation method, the equation method is suitable for most concentration problems, and has the characteristics of simple thinking process. Generally speaking, there are two elements of the equation method, the first is to set the unknowns, and the abscurity requires that it is easy to solve; The second is to find the equivalence relation and list the equations. In concentration problems, concentration is often used as an unknown variable, so that the equivalence relationship is easy to express macro burial, but it is also accompanied by the disadvantage that most of the concentration values are decimals that are difficult to calculate, and it needs to be carefully experienced in the actual problem.

2. For those concentration problems with very clear proportions, we can use the special value method to avoid the appearance of fractions, so as to simplify the calculation steps.

3. For the result of mixing two solutions: the solute of a certain solution increases relative to the mixed solution; The other solution has a reduced solute relative to the mixed solution. Since the total solute does not change, the increased solute is equal to the decrease in the solute of the mountain search, which is the principle of the cross method.

The concentration is calculated by the solute mass of the solution 100%.

where solution mass = solute mass + solvent mass.

The amount of solute contained in a unit solution is called the concentration of that solution. The higher the solute content, the greater the concentration.

Calculation formula for concentration problem:

1. Solute quality, solvent quality, solution quality.

2. The weight of the solute is 100% concentration by the weight of the solution.

3. Solution weight, concentration, solute weight.

4. Solute weight, concentration, solution weight.

The concentration formula is c=n v.

The formula for calculating the quantity concentration of a substance is a formula used to calculate the quantity concentration of a substance. The content of the formula is: the amount of the solute The amount of the solute x the volume of the solution, that is, n = c · v, the formula is also called Moore's law.

1. At the same temperature and pressure: v1 v2 = n1 n2 = n1 n2.

2. Same temperature and same volume: p1 p2 = n1 n2 = n1 n2.

3. The amount of the same pressure and the same substance: v1 v2 = t1 t2.

4. The amount of the same temperature and substance: v1 v2 = p2 p1.

Commonly used concentration notations are:

Mass Percent Concentration (mass fraction, mm): Most commonly used. Refers to the mass of solute per 100 grams of solution, in grams.

Mass percentage = (solute mass (g)) solution mass (g)) 100% = solute mass (g)) solute mass (g) + solvent mass (g)) 100%.

Percent by volume (volume fraction, v v): Commonly used in alcohol. Refers to the volume of solute per 100 ml of solution, in milliliters.

Volume percentage concentration = (solute volume (ml) solution volume (ml)) 100% = solute volume (ml) (solute volume (ml) + solvent volume (ml)) 100%.

Parts per million (ppm): Refers to the mass of solute (in milligrams) per kilogram of solution.

Parts per million concentration = mass of solute (mg) mass of solution (kg).

The amount of substance in the original solution is.

The amount of chlorine-trapped ammonium substance is.

The literal weight of the substance in the solution is large.

Solution concentration =

The quality of the solution refers to what the liquid is.

The mass of the g-solute is.

The quantitative concentration of GH2SO4 is.

mol orange only l selection. a

Solution Mass =

g mass of solute =

The amount of the substance of GH2SO4 occlusion envy concentration car beat =

The source of mol l withered to a

aml=a/1000lb

ml=b/1000

The amount of nitric acid disturbing magnesium is a 1000 times 5mol l=a 200mol

Magnesium nitrate ionizes one magnesium ion and two nitrate ions in water.

Then the amount of nitrate substance is a 200mol multiplied by nucleus friend to 2 = a 100mol according to the amount concentration = the amount of substance divided by the volume of solution.

It can be calculated: A 100mol multiplied by b 1000L equals 10A B

This is the concentration of nitrate.

The concentration of sulfur ions in the saturated hydrogen sulfide solution is to avoid sulfidation and residual silver precipitation.

q<=ksp is required

Covered by Li Yunyi.

ag+] 2*The [ag+] corresponding beam should be the largest.

The level is limited. I hope you're on the list.

c 100 g of ammonia solution with a concentration of 10mol l The volume of 1000 L ammonia solution is (100) 1000 L

100 g of ammonia at a concentration of 10mol l The amount of solute is 10 * 100 ) 1000 = 1 mol

Set to add water x ml

1/ρ ÷x+(100/ρ)/1000) = 5x = 1 / (10ρ)

Because ammonia is less dense than water, i.e. 1, c is correct.