Chemistry questions 2 in the first year of high school. Urgently. To the process

Idea: If there are 4 parts of chlorine atoms 35Cl and 37Cl in a ratio of 3:1, then the probability of 35Cl and 35Cl combining into molecules is 3 4*3 4, and so on 35Cl and 37Cl The probability of combining 35Cl and 37Cl is 3 4*1 4, and the probability of combining 37Cl and 37Cl is 1 4:

9: The probability that you calculated earlier was 75% correct, but the mass ratio of sodium chloride containing 37cl was wrong.

x=1*(37+23) [1*(37+23)+3*(35+23)], calculate the mass ratio x and then multiply it by 10, then the final mass is 10*x and is rounded to two decimal places).

In fact, it is very simple, chlorine 35 in the gas accounts for 3 4, chlorine 37 accounts for 1 4, the chlorine gas relative to molecule 70 is a combination of chlorine 35 and chlorine 35, the combination probability accounts for 3 4 * 3 4 = 9 16, the relative molecular mass of 72 is a combination of chlorine 35 and chlorine 37, the combination probability accounts for 3 4 * 1 4 = 3 16, the mass fraction of 74 is the combination of chlorine 37 and chlorine 37, the combination probability accounts for 1 4 * 1 4 = 1 16, so the ratio of the three is 9 16 : 3 16 : 1 16 = 9:

3:1, so choose C

probability. 75*。75=9/16

35cl=。75.

Satisfied, no calculator.

The application of conservation ideas is solved by a system of equations.

Let sodium carbonate x, sodium bicarbonate y, just complete reaction, molar hydrochloride.

2x+y=106x+84y=19

The precipitate comes from carbonate, that is, the mixture of carbonate ions has 2 moles, let sodium carbonate x, let sodium bicarbonate yx+y=2106x+84y=19

Raw solid: Fe

Residual solids: Fe, Cu are equal in mass.

Changed n(fe): n(cu) = 64:56 = 8:7 changed n(fe) = n(so4+).

n(h2):n(so4+)=(8-7):8=1:8

Solution: Cuo + H2SO4 CuSO4 + H2OFE + CuSO4 FeSO4 + CuFe: 56 Cu: 64

n(cu)= mol

m(cuo)= g

The mass of copper oxide powder is ( ) grams.

As for hydrogen (Fe + H2SO4 FeSO4 + H2 ) I think it doesn't matter at all = =;

Moreover. This seems to be the title of the third year of junior high school.

Looking only at the transformation of N, the amount of N2 substances actually involved in the reaction of N2 2NH3 2HNO3: 85%*200*1000 28=170000 28 mol

The amount of nitric acid generated: 2*N(N2)=170000 14 mol Mass: 170000 14 *63=765000g=765kg

40% nitric acid mass: 765 40% =

Solution: A hydrocarbon with this chemical property, presumed to be a homolog of benzene.

The relative molecular mass is 2x60=120

The amount of 12 g of the hydrocarbon is 12 120 =

The amount of the substance of hydrogen is:

The hydrocarbon energy is completely added to hydrogen, and the structure of benzene ring is also verified.

Therefore, it is confirmed that it is a homolog of benzene, and the molecular formula can be set to CXH2X-6, and 12X+2X-6 can be solved to obtain X

Therefore, its molecular formula is.

c9h12m=9

n=12 is completely added to give naphthenes.

c9h18，p

3.Solution: Let the mass of the aluminum in the mixture be x

2al + 2naoh + 2h2o = 2naalo2 + 3h254 3mol

x54:3= x=

So the mass of iron in the stove-pin mixture is:

The mass fraction of aluminum in the mixture is:

Therefore, the mass fraction of iron in the mixed and deficient free complexes is.