A chemical calculation problem needs to be explained and analyzed in detail.

The mass ratio of hydrogen to carbon is greater than 1:6, that is, the ratio of the number of atoms of hydrogen to carbon is greater than (1 1) :(6 12), that is, greater than 2:1

The only hydrocarbons with H atomic number greater than 2 times the number of C atoms are alkanes. The general formula is CXH (2x+2).

Combustion formula: CXH(2x+2) +3x 2+1 2)O2 == XCo2 + X+1)H2O

The products of complete combustion pass through soda lime, which increases weight.

i.e.: +18(x+1)] = , and the solution is x = 3

Therefore, hydrocarbon A belongs to [alkanes] with the molecular formula [C3H8].

After combustion, it returns to its original state, and the water produced is in a gaseous state.

The equation for the complete reaction to produce CO2 and H2O: C3H8 + 5O2 == 3CO2 + 4H2O

The equation for complete combustion to produce CO and H2O: 2C3H8 + 7O2 == 6CO + 8H2O

It can be seen that when Co and H2O are completely combusted, the volume of gas after the reaction - the volume of gas before the reaction = (6+8)-(1+7) = 5 satisfies m-n = 5

Therefore, if m-n=5, the composition of NL gas can only be [CO and H2O], in this case, the combustion equation is: [2C3H8 + 7O2 == 6CO + 8H2O].

If it is always maintained before and after combustion, it means that the gas volume remains unchanged before and after the reaction.

The equation is: XC3H8 + YO2 == (10X-2Y)CO + 2Y-7X)CO2 + 4XH2O (trimmed).

So there is: 1 + y = 10x-2y + 2y-7x + 4x

This results in 7x = 1+y

So m-n = (2y-7x + 4x) x = (11x - 2) x = 11 - 2 x (since m-n corresponds to 1l of c3h8 combustion, divide 11x - 2 by the coefficient x before c3h8 in the equation).

When x = 1, the maximum value of m-n is .

If the mass ratio of hydrogen to carbon is greater than 1:6, the mass ratio of hydrogen to carbon is greater than 1 1:6 12=2:1, so the hydrocarbon can only be alkanes; This alkane is completely burned to produce, made of:

n=3 and the hydrocarbon formula is C3H8

2) At 120, 1LA and some oxygen are mixed and ignited, after combustion, it is restored to the original state, the volume of the mixed gas is ml, and the mixed gas is passed through the soda lime, and the gas remains NL. Since soda lime can absorb both CO2 (G) and H2O (G), the composition of NL gas can only be burned to generate 4LH2O (G), and M-N=5, so 1LCO2 is generated, and 2LCO2 is generated by the conservation of carbon mass, in this case, the combustion equation is: C3H8 + 4O2 = CO2 + 2CO + 4H2O

If 120 is always maintained before and after combustion, the maximum value of m-n is when there is no gas left, that is, when carbon completely generates CO2, so the maximum value of m-n is 3lCO2 + 4lh2o = 7

Known: c = mol l; ao =

Solution] -1: The percentage transmittance tConvert to absorbance a:

A = 2-LG100T = 2-LG20 = 2- = solution]-2: Solution concentration c = CAO) a = mol l (mol l

na2co3 + 2hci = 2naci + co2 + h20106---44

x = mass fraction of Na2CO3 in the mixture = =

Solution: Let the mass of alumina be m. The EDTA mass of the reaction with aluminum ions is 25*10 to the minus 3 power* and the minus 3 power* is equivalent to the EDTA solution.

Because the aluminum ions react 1:1 with EDTA, M = the percentage of alumina content is.

If it's not well written, the general idea is like this, I hope you can understand it!

Acid-base titration questions are not difficult, answer carefully, you need to be able to calculate acetic acid, buffer solution, and pH of alkali. This is the acid base.

This skill is examined, and attention is also paid to be sure to use balance.

Calculate, otherwise there is an error, and the chemical equilibrium is moving.

state equilibrium, the specific process is shown in the figure below.

Giving a ** may be clearer than saying.

(1) s=(ksp)^

2) First, find the distribution fraction of oxalic acid in solution when pH=δ=(ka1ka2) ([h+] 2+[h+]ka1+ka1+ka2)=10

ksp'=ksp/δ=

s=(ksp')^

3) Because the amount of oxalate is much greater than the amount required to form a precipitate, the solubility formula is:

s=(ksp)/[c2o42-]

In order to obtain the equilibrium concentration of oxalate, it is also necessary to use the distribution fraction.

When pH=, δ=(ka1ka2) ([h+] 2+[h+]ka1+ka1ka2)=10

c2o42-]=c×δ= mol/l

s=(ksp)/[c2o42-]=

The conditional stability constant of zny=lgkzny-lg y(h)ph=2

Conditional stability constant of zny=lgkzny-lg y(h)=ph=5

Conditional stability constant of zny=lgkzny-lg y(h)=