Who told me how to do these chemistry problems

1) The resulting white precipitate may be BaCO3 or CaCO3Ba2 + CO3 2 -==BaCO3

Ca2+ +CO3 2-==CaCO3 2) The reaction of the disappearance of the precipitate is .

BaCO3+2H+==Ba2+ +CO2 +H2OCAco3+2H+==Ca2+ +CO2 +H2O3) precipitated when agno3 solution was added to the acidic solution, indicating that there was Cl- in the solution.

cl-+ ag+==agcl↓

Therefore, there must be K2CO3 and CaCl2 in the original white powder, and there may be Ba(NO3)2

As can be seen from (1).

There is a white precipitate produced. This indicates that Ba(NO3)2 and K2CO3 react. After the two substances form a solution, barium ions and carbonate ions combine to produce precipitated BaCO3, in the same way. It may also be that Ca and CO32- react to form CaCO3

The ion equation is hard to play.

Ba2+ (barium ion) + CO3 2- (carbonate ion) == BaCO3

Ca2+ (calcium) + CO2- (carbonate ion) ==CaCO3

2) The hydrogen ions in dilute nitric acid react with BaCO3 or CaCO3. Converts carbonate ions into CO2 and H2O

2H+ (Hydrogen Ion) + CO3 2- (Carbonate Ion) ==CO2 + H2O

3) Ag+ in Agno3 to produce a white precipitate. (Because there is an excess of nitric acid in (2).) So it can be determined that all the carbonate ions are converted to CO2 and released, so the precipitate produced now is insoluble in nitric acid.

There are only two middle schools in the junior high school range. Because of the presence of AG+, we judged it to be AGCL).

AG+(Silver ion)+Cl- (chloride ion)==AGCL

To sum up, we can conclude that CaCl2 and K2CO3 must be present. ba(no3)2 does not necessarily exist.

There is CaCl2, K2CO3, and possibly Ba(NO3)2

It's best to write the ion equation by yourself, and it's very troublesome to use the computer to type, if you don't understand it, 934883137 add this q, as long as it's the knowledge below the second year of high school, it should be no problem k2co3

Must be barium nitrate, possibly calcium chloride. A lot of knowledge has been forgotten, so it is recommended that you take a look at the relevant knowledge.

1 8 120

159 159 to the 23rd power (2 3mol) to the 23rd power one-third of one to two ccbab

1. 200 grams of potassium nitrate solution at t, after evaporating 20 grams of water, 8 grams of precipitated crystals are precipitated, indicating that the remaining solution at this time is saturated solution, on this basis, 12 grams are precipitated after evaporating 20 grams of water, where 12 grams of solute is the maximum mass dissolved in 20 grams of water in saturated solution at t, and the solubility is the maximum mass dissolved in 100 grams of water at t, that is: 20g 12g = 100g s

s=60g2, saturated solution: m (solution) = p v20, mg (p v) g = s (100 + s) s = 100m (pv-m).

3. If you guess correctly, the question should be a solid mixture composed of sodium chloride and sodium nitrate.

Solution: Let the mass of sodium chloride in the reactant be X, and the mass of sodium nitrate generated is YagnO3 + NaCl = AgCl + Nano 3x Y

x/ x =

y = reaction mixture of substances for the amount of chaos, so m (sodium nitrate) =

So after the reaction, m (sodium nitrate) =

Post-reaction solution: m(solution) =

100g/40g=15g/m m=6g

So: 10 o'clock (40g—.)

s=80g

The two-step oxidation method of alcohol to form carboxylic acid refers to the oxidation of alcohol (-CH-OH) to aldehyde (-CH=O), and the re-oxidation of aldehyde to carboxylic acid (-COOH).

It can be seen that the alcohol that can form carboxylic acid must have two hydrogens on the carbon where the hydroxyl group is located, that is, there must be a -ch -oh structure.

The isomers of such alcohols are (without mirror isomerism):

n-hexanol: ch -ch -ch -ch -ch -ch-oh

2-Methylpentanol: ch-ch -ch-ch(ch)-ch-oh

3-Methylpentanol: ch-ch-ch(ch)-ch-ch-oh

4-Methylpentanol: ch-ch(ch)-ch-ch-ch-ch-oh

2,2-Dimethylbutanol: CH-CH-C(CH) CH-OH

2,3-Dimethylbutanol: CH-CH(CH)-CH(CH)-CH-OH

3,3-Dimethylbutanol: CH-C(CH) CH-CH-OH

2-Ethylbutanol: ch-ch-ch(ch-ch)-ch-oh

There are eight in total. Addendum: This question is actually a -ch-oh connecting an amyl group, and how many kinds of amyl groups there are, then there are as many kinds of alcohols.

K1=,K2=,K2=,K2=,Ignore secondary ionization according to the known pH,CH+=10 -4moll-,H2S= reversible=H++HS-,CH+ CHS-=10 -4moll-

1)2x50=2x14/(18+62)x mm=

2x50=14/(18+61)x n

n=2)56x2x m /120=56x3x n/232，m:n=45:58

4) Nxoy, 14x:16y=7:4, x:y=2:1, the chemical formula of oxide is N2O.

kg*5 35%=kg, 2kg*5kg, kg*35% kg.

2*x/120=56*3*y/232，x:y=45:58。

4. The chemical formula of oxide is NxOy, 14X:16Y=7:4, X:Y=2:1, and the chemical formula of oxide is N2O.

I don't know this kind of question.

1. According to.

g=△h-t△s；Subtract to get h and s respectively.

You can find it.

lnk=n*e mf, it can be calculated.

2、（-pt | i2(s) |2i-(aq) |fe2+（aq），fe3(aq) |pt(+)

e mf= (positive)- negative)=;

rgmθ=-nef=-2*。

fe+h2so4=feso4+h2

The mass of ferrous sulfate per day: 152 * 560 56 = 1520kg = tons.

The mass of 30% sulfuric acid is required: 98*560 (56*30%)=3267kg=ton.

koh+h2so4=k2so4+2h2oxx=

k2co3=

2koh+h2so4=k2so4+2h2o20*10% y

y=1 happens to be neutral, then sulfuric acid and potassium hydroxide are both inversely deregistered, and the previous data of 98 grams of dilute sulfuric acid is useless.

The last solid mass bit (

7. The consumed Hno3 is converted into Cu(No3)2 and NoX, so there is:

n(hno3) consumption = 2n(cu) + n(nox)2*+

8、.(1) Analysis: N in nitric acid is reduced to None or None 2, and the amount of N is 1 to 1

Amount of reduced nitric acid = n (gas) =

2) Analysis: The mass fraction of silver is required, and the total mass of copper and silver alloys is known, so it is also necessary to know another relationship between copper and silver. When the amount of nitric acid is excessive, all copper and silver are converted into the corresponding nitrate, so it can be found according to the amount of nitric acid.

n (excess nitric acid) = n (h+) = 1

Conservation according to n:

The amount of nitrate in nitrate = n (total nitric acid) - n (excess nitric acid) - n (reduced nitric acid) =

Suppose: The quantities of copper and silver substances in the alloy are x and y respectively

cu---cu(no3)2---2( no3ˉ)x---2x

ag---agno3---no3ˉ

y---y2x+y=

64x+108y=30 ..

Mass fraction of silver = 36%.

These are the things I found in it.