A chemistry problem A master, a chemistry problem... Come on

Use the criss-cross method.

The same is obtained with NaHCO3 g, Na2CO3 g.

So the ratio of the amount of matter between the two is 1:1

The masses are grams, grams, respectively.

A minimum of hydrochloric acid, i.e. 600ml, is required

Can't do such a simple chemistry problem? Then you've thought about it for the exam. Go back and look at the example problems in the textbook.

Since the hydrochloric acid is sufficient, the reaction produces CO2

CO2 = 19gnaHCO3 and Na2CO3

So let nahco3=x and na2co3=y

84x+106y=19

x+y=so x=,y=

Na has based on the last conservation you can know that the molar mass of Cl and the molar mass of Na are equal. So the molar mass of h is also .

600ml of hydrochloric acid is required

(1) First of all, the powder is colorless when dissolved in water, and it can be seen that there is no cuso4 cuso4

Secondly, there is no obvious phenomenon after adding dilute hydrochloric acid, and it can be seen that there is no na2co3 na2co3

2) May contain two remaining substances: NaCl and Na2SO4 NaCl Na2SO4

3) If A is AgNO3, then B is BA(NO3)2 (it can't be BACl2 because A is excessive).

Then the original powder is NaCl

If A is BA2(NO3)2 and B is AGNo3, then the original powder is Na2SO4

A can't be BACL2 because A is in excess.

1) Possibly.

Composition NACL, A: AgNO3 B: BA(NO3)22) possible.

Neither of the components Na2SO4, A:BA(NO3)2 B:AgNO3 can use BACL2 because chloride ions can have an effect on the detection with silver ions.

In the first case, the powder composition is Na2SO4

A is BA(NO3)2

b is agno3

First of all, through the question, no CuSO4 and Na2CO3 can be introduced, because the solution is colorless, and no gas is formed when excessive hydrochloric acid is added.

Assuming that the powder composition is Na2SO4, then A and the powder form a precipitate, then A is BA(NO3)2, and nano3 is formed after the reaction, which does not react with AgNO3, which is in line with the topic.

In the second case, if the powder is NaCl, then A is Agno3, which generates AgCl precipitate and Nano3 with the powder, and it does not react with Ba(No3)2.

2NaHCO3 = Heating = CO2 + Na2CO3 + H2OC2 + Ca(OH)2==CaCO3 + H2O The amount of sodium bicarbonate in the original mixture is x, and the amount of sodium carbonate is y.

According to the equation, the relationship between the quantities of substances can be found, the gas released by 2NaHCO3---CO2--- Na2CO3---CaCO3x x 2 x 2 is passed into sufficient saturated lime water, and the precipitated calcium carbonate is obtained after full reaction, i.e., x 2== So x== that is, there is sodium bicarbonate in the original mixture.

The solid mass obtained after heating reacts with a sufficient amount of hydrochloric acid, and the gas (, according to the conservation of carbon atoms, there is: x+y==, and because x==, so y==, that is, there is sodium carbonate in the original mixture.

In summary, the ratio of the amount of sodium carbonate and sodium bicarbonate in the original mixture is::2

na2co3+2hcl=2nacl+h2o+co21/100g/mol=

Therefore, the ratio of the amount of sodium carbonate and sodium bicarbonate in the original mixture is 4 to 1

1 For all acids, H+ can be ionized, and they have reacted with more reactive metals (MG, AL, ZN, Fe, etc.) to release hydrogen. can make purple litmus red, and can neutralize with alkali. For common acids, sulfuric acid, hydrochloric acid, etc. can remove deductive oxides such as rust (NaO, Al2O3, MGO, etc.) and can react with limestone (CaCO3) to release CO2

It is more acidic than H2CO3).

2 is not a quantity-related problem, so the equation is 2NaOH+CuSO4==Cu(OH)2+Na2SO4 (the precipitation sign is not indicated, the blue precipitate is Cu(OH)2).

3 (1) Nitrogen content is to divide the total relative atomic mass of nitrogen in ammonium nitrate by the relative molecular mass of ammonium nitrate, i.e.

14 2) (14 1+1 4+14 1+16 3) 100 =35 Ionization of NH4+ and NO3- Plants absorb the element N.

2) Because it contains carbon, it is an organic substance. Organic matter must contain C, but the one containing C is not necessarily organic matter (such as carbonate, CO CO2, CN- (cyanide), SCN- (thiocyanide), etc.) The ratio of the number of atoms of C, O, N, and H contained in the urea molecule is 1:1:

2：44(1)fe2o3+6hcl==2fecl3+3h2o

2)naoh+hcl==nacl+h2o

3) 2kmNO4==K2MNO4+MNO2+H2O (need to be heated).

H+ are both free-moving ions.

1，h+;It can make litmus solution red, can react with alkali, can react with metal oxides, can react with salts, and can react with metals before h (in the order table of metal activity);

2,2naoh+cuso4=cu(oh)2↓+na2so4；

3,35%；no3-；Carbon; 1:1:2:4

4,6hcl+fe2o3=2fecl3+3h2o;naoh+hcl=nacl+h2o;2kmNO4 = K2mNO4 + O2 + MNO2 (heating conditions).

and neutralize the bai water reaction.

Naoh is generated, and the yin is OH-

Then according to the reaction formula, the ratio of Na2O2 to H2O is DAO1:1, and the internal ratio of Na2O and H2O reaction is 1:1

That is, the amount of Na2O2 and Na2O react with the same amount of substance, and the amount of NaOH produced is also equal.

i.e. the last solution generated when the same solution, i.e. the anion concentration is the same because the anion is generated.

Sodium carbonate is soluble in water, the anion is CO32-, and hydrolysis produces OH- and HCO3- a little bit, so it is a little out.

The anion in sodium chloride is. So

1=2>3=4

Since C = N V, the volume does not change because the volume change of the solution is negligible, so the size of C depends on N1 and 2, both substances react with water, and the amount of water is excessive, so the amount of equal substances is generated. Because the amount of Na+ matter contained in both is equal, the amount of Na+ matter is conserved, and the amount of Na+ substance is. The concentration of oh- is also equal for.

In 3 and 4, the substance does not react with water, the anions are CO3- and Cl-, respectively, and the amount of the substance is, so the concentration is.

So (1) = (2) > (3) = (4).

This question is not difficult, the main points (1) master the conservation of matter (2) read the question accurately, it is anions (3) be proficient in chemical formulas.

Hope it helps.

It should be 5:1

1 Add 10 grams of each of A, B, and C and mix them to generate 3 grams of D and 9 grams of C. (It is explained that substance C does not react, and substances A and B participating in the reflection have a total of 3+9=12g--conservation of mass).

2 If you add another 40 grams of A. then b can be fully reacted. (It is explained that in order to make the B substance completely react, there must be 10 + 40 = 50 Ga substance, because there is 10 GB of substance and 10 Ga of material in the reaction, so the 10 Ga substance must be completely reacted, then there is 12 G-10 g = 2 GB of material to participate in the reaction).

a + b = c + d

9g 3g and then add 40g a, it means that a has reacted 10g at the beginning, then b is 3g + 9g - 10g = 2g

The mass ratio of A to B in this reaction is 10g:2g=5:1

Generation C means that C does not participate in the reaction.

It is necessary to add 40ga to prove that A reacts completely, and C and D are generated in total 12G, indicating that B participating in the reaction only reacts 2G, so A:B=10G:2G=5:1

4oh- +4e-=o2+2h2o

The increase in pH is essentially equal to the electrolysis of water with anode hydroxide and cathode hydrogen ion discharge. Therefore, the concentration of sodium hydroxide increases.

I won't repeat it here.

w>1, then the mixture, the substance that reacted with water for the first time has a surplus.

w=1, then it is a mixture or a pure substance, and the substance in the solid powder has a residue in the first reaction with water.

I choose a CD. It may be one element or several, and there is a surplus in excess.

This question should test the knowledge of solubility. "Put the WG solids in hot water again", isn't it mentioned here how much hot water there is? lz Let's take a look at the next topic. I can't do this with the conditions given now.

When it is mg, the problem is hard to do, and solubility ......