A beginner 3 chemistry problem. A chemistry problem for junior high school

Answer: A solution with dph=1 is an acidic solution and should be considered with the addition of hydrogen ions.

Koh MgCl2 cannot coexist in large quantities in A, and Koh cannot coexist in large quantities in acidic solutions.

The Cu(NO3)2 solution in b is colored, and the colorless transparent solution cannot be obtained.

Na2CO3 in C cannot coexist in large quantities in acidic solutions.

11 The answer is that there must be copper in the insoluble matter filtered out by b, and there may also be iron filtered out after the reaction stops, and then a thin iron sheet is added to the filtrate, and after enough time, there is no change on the iron sheet. The results indicated that the dilute H2SO4 reaction had been completed, and no Cu2+ was present in the filtrate.

12 The answer is c, zinc and copper.

Due to the addition of excessive zinc powder, the Cu2+ of copper chloride is replaced by Zn to Cu after full reaction, and the zinc powder is excessive, and the substances left on the filter paper are zinc and copper.

i.e. strongly acidic conditions. Alkali, precipitate dissolved in acid, reaction to produce gas can not coexist with acid.

A koh can not be acid-base neutralized.

The bCu ion is blue.

C Na2CO3 meets acid to form CO2

d Yes. After reacting with Fe and acid, CuSO4 and FeSO4 are insoluble, and there must be Cu. (If the acid is insufficient, there may also be Fe) There is no change in the insertion of Fe, that is, there is no displacement of copper. Therefore, there is no CuSo4 in the solution and no Cuo in the solid

Zn can replace Cu but not Al, and it is due to the excess of Zn.

So the precipitation is zn cu choose c'

9:d acidic solution, a, with hydroxide ions, b, with cu ions is green, c, with carbonate ions.

11:B, there is no reaction when adding iron to the filtrate, naturally there is no CuSO4 and H2SO4 in it. As for iron, there may be a surplus, or it may have just finished reacting.

12:c, excess zinc must be left, and the aluminum row in front of the zinc naturally cannot replace the aluminum.

The coexistence of ions mainly depends on whether they will react (the question talks about color should be considered separately), and there is a hidden H ion in question 9. So a, there is a koh and h reaction; b, copper ion blue; c. Carbonate and hydrogen ions react to form water and CO2, d correctly.

11. Question B, because a thin iron sheet is added, after enough time, there is no change on the iron sheet, which means that there must be no copper ions and sulfuric acid in the filtrate, otherwise there will be copper displacement or bubbles, and CD can be excluded. From the above, it can be judged that sulfuric acid preferentially reacts with Cuo (otherwise, no iron replaces copper), and there is no sulfuric acid left, so it is impossible to judge whether there is a surplus of iron powder, so B is selected

12. According to the order of metal activities, zinc can only replace copper, and zinc powder is excessive, so choose C

The solution is acidic, containing hydrogen ions, OH-C in A, and carbonate ions in A, and the solution is colorless. The copper ions in b are colored, and d is selected

Select c by the chemical equation of "carbon dioxide, carbon monoxide, carbon dioxide" is:

CO2+C=2CO (reaction conditions: high temperature), it can be seen that the carbon element increases to 2 times, and according to the carbon mass before and after the reaction, the carbon element in the precipitated calcium carbonate generated also increases by 2 times, so n = 2m

Remove the copper powder? After the correction hi me.

According to the title, the mass of the mass of hydrogen can be calculated according to the conservation of mass 7g+fe+2HCl=FeCl2+H2

56 2x56/x=2/

x = mass fraction of copper powder in the original mixture (

As the question: What is produced by the combustion of hydrogen.

Except AL or FE or CU

One copper powder in the mixture, what about the others? What is it?

According to the title, the mass of hydrogen can be calculated according to the conservation of mass 7g+

Mass of hydrogen generated: 7g+

But I don't know what the impurities are, so I can't continue to count them, and if you add them then I can complete the answer.

There is NH4HCO3 ammonium bicarbonate.

NH4)2CO3 ammonium carbonate.

nh4hco3

1.Introduction: Ammonium bicarbonate, also known as ammonium carbonate, is a carbonate with about nitrogen. It can be used as nitrogen fertilizer, because it can be decomposed into NH3, CO2 and H2O gases and disappear, so it is also called gas fertilizer. The raw materials for the production of ammonium carbonate are ammonia, carbon dioxide and water.

Ammonium bicarbonate is colorless or shallowly granular, tabular or columnar crystal, ammonium carbonate is no (sulfur) acid nitrogen fertilizer, its three components are nutrients of crops, do not contain harmful intermediate products and final decomposition products, long-term application does not affect soil quality, is one of the safest nitrogen fertilizer varieties.

2.Definition. Ammonium bicarbonate is a carbonate with the chemical formula NH4HCO3, a relative molecular weight of 79, and a nitrogen content of about 19. It can be used as a nitrogen fertilizer. The raw materials for the production of ammonium carbonate are ammonia, carbon dioxide and water, and the reaction formula is:

NH3+H2O NH3*H2O+ Calories NH4OH+CO2 NH4HCO3+ Calories The chemical equation for the reaction is as follows: NH4HCO3+HC*****NH4Cl+H2O+CO2 NH4HCO3== Heating==NH3+H2O+CO2 H2SO4+NH4HCO3===H2O+CO2 +NH4HSO4 (when there is an excess of ammonium bicarbonate) H2SO4+2NH4HCO3==2H2O+2CO2 + (NH4)2SO4 (in excess of dilute sulfuric acid) 2NH4HCO3 + 2NAOH (a small amount) (NH4)2CO3 +NA2CO3 2H2O NH4HCO3+2NAOH (mass) NA2CO3 H2O+NH3:H2O NH4HCO3+2NAoh==NH3 +NA2CO3+2H2O

NH4HCO3 or (NH4)2CO3 ammonium carbonate or ammonium bicarbonate.

You say junior high school chemistry that is "ammonia bicarbonate, ammonia carbonate", the chemical formula NH4HCO3, (NH4)2CO3

Urea is also known as carbodiamine, carbaamide, urea. It is an organic compound composed of carbon, nitrogen, oxygen, and hydrogen, also known as urea (homonym with urine). Its chemical formula is Con2H4, (NH2)2CO, or CN2H4O

There are two possibilities, a small amount of CO2, or an excess of CO2.

m(caco3) = 10g

1) A small amount of CO2:

co2 + ca(oh)2 == caco3 + h2om1(co2)..10g

The excess of M1(CO2) = 44*10 100 = 2) CO2 was obtained, and all Ca(OH)2 was first generated to CaCO3, and Ca(HCO3)2 was formed by reacting with part CaCO3

Ca(OH)2 energy sum * 44 = reaction to produce * 100 = 20g CaCO3

So there are 20-10 = 10g of CaCO3 that reacts with CO2 to form Ca(HCO3)2

co2 + caco3 + h2o == ca(hco3)2m'...10g

Solve m' = 44*10/100 =

So m2(co2) = + =

So the quality of the carbon dioxide that is introduced may be or.

Option B First of all, say that small A is because the 100g solution is saturated, and the added salt can no longer be dissolved, so it cannot become a part of the solution, so the quality of the solution remains unchanged.

Small B with a solution saturated and the concentration will not change.

Small C From the above content, it can be seen that the solution part is only the original 100g of saturated saline, so the quality of the solute remains unchanged.

Small d solubility is an intrinsic property of a substance, and under normal conditions, it will not change with the change of the external environment, so it is wrong.

In saturated salt water, no matter how you add salt (at the same temperature).

The quality of the solution, the quality of the solute, the quality of the solvent, the solubility, the concentration, and the amount and concentration of the substance do not change.

Therefore, add 10g of salt to 100g of saturated salt water and stir thoroughly. Since it has been saturated, the temperature will not be dissolved if it does not change, and the quality of the solution will not change, and the small A error. The concentration of the solution does not change, and the small B is wrong.

The quality of the solute in the solution does not change, and the small C is wrong. The solubility of salt does not change when the temperature does not change.

Choose B small A: because of saturation, NaCl can no longer be dissolved, the mass of the solution = water plus the mass of the original NaCl, small B: saturated solution concentration = solubility (solubility + 100), small C: the solute is not dissolved, = the mass of the original NaCl, small D: the temperature is unchanged, the solubility of salt is unchanged.

Answer B is wrong, since it is already a saturated solution, so the salt can no longer be dissolved, and the added 10 table salt will form crystals, so the mass of the solution changes, and the concentration does not change The solute mass changes, the temperature does not change, and the solubility does not change.

Little A's words must be wrong. Needless to say. The problem is to choose between B and D.