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1. Sodium hydroxide.

cuso4+2naoh=na2so4+cu(oh)2↓1.Add water to dissolve.

2.Excess Na2CO3 is added to allow Ca2+ to precipitate completely.

3.Filtration. 4.Add excess hydrochloric acid.

5.Evaporation crystallization, drying (excess HCl will volatilize during evaporation) Attached: If "stop adding Na2CO3 when the pH test paper shows neutrality", due to the dissolution equilibrium of solid CaCO3, there is still a trace amount of CaCO3 in the solid obtained by evaporation crystallization.

Add BACL2 first

CuSO4 + BACL2 = BaSO4 + CuCl2 BaSO4 precipitation.

Add na2CO3 again

Na2CO3+CuCl2=CuCO3+2NaClCuCO3 precipitation.

1.Add an appropriate amount of sodium hydroxide.

2.Add an appropriate amount of sodium carbonate.

3.Add an appropriate amount of barium hydroxide.

Chemical equations, please write them yourself, ok!

How to remove NaSO4 when it is doped with CuSO4 - add sodium hydroxide to generate copper hydroxide precipitate.

How to remove CaCl2 when NaCl is doped - sodium carbonate is added to form calcium carbonate precipitate.

How to remove CuSO4 when NaCl is doped - sodium hydroxide is added first to generate copper hydroxide precipitation; Then add barium chloride to form barium sulfate precipitate.

1. Sodium sulfate should be Na2SO4. Add Na2CO3, CuSO4 + Na2CO3--- CuCO3 (remember the down arrow, precipitate) + Na2SO4.

2. Same as above, add >> 2NaCl+CaCO3 (precipitation) 3, first add Na2CO3 to remove Cu ions, and then add BaCl2 to remove sulfate ions.

BaCl2 + Na2SO4 --> BaSO4 (precipitation) + 2NaCl or above, for reference.

1. Add sodium hydroxide to filter 2. Add sodium carbonate to filter 3. Add an appropriate amount of barium chloride and then add sodium hydroxide.

1) From the density is a multiple of hydrogen, it can be seen that the average relative molecular mass of N2 and H2 is 2*The ratio of N2 and H2 is found by the cross method.

Therefore, the volume ratio of nitrogen and hydrogen is 1:4, that is, n2 is and h2 is 2) From the pressure in the equilibrium vessel is multiple of the pressure before the reaction under the same conditions, it can be seen that the total amount of gas after equilibrium is reduced.

n2 + 3h2 = 2nh3 △v

So the amount of nitrogen in the gas mixture when equilibrium is reached is.

3) The conversion rate of nitrogen at equilibrium is .

1) Nitrogen and hydrogen have an average molecular weight of 2

It is easy to calculate the volume ratio of nitrogen to hydrogen by cross multiplication as 13:172)n2+3h2=2nh3 v

The relative molecular mass of KiO3 is m=39+127+16*3=214g, which contains i, and the mass is 127g

Add iodine.

So add m(kio3)=

The title is wrong, is the quality of potassium iodate that needs to be added, not the quality of sodium iodate?

m (potassium iodate) = g

The mass fraction of iodine in potassium iodate is: 127 divided by 214 equals 0.593, and then 25.4 divided by 0.593 and so on to 42.83 is the amount of potassium iodate added per kilogram.

1) 2h2o2 mno2==2h2o+o2 shirt model shout code Jane 17:9:8

2) 2H2O energized ==2H2 +O2 9:1:83) 2Na2O2+2CO2==2Na2CO3+O2 39:22:53:8

4) OR WILD 2HGO =2HG+O2 :16

2H2O2=2H2O+O2 is the catalyst on the empty number, and the gas has a directional deficit and the upward arrow is open.

2H2O=2H2+O2 gas arrow, and reaction conditions: 2Na2O2+2CO2=2Na2CO3+O22Hgo=2Hg+O2 equal sign listed heating symbol.

I don't know the amount of matter and I can't do the ball mass.

2H2O2 Condition Disturbance: Manganese Dioxide as Catalyst = 2H2O +O2 2H2O Condition: Energized Liquid = 2H2 + O2 2Na2O2 + 2 CO2 = 2Na2CO3 + O2 2Hgo Conditions:

Heating = 2 hg + o2

A is a funnel or a separating funnel (** is not very clear), while B is an Erlenmeyer flask, and the liquid contained in B device should be concentrated sulfuric acid, which is used to dry hydrogen, that is, to absorb the water mixed in hydrogen.

The phenomenon observed in the c-device is the gradual change of the powder from black to purplish-red.

The function of the E device is to absorb the water generated by the reaction of copper oxide and hydrogen, and if the E device is not connected, the mass ratio of hydrogen and oxygen in the experimental conclusion is greater than the theoretical value.

1. Long-necked funnel Erlenmeyer flask.

2. Concentrated sulfuric acid and dry hydrogen.

3. The powder turns from black to red.

4. Prevent water vapor in the air from entering the U-shaped tube than a very simple problem, just do more.

(1) a: long-necked funnel b; Erlenmeyer flasks.

2. Concentrated sulfuric acid and dry hydrogen.

3) The black cuo gradually turns red (cu).

4) Prevent the entry of external water vapor, which will affect the test results greater.

(1) A long-neck funnel, B Erlenmeyer flask.

2) Concentrated sulfuric acid desiccant absorbs water.

3) The black powder turns red.

4) The effect of isolating the outside air on the experiment is greater.

1. C2, purple litmus solution turns red.

3. Fe2O3 + 3Co = high temperature = 2Fe + 3Co2