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The phenomenon of dropping and adding to each other is mainly related to the amount. Two different reactions may occur in one reactant from less to more, so the phenomenon will be different.

For exampleThe dilute sulfuric acid is added to the sodium carbonate solution drop by drop, at the beginning the sodium carbonate is excessive, and the sulfuric acid is insufficient, the reaction CO32- +H+ = HCO3-, phenomenon: no gas is generated;

If the sulfuric acid is added dropwise, the amount of sulfuric acid may become excessive, and the second step of the reaction H+ +HCO3- = H2O + CO2 will occur: there is a formation of colorless gas (colorless bubbles).

In turn, the sodium carbonate solution is added to the dilute sulfuric acid drop by drop, and the sulfuric acid is excessive at the beginning, and the reaction occurs 2H+ +CO32- = H2O + CO2, and the phenomenon: there is a colorless gas generated.

Add dilute sulfuric acid drop by drop to sodium carbonate solution:

There are no bubbles at the beginning, CO32- +H+ = HCO3-

Later, bubbles formed. h+ +hco3- = h2o + co2↑

Add sodium carbonate drop by drop to dilute sulfuric acid:

Bubbles are formed at the beginning. 2H+ +CO32- = H2O + CO2, zinc sulfate is added dropwise to the sodium hydroxide solution:

There is no obvious phenomenon at the beginning, Zn2+ +4oh- = ZnO22- +2H2O

Later, a white precipitate appeared. zn2+ +zno22- +2h2o = 2zn(oh)2↓

Add sodium hydroxide drop by drop to the zinc sulfate solution:

A white precipitate begins to form, Zn2+ +2Oh- = Zn(Oh)2

Later the precipitate dissolves. zn(oh)2 + 2oh- = zno22- +2h2o

The phenomenon is the same as .

Zinc hydroxide is also soluble in ammonia. Therefore, the detailed process is omitted.

1. Dilute sulfuric acid is added to sodium carbonate, and there are no bubbles at first, and bubbles will be generated after a while. The addition of sodium carbonate to dilute sulfuric acid immediately produces bubbles.

2. Zinc sulfate is added to sodium hydroxide, which first produces precipitation and then dissolves. Zinc hydroxide is an amphoteric precipitate that is soluble in strong bases. Sodium hydroxide is added to zinc sulfate to form a precipitate that is insoluble.

3. Zinc sulfate drops into ammonia, and there will be no precipitation at this time, because a large amount of ammonia exists to generate ligand ions, and when there is excess zinc, there is a white precipitate generated. When ammonia drops into zinc sulfate, a white zinc hydroxide precipitate is formed at the beginning, and when the ammonia is excessive, the zinc hydroxide dissolves again.

1) Dilute sulfuric acid drops into dilute sodium carbonate solution:

2Na2CO3 + H2SO4 = Na2SO4 + 2NaHCO3, 2NaHCO3 + H2SO4 = Na2SO4 + 2CO2 + 2H2O (the solution must be very thin, no bubbles at the beginning, and bubbles will start to produce after dropwise addition for a period of time).

2) Sodium carbonate solution drops into sulfuric acid:

H2SO4 + Na2CO3 = Na2SO4 + CO2 + H2O (bubbles at the beginning).

1) Sodium hydroxide solution drops into zinc sulfate solution:

ZnSO4 + 2NaOH = Zn(OH)2 + Na2SO4, Zn(OH)2 + 2NaOH = Na2ZNO2 (sodium zincate) + 2H2O (white precipitate is produced at the beginning, and after some time the precipitate begins to dissolve).

2) Zinc sulfate solution drops into sodium hydroxide solution:

ZnSO4 + 4NaOH = Na2ZNO2 + Na2SO4 + 2H2O, Na2ZNO2 + ZNSO4 + 2H2O = 2ZN(OH)2 + Na2SO4 (no precipitation at first, precipitation begins after a period of time, similar to aluminum sulfate).

1) Drop ammonia into zinc sulfate solution:

ZnSO4 + 2NH3·H2O = Zn(OH)2 + NH4)2SO4, Zn(OH)2 + 4NH3·H2O = ZN(NH3)4(OH)2 + 4H2O, ZN(NH3)4(OH)2 + NH4)2SO4 = ZN(NH3)4SO4 + 2NH3·H2O (precipitation at the beginning, followed by the formation of zinc ammonia ions and dissolution).

2) Zinc sulfate solution dropped into ammonia:

ZnSO4 + 4NH3·H2O = ZN(NH3)4SO4 + 4H2O (no precipitation at the beginning, precipitation begins after a period of time).

1.There is no phenomenon when dilute sulfuric acid drops into sodium carbonate, and sodium bicarbonate is produced, and bubbles are produced by continuing to be added dropwise. Sodium carbonate drops into dilute sulfuric acid and immediately produces bubbles.

2.There is no phenomenon of zinc sulfate dropping into sodium hydroxide, and the complex is formed, and the precipitation will be generated if the droplet is continued. Reverse drip will produce precipitate first, and continue to dissolve the precipitate dropwise.

3.The phenomenon is the same as 2

The first should be bubble-producing, the second should be solid, and the third should be solid.

1. Ammonia and phenolphthalein reagents. Reason: Ammonia is volatilized by heating ammonia, and phenolphthalein turns colorless from red. After cooling, the ammonia gas redissolves and the phenolphthalein turns red.

2. SO2 and magenta solution. Reason: Colorless before heating, SO2 bleaches magenta. When heated, the SO2 volatilizes and the solution turns red. When the cold is gone, the SO2 dissolves and the magenta is re-bleached.

No, unless the reaction is reversible.

(1) Suppose that O2 has 1mol at the beginning

3o2====2o3

Start with 1 0 reactions.

Equilibrium, so the average molar mass m=(

2) 3O2====2O3 Volume difference 3 2 1x

x=2*, so ozone is 3l

3) The quality of the gas mixture is defined by the title.

The amount of gaseous substances mixed is.

So the average molar mass of the gas mixture m=

O2 32 840O3 48 8 So the volume fraction of ozone in the gas mixture is 50%.

(1) Suppose the original oxygen has 10 mol

3o2==2o3

30%×10 n

n（o3）=20%×10=2 mol

m=（32×7+2×48)÷（10-3+2）=(2) 3o2==2o3 δv3 2 1v

v = 2 (3) n (gas) =

n(o)=(

Let oxygen have x mol and ozone y mol

Available: x+y= 2x+3y=

Solution x=y=

v(o3)%=

m: vertical brigade 4o: 8 * 1 8 + 12 * 1 round fiber burning 4 + 6 * 1 2 + 1 = 1 + 3 + 3 + 1 = 8

Sola's hypocrisy is mo2

I forgot the formula.

(1) The amount and concentration of the residual chlorine substance is .

710mg/

2) b3)1.Chemical Equation:

5nac1o2 + 4hcl = 5nacl + 2h2o +4 clo2

Ionic equation: 2 clo2 + 10 i-1 + 8 h+ = 5 i2 + 2 cl-1 + 4 h2o

2.2 clo2 --5 i2---2 s2o32 - So the amount of sodium thiosulfate consumed is the amount of chlorine dioxide produced, n=cv mass fraction w=mn m x 100%=103cv m x 100%=

(1) The chemical formula of liquid chlorine is Cl2, and the formula amount is 71

n=m/m c=n/v=(m/m)/v=(m/v)/m=710/71=10mol/m3=10÷1000=

Explanation: Quantity concentration of a substance = mass concentration of molar mass. where the relative molar mass is equal to the formula].

Therefore, the amount concentration of the substance of residual chlorine is .

2）b3）5naclo2+4hcl=4clo2↑ +5nacl + 2h2o

2clo2+10ki+4h2so4=2kcl+4k2so4+5 i2+4h2o

2clo2+10i- +8h+ =2cl- +5 i2+ 4h2o

The last question is counting.

（1） (2) b

3)1.Chemical Equation:

5nac1o2 + 4hcl = 5nacl + 2h2o +4 clo2

Ionic equation: 2 clo2 + 10 i-1 + 8 h+ = 5 i2 + 2 cl-1 + 4 h2o

2. 2 clo2 --5 i2---2 s2o32-

From the title, x::s, z:n, w:

1) Second period, IVA group 2) C 2 H2SO4 (concentrated) = CO2 2 SO2 2 H2O3) 2Na2O2 + 2CO2 2Na2CO3 + O22Na2O2 + 2CO2 = 2Na2CO3 + O2 analytical valency,

Two oxygen (-1 valence) in each sodium peroxide participate in the reaction, one of which becomes an oxygen atom in oxygen (0) and the other becomes an oxygen atom (-2 valence) in sodium carbonate

Therefore, according to the conservation of valency, the electron transfer is only based on the number of rises of the party with increased valency (or only the number of decreases of the side with decreased valency).

4) Add hydrochloric acid and BACL2 solution, if a white precipitate is produced, then 1 electron transfer per sodium peroxide.

78 Na2O2 reacts with a sufficient amount of CO2 and the number of electrons transferred is 1Na, then the gas is formed when it is transferred.

Second Cycle IIA Family.

C + 2H2SO4 = (heating) = CO2 + 2SO2 + 2H2O Add hydrochloric acid acidified barium chloride, if there is a white precipitate.

The second cycle is the ivth main family.

2H2SO4 (concentrated) + C== CO2 +2SO2 +2H2O (to be heated).

The addition of BACL2 yields a white precipitate.

(1) The ratio of O2 and C and S to form CO2 and SO2 gas is 1:1, so the amount of O2 is equal to the amount of mixed gas, which is 2mol

2) The mass and quantity of gas (a) are only the sum of the other two gases, so the relative molecule of a is between the other two gases, so gas a is CO2

3) The average molar mass is equal to the molar mass of CO2, which is equal to 44g mol (4) The ratio of the amount of remaining O2 and SO2 is 5:3 by the cross method, and if O2 is 5xmol, then SO2 is 3xmol, CO2 is 8xmol, and there are 3x+5x+8x=2mol, so x=1 8mol

n(co2)/n(so2)=8x：3x=8:3（5）q（co2)=

q(so2)=

1mol of sulfur is completely burned and emits heat q=

c+o2=co2

s+o2=so2

Regardless of whether oxygen is surplus or not, the volume of the gas mixture is extremely 2mol, so there is a volume of 1, the total volume of O2 under standard conditions.

2. SO2mol (isobaric change, volume unchanged, average molar is 2mol:1 (it is known that the mass and quantity of a gas (a) in the gas mixture after the reaction are the sum of the other two gases) 5

From the meaning of the title and the chemical equation, it can be seen that the amount of oxygen consumed is the same as the amount of SO2 and CO2 produced, so the amount of oxygen is also 2 moles, assuming that A is CO2, the mass of a gas (A) and the amount of matter are the sum of the other two gases, C(S) + O2 (G) = CO2 S(S) + O2 (G) = SO2

x x y y let the amount of the substance remaining O2 be z and a be CO2

x=y+z , 44x=64y+32z

Then the ratio of the amount of CO2 to SO2 is 8 to 3, the amount of residual oxygen is 5, and the amount of total oxygen is 8+3+5=16

The average molar mass is (8 16x2x44+3 16x2x64+5 16x2x32) 2=49mol l

multiply by 8 16x2mol) divided by 3 8 = 537kj

x is the first main group element, and the relative atomic mass is a

x2o + h2o===2xoh

2a+16 2(a+17)[

a=39 then x is potassium, y is in the same group as x, and y is sulfur.

ZO2 (2 is the subscript) The amount of the gas is a substance: , the molar mass of the substance is: 16 divided by, and Z is sulfur.

The highest oxide of this main group compound is X2O (2 is the subscript) indicating that it is the first main group element, and it is known that it is Naoh, so that X is potassium.