1 High School 1 Chemistry Topic A high school chemistry topic

15 In the laboratory, 480 ml mol l of copper sulfate solution is used, and 500 ml volumetric flasks are used for preparation. The following is correct for B

A Weigh g of copper sulfate and add 500 ml of water.

b Weigh g of gallium to make a 500 ml solution.

c Weigh g of copper sulfate and add 500 ml of water.

d Weigh g of gallium to make a 500 ml solution.

16 The following solution contains the same amount of potassium ion as 100 ml mol l of potassium sulfate solution, and the concentration of potassium ions is a b

A 50 ml moll potassium chloride solution.

b 200 ml moll potassium sulfite solution.

c 100 ml moll potassium nitrate solution.

D ml moll potassium bisulfate solution.

17 Add a sufficient amount of copper sheet to 50 ml of 18 mot H2SO4 solution and heat, after a full reaction, the amount of H2SO4 substance is reduced b

A is less than Mo B is equal to Mo.

c Between Mo and Mo d is greater than Mo.

cu+2h2so4=cuso4+2h2o+so2

19 Burning a compound composed of g sulfur and iron in oxygen to convert all the sulfur in it to SO2, and oxidizing all of these SOs to H2SO4. These H2SO4s can be completely neutralized with 20 ml mol L NaOH solution. Then the mass fraction of sulfur in the original compound is b

a．18％ b．36％ c．46％ d．53％

20 In an acidic solution containing FeCl3 and BaCl2, a sufficient amount of SO2 gas is introduced, and a white precipitate is observed, and the solution turns light green, from which the result CD can be obtained

A white precipitate is sulfur b white precipitate is a mixture of baso3 and baso4.

c The white precipitate is BaSO3 D FeCl3 is reduced to FeCl2 by SO2

First: The answer in the stem you give should be "H:C:O mass ratio 1:6:16" and the answer is formic acid (H2CO2).

Second: hybridization mode: central c atom sp2 hybridization: because the number of δ is 3 and the number of lone electron logarithms is 0, in the VSEPR model, the number of valence shell electron pairs ==δ + the number of lone electron pairs ==3, that is, there are 3 hybrid orbitals so the central c atom sp2 is hybridized.

Oxygen:sp2 hybridization in carbon-oxygen double bond: Because the number of δ is 1, the number of lone electron pairs is 2, the number of valence shell electron pairs ==δ number + the number of lone electron pairs ==3, that is, there are 3 hybrid orbitals, so the O atom sp2 is hybridized.

Single bond O: Because the number of δ is 2, the number of lone electron pairs is 2, the number of valence shell electron pairs ==δ number + the number of lone electron pairs = = 4, that is, there are 4 hybrid orbitals, so the oxygen atom sp3 is hybridized.

To put it simply, each bond of an atom is a single bond, which is sp3, and obviously h and one of the oxygen is sp3

A double bond is sp2 c and another o is sp2

Specifically, you can look at the hybrid theory.

The question is about hybridization.

kclo3 + mno2 — kmno4 + cl2↑ +o2↑

cl: +5 0, get 5e-

mn: +4 +7, lose 3e-

o: -2 0, lose 2e-

Since the K atom and the Cl atom in Kclo3 are 1:1, and the K atom and Mn atom in KMno4 are also 1:1, it can be seen that the Cl atom and Mn atom in the reaction must be 1:1.

cl gets 5e-, plus mn loses 3e-, and the total is 2e-; And o lose 2e-. So the ratio of cl : mn : o is 1:1:1.

After trimming: 2kClO3 + 2mNO2 = 2kmNO4 + Cl2 + O2.

Equation trim is no problem, you're doing it right.

In this equation, 2molKCLO3 gives 10 mole-, 2 molmno2 loses 6 mole-, and the 2molo atom of these two species loses 4 mole-, so the total number of electrons transferred is 10 mol, and the simultaneous formation of cl2 is 1mol. If there is a transfer of 3 mol of electrons in the above reaction, chlorine gas is generated. The answer is also correct.

The average molar mass of the mixture is 2g mol = .

Criss-cross method:

= -- This step is wrong, mistaking the molar mass of Cl2 for is, it should be 71g mol).

The amount of Cl2 and O2 in the mixture is 7:1553.

According to 2kclo3 = 2kCl + 3O2 and 2kClO3 + 2mno2 = 2kmNO4 + Cl2 +O2, the ratio of kclo3 participating in the reaction to the amount of kclo3 participating in the reaction is: [(1553 - 7) 2 3] :

The mass of potassium chlorate reacted according to the above equation accounts for the total mass of potassium chlorate reacted: 21 21 + 1546) 100% = .

You're smart, just be careful.

CO2 + Ca(OH)2==CaCO3+H2O2CO2+Ca(OH)2==Ca(HCO3)2 It can be seen from the equation that x:y<1 is precipitated, when it is greater than 1 and less than 2, the precipitate is partially dissolved, and when it is greater than 2, the precipitate is all dissolved without precipitation.

A True B False The precipitate will partially dissolve.

c to d to d.

When CO2 is excessive, it will continue to react with CaCO3 to form soluble Ca(HCO3)2, so the mass of the precipitate will be reduced instead of 100yg

When there is an excess of carbon dioxide, it will continue to react with the formation of calcium carbonate to form calcium bicarbonate that is soluble in water, which is a decrease in the precipitate mass.

Calcium hydroxide is introduced into carbon dioxide, which is precipitated into calcium carbonate, and excessive carbon dioxide will produce calcium bicarbonate. When x is greater than y, a certain amount of calcium carbonate will be dissolved to form calcium bicarbonate, so the amount of precipitation cannot be expressed as 100y g.

Solution: 2Na+O2=Na2O2 2Hgo===2Hg O2 (The reaction condition is heating).

x x= y y=

The idea of this question is mainly to explain that all oxygen generation is used to produce sodium peroxide according to the constant composition of air.

Hope for thanks.

Bubbles are formed, and a blue precipitate is formed.

It can be seen that sodium cannot replace the Cu element in the salt solution.

Analysis: When sodium is put into the copper sulfate solution, the sodium element actually reacts with water to produce sodium hydroxide and hydrogen gas, which produces bubbles because of the formation of gas.

The sodium hydroxide solution formed by the reaction of sodium and water will immediately react with the copper sulfate solution to form copper hydroxide precipitate and sodium sulfate solution, so there will be blue precipitate.

Because sodium is put into any solution, the first thing that happens is the reaction of sodium with water in the solution to produce sodium hydroxide and hydrogen gas, so sodium cannot displace the elements in the salt solution.

(colorless bubbles) and (blue flocculent precipitates) are generated, so it can be seen that sodium cannot displace the (copper) element in the salt solution.

Explanation: na threw herself into the water and na was lively (immediately!!) ) reacts with water to form NaOH and hydrogen, and NaOH reacts with CuSO4 to form Cu(OH)2 blue flocculent precipitate, instead of Na displacing copper sulfate with copper, so sodium cannot replace the Cu element in the salt solution!

The density is grams per cubic centimeter, which is g ml).

The mass of the solution is 100ml g ml 119 g, and the solute mass cvm (100ml 1000) is 12 mol l g mol g, so the solvent mass solution mass solute mass 119 g, which is the mass of water.

Whereas the density of water is 1 g ml, so the volume is.

In this problem, the sum of the separate volumes of solute and solvent is not equal to the volume of the solution because the concentration is too large. And HCl is a gas, and the volume in solution cannot be found.