Chemistry Problems Let s Help Everyone Help Two chemistry problems

1. Add silver nitrate solution to generate white precipitate, add dilute hydrochloric acid precipitate and do not dissolve, and the presence of chloride ions can be determined.

Add dilute hydrochloric acid to introduce chloride ions, and the results are of course inaccurate.

Barium nitrate solution is added to generate a white precipitate, and the precipitate does not disappear with dilute hydrochloric acid, which can determine the presence of SO42- ions.

There is nitrate, and the sulfurous acid heel can be oxidized under acidic conditions, so that the interference of the sulfurous acid heel cannot be ruled out.

2 plus naoh

Add ammonia. 3co2

NH3 reacts with H2SO3 to form sulfurous acid.

NO2 is dissolved in water to form HNO3 sulfurous acid oxide.

Cl2 also plays an oxidizing role.

Sulfurous acid produces sulfate ions under the action of H+ and NO3-. Phenol is removed with NaOH. The introduced gases are ammonia, chlorine, NO2, carbon dioxide and ammonia to produce SO32-, and chlorine and nitrogen dioxide to obtain SO42-.

The first question is based on the law of conservation of mass, which can be found every 10g

A reaction consumes 2 g

b, so, a with b

The mass ratio is 5:1, so choose B

The second question is based on the law of conservation of mass, which can be found every 30g

aReaction generation.

40GD and 6G

c.So according to the proportional relationship, there is 10g

d generates and becomes there.

c is generated, so D is selected for this question

(1) Fe3+-3Cl- so FeCl3 in m(Cl-) g.

56 so kcl in m(cl-) grams.

28 m(cl-)

k+ -cl- so m(k+)=39*g.

m(k+)2) n(cl-)=71 in the original solution, n(cl-)=2 4= in 250ml of mixed solution

So after dilution, n(cl-)= so c(cl-)=

Fe (3-valent) has a mass of 28g for moles.

Cl has a mass of 71g and 2 moles.

Because of FeCl3, the ratio of Fe to Cl is 1:3, and the remaining moles are matched with K+.

The mass can be obtained by the formula of mol * k+.

The concentration of Cl in the original solution is 2 mol liters, and now 250 ml is diluted to 1 liter to 2*250 1000, and the result is calculated.

2Al+3H2SO4=Al2(SO4)3+3H2 The balance is still balanced after the experiment, indicating that the mass change of the left and right sides is equal, that is, the amount of H2 escaping from the left and right ends is equal.

According to the equation, it can be seen that in order to make the production of H2 equal, then the ratio of iron to aluminum = 3:2, so iron has (3a*56) g, aluminum has (2a*27) g, because iron and aluminum-copper alloys have equal mass, copper has (3a*56)-(2a*27)=114a g

Mass ratio of aluminum and copper = 54a:114a = 9:19

2.If the pure metal reacts with the acid to generate 1 g of hydrogen, Zn 32 5 g is required, mg needs 12g, Fe needs 28 g, Cu does not react with dilute acid, because the original mixture is 20g in total to generate 1g of hydrogen, and the mass of the original mixture required should be greater than 20g, less than 20g, or a mixture of less than 20g of metal and copper that does not generate hydrogen, so C should be selected

Too long lazy answer.

But you don't have to bother to bring it up.

Then I'll answer it for you.

Tell you the process and count it yourself.

1 Equal mass of iron and aluminum-copper alloys, why the balance is balanced, indicating that the two of them emit equal gases.

In aluminum-copper alloys, copper does not react with dilute sulfuric acid.

That is, iron and aluminum emit the same mass of hydrogen.

After calculating the mass ratio of iron and aluminum, if it is a:b, the ratio of iron to copper is a:a-b2, and this problem is not troublesome.

You can set the average relative molecular mass to release 1g of hydrogen

They are all +2 valence metals, and the average relative molecular mass is 40, and a relative molecular mass greater than 40 is less than 40.