High School Chemistry, 10 points for two questions 10 points for Senior 2 Chemistry

1) C Reason: Rice soup contains starch, starch will only turn blue when it encounters elemental iodine, and the substance given by the title is potassium iodate (kio3), iodine is not a simple state, so it will not change color, to identify whether salt is iodized, potassium iodide (ki) solution can be added to a small amount of salt solution, and a small amount of rice soup can be added later. If it changes color, it is iodized salt, otherwise it is not.

2) BC How to judge whether the ion equation is correct or not:

a.First of all, check whether the reaction is in line with the objective facts, that is, there is no ionic equation for the reaction that cannot happen objectively.

b.For substances that are prone to ionization, it should be written as ionic. Substances with small solubility are not necessarily difficult to ionize, and substances with high solubility are not necessarily easy to ionize, which should be accumulated in daily study.

c.Check if the charge is conserved before and after the reaction.

For example: option a: the reaction between the aluminum sheet and the sodium hydroxide solution: al+2oh = alo2 +h2 There are 2 negative charges before the reaction, and only 1 negative charge after the reaction, and the charges are not conserved before and after the reaction, so it is wrong.

Option D: Copper sheet reaction with dilute nitric acid: Cu + NO3 + 4H+ = Cu2+ +No +2H2O There are 4 hydrogen ions before the reaction, there should also be 4 nitrate ions, copper nitrate is generated after the reaction, copper nitrate is a substance that is easy to ionize, two copper nitrate ions are generated after ionization, and two copper nitrate ions are canceled before the reaction, and there are two nitrate ions left before the reaction, which does not conform to objective facts, so it is wrong.

There is also the fact that the charge before and after the reaction is not conserved, which is also wrong.

This is my superficial conclusion, I hope it can help you, if you have any questions, you can send e-mail:

1) B, rice soup contains starch, starch turns blue when iodine turns blue when potassium iodate (kio3) does not change color.

2) C, do more questions, after getting used to it, you will understand that the flipping is those questions, don't worry!

Just remember the common ones.

For example, 2) the reaction between aluminum and caustic soda requires four hydroxides.

The magnesium in b also reacts with ho.

d You should memorize the coefficients of the chemical equation.

This is the most basic requirement for the college entrance examination.

bc first counts whether the electron is conserved or not, and then the element, so I have to write it out and look at it.

Heat and evaporate to half the volume;

When the hydrochloric acid solution evaporates, the solute HCl is basically volatilized.

Add a small amount of CH3Coona solids (still acidic after addition);

The solution becomes a mixture of CH3COOH and HCl, and when titrated with phenolphthalein as an indicator to phenolphthalein discoloration, the pH should be slightly greater than 8 and will not be affected by CH3Coona solids.

Nothing is changed.

So the answer is C ③

PS: If methyl orange is selected as the indicator, the amount of NaOH will be slightly reduced due to the presence of a small amount of CH3COOH, then > is selected

According to the essence of neutralization titration, the hydrogen ion is equal to the hydroxide ion in the end, and the first bottle is heated and evaporated, and the hydrochloric acid is volatile, so the hydrogen ion of the first bottle is reduced, and the addition of sodium acetate to the second bottle will not reduce the hydrogen ion, and after adding sodium hydroxide, it will eventually become sodium acetate and sodium chloride, if you are worried about the hydrolysis of sodium acetate is completely unnecessary, the hydrolysis is very weak, so the third bottle is equal to the second bottle.

Select C, according to the essence of neutralization titration, the last hydrogen ion is equal to the hydroxide ion, the first bottle is heated, the hydrochloric acid is volatile, the hydrogen ion of the first bottle is reduced, and the sodium hydroxide solution consumed is also reduced. The addition of sodium acetate to bottle No. 2 does not increase the concentration of hydrogen ions, so No. 2 and No. 3 are the same.

Choose D, don't listen to their "HCl volatilizes after heating", this question is not on this little trick at all.

This question is to test whether you know the method of titration, the corresponding number of moles of acid (H ion) is the same (concentration and volume are the same), so the NaOH that needs to be titrated is the same. over!

1.When Fe2 is generated

When the mass ratio to H2 is 112:1, the mass fraction of Fe is 56, and the relative auspicious molecular mass of H2 is 2, so the molar ratio of the two substances is n(Fe2

n（h2）=（112/56）：（1/2）=4：

1. Because a sufficient amount of dilute H2SO4 is added, a FeSO4 solution is formed after a full reaction, so neither Fe nor Fe2O3 remains, and because H2 is generated, it means that Fe is excessive, which is enough to convert Fe3

The ion is reduced to Fe2

Ion. Fe H2, Fe Fe2

fe2o3～2fe3

2Fe2 Let n(fe)=x, n(fe2O3)=y, so there is x2y=4

x=1 gives x=1

y=3/2m（fe2o3）：m（fe）=3/2×160：1×56=30：7

2.All the solid substances can be dissolved and collected into H2 under standard conditions, and the KSCN solution is added dropwise to the reaction solution without changing the red color. All indicate that there are no Fe3 ions in the solution. Fe will Fe3 in Fe2O3

The ion completely changes to Fe2

Ions i.e. FE quietly known as 2FE3

3fe2 is collected to H2 under standard condition

i.e. n(h2)=

From fe h2 to fe3

The remaining FE after the reaction is still there.

Let the Fe reacted with Fe3 ions be Xmol, which consists of Fe 2Fe33Fe22Fe3

Fe2O3, knowing that Fe2O3 is also Xmol; Hence 56x160x=

The solution is x=so (1) the mass of Fe in the original mixture is.

Fe2O3 has a mass of .

2) The amount of FeSo4 substance in the solution is:

（1）。Find the amount of SO2 first: formula: v vm=n. According to the equation Cu+2H2SO4=CuSO4+SO2+2H20, the amount of the substance of the oxidized copper is then multiplied by the molar mass of the copper.

2)400ml.The proportional equations are then listed according to the equations for SO2 and NAOH. The amount of substance to which sodium hydroxide is obtained is. Then divide the amount of NaOH by the amount of NaOH according to the formula C=N V. Get.

3) Divide 64g by 64 (the amount of the formula substance n=m m) to get the total amount of Cu added as 1mol, and then use the amount of copper that is oxidized in the first question) = the amount of remaining copper substance) and then according to the equation Cu + 4Hno3 = Cu (No3) 2 + 4 No + 2H2O list the proportional formula to get the amount of Hno3 substance as.

Evaluation: [This question is a simple question, if you are studying liberal arts, this question is required for the entrance examination, and it is even more necessary to read science].

cu + 2h2so4 ********** cuso4 + so2 + 2h2o

64g 2mol 36g

Therefore, if there is an excess of sulfuric acid, assuming that copper is a complete reaction (involving the ability of copper oxide after a certain degree of sulfuric acid reaction), after the consumption of sulfuric acid, the sulfuric acid concentration is about 12mol l (including the generation of water), so it is still concentrated sulfuric acid, so it can be considered that copper is a complete reaction.

So the generated SO2 volume is.

A little reminder.

The minimum amount of NaOH is to generate NaHCO3So it's 200ml and scared to pee when I see the first floor.

First of all, the consumption ratio of each substance = the ratio of change in concentration = the ratio of coefficients in the equation.

From the question condition, we can know that the rate is obtained by C, so the rate of A is the same as C, and circle 1 is correct. b according to the coefficient ratio its rate should be, circle 2 wrong. It can be seen that C is generated, so A is consumed, the conversion rate is 30%, and the circle 3 is wrong.

C generated, then B consumed, the original 2mol, so the remainder, concentration, circle 4 wrong. Pick B

5min generates w, the rate is, and the z rate is, so its coefficient ratio is 1:2, so n=1, choose a

The concentration changes from 20s to 20s, which is equivalent to 12s change. Now it is required to continue to decrease, because the concentration decreases, the rate will also decrease, so the time increases, so it should be greater than 12s, choose C

From the question, the rate of matter d is aq p, and the increase in the amount of matter = rate * volume * time = (aq p) * t * v, choose c

14,A w at a rate of .

Because the volume does not change, the concentration becomes smaller, the effective collision of the reaction decreases, and the reaction rate slows down by 16 C

Four questions? Hi me, answer o( o

It's the topic of chemical equilibrium!

In this kind of question, you can list all the conditions, before the reaction, after the reaction, and the mass of the reaction are all expressed, and they are clearly listed on the straw paper. The reaction rate or the amount of product generated is then calculated based on the amount of reactant consumed. The relationship between the reaction speed expressed by various substances depends on the coefficient of the reaction.

If it is a closed vessel, the change in volume before the reaction should also be considered, and the differential method can be considered. If you really don't understand it, you can give yourself an example, which is simple and easy to understand. Believe in yourself, you can.

1. The concentration of concentrated nitric acid is about 16mol l, so 1mol l nitric acid belongs to dilute nitric acid. The reaction formula is as follows:

8Hno3 (dilute) + 3Cu = 2 No 3 Cu (No3) 2 + 4H2ONo3 - the amount of the descending substance n =

Then the amount of substance that produces no is n=

1) The amount of substances that h+ decreases is n=

The concentration of hydrogen ions in the solution decreases.

2) The volume of the production of no is v= under the standard condition

The amount of HCl gaseous substances under the standard condition is n(HCl) = 10mol, and the mass of the solution obtained by dissolving in 635ml of water is: m = 635 + 10 * solution volume is v = 1000

Take the amount of HCL contained in 10ml solution as N2 (HCL) = 10 The amount of dilute hydrochloric acid obtained after dilution The concentration is: C2 (HCL) = Happy learning!

So easy is mainly about mastering the equations.