Ask for the topic of the second semester of the first year of high school in chemistry! Additional p

What is the order in which the following oxides are listed in order of increasing acidity of the oxygenated acids they form?

What is a hydroxide with a basic strength between KOH and Mg(OH)2?

First question.

According to the position of these four elements in the periodic table, we can deduce their oxidation Knowing their oxidation, we can know its most ** acidic element of oxygenated acid The position of the element is.

c si p s

According to the top-down oxidation of the elements and the main group, the oxidation is weakened, so c>si is oxidized from left to right in the same period, so s>p>si

So the answer to this question is b

The second question is also a problem with the periodic table.

na mgk These are the three elements of the first and second main groups.

According to the above, the top-down reducibility of the same main group is enhanced, and the reducibility of the same period is weakened from left to right, and it can be known that the reducibility of Na is weaker than K and stronger than Mg.

So the answer to this question is A

From these two questions, it can be seen that the acidity of oxygenated acid depends on the oxidation of the anhydride, and the alkalinity depends on the reducing property.

In fact, this kind of question is mainly to examine the memorization of the periodic table, and can memorize the short period of the periodic table and the first 20 elements, and this kind of question is very easy to do even if it comes out.

Upstairs is right!

Being the periodic table is the most basic! It's been since the third year of junior high school! High school needs more!

1.(NH4)2SO4 + CA(OH)2 = (heating) 2H2O + CASO4 + 2NH3 NH3 +H20 =

BaCl2+(NH4)2SO4=BASO4(precipitation)+2NH4ClagNO3+NH4Cl=AGCL(precipitation)+NH4NO32 agcl

Four packets of powder are dissolved in water, and four test tubes are taken, and a small amount of four solutions are added respectively.

1.Add BACL2 dropwise, there is precipitation production, filtration, precipitation dropwise nitric acid, there is gas released is (NH4)2CO3, insoluble is (NH4)2SO4

2.Add Na2SO4 dropwise, and there is a precipitate insoluble in nitric acid is BACL23Add AgNO3 dropwise, there is a precipitate, insoluble in nitric acid is KCL, and there is no reaction in Ca(NO3)2

1) Experiment 1: NH4+ +OH- = NH3 +H20 NH3 +H20 = alkaline).

Experiment 2: Ba2+ +SO42- = BaSO4 Experiment 3: AG+ +Cl- =AGCL

2) The chloride ions in the solution react with the silver ions in the silver nitrate to produce precipitation.

3) KCL: Cl- was tested with silver nitrate, and potassium ions were reacted with flame color; (NH4)2SO4: strong alkali and slight heat for ammonium ions, soluble barium salt for sulfate; (nh4)2co3：

Ammonium ions with strong alkali and slightly heated, carbonate with soluble calcium salt, hydrochloric acid and clarified lime water;

s~so2~caco3

m(s) = 1000000*1% = 10000g so n(s) =

So n(caco3) =

The amount of limestone = 100* = 31250g = calcium sulfate can be obtained after the reaction = 136* = 42500g =

(1)(112 56):1 2)=4:1(2) 1 part is from the reaction between Fe and H+, and 3 parts are reduced products of Fe3+.

Note: The amount ratio of Fe2O3 to Fe in the original mixture is 1:2Fe half reacts with Fe3+ (Fe2O3 and dilute sulfuric acid reaction) to obtain Fe2+, and the other half reacts with dilute sulfuric acid to form Fe2+ and H2

The answer upstairs is very good, but the question "just enough to neutralize hydrochloric acid containing grams of HCl" should be changed to "just enough to neutralize hydrochloric acid containing HCL".

So that the relative molecular masses of a, b, and c are respectively . i.e. lioh naoh koh

Question 1, according to the meaning of the question, it can be known that the mass before and after the reaction is conserved, and the state of matter is its question. According to density = mass

volume, the volume after the reaction is 5, the density is d, and the volume before the reaction is 2, because the mass does not change, the density of a is .

2.Let the molecular weights of A, B, and C be 3x, 5x, and 7x, respectively.

Let the quantities of a, b, and c be 7n, 5n, and 3n, respectively.

Exactly neutralized, the amount of hydrogen ions is equal to the amount of hydroxide ions.

And because acids and bases are both monobasic acids and monoalkalis.

The amount of substance of hydrogen ions =

The amount of hydroxide ion of the substance is 15n =

It can be found that n = has a mass relation 3x*7n + 5x*5n + 7x*3n = 67nx=

Substitute the n-value.

x=292

The relative molecular weight of a is 3x=3*292=876

The rest is 5x=1460, 7x=2044, and so on

Select D, when it is an insulator, only Fe reacts with the solution CuSo4, Fe+CuSo4=FeSo4+Cu, iron replaces copper, and the mass of the replaced copper is greater than that of iron, so the B end is lower.

When it is a conductor, Fe and Cu and CuSO4 constitute a galvanic cell, Fe is the negative electrode, Cu is the positive electrode, and CuSO4 is the electrolyte solution, and the reaction that occurs at this time is also Fe + CuSO4 = Feso4 + Cu, but at this time Fe is the negative electrode and loses electrons, and the electrons flow along the conductor to the positive electrode, and the Cu2+ ions in the solution are at the positive electrode, that is, the Cu pole gets electrons, becomes Cu, and attaches to the surface of the positive electrode, so that the A terminal is low.

Analysis: (1) When the lever is an insulator, only the Fe pole reacts with the solution CuSO4, Fe+CuSo4=FeSo4+Cu, 1molFe generates 1molCu, and the mass of 1molCu is greater than the mass of 1molFe, so at this time, the B end sinks, that is, the B end is low.

2) When the lever is a conductor, Fe and Cu and CuSO4 constitute the galvanic cell, Fe is the negative electrode, Cu is the positive electrode, CuSO4 is the electrolyte solution, the battery reaction is Fe + CuSO4 = FeSo4 + Cu, and the Fe electrode loses electrons due to being the negative electrode, and the mass decreases. The Cu electrode is positive, where the Cu2+ ions in the solution become Cu and attach to the surface of the Cu electrode, thus increasing the mass.

Therefore, at this time, the A end sinks.

To sum up, choose D.

Explanation: Choose B

Analysis: Neutralization titration to determine the concentration of acetic acid solution (starting total concentration) Determination of the concentration of H+ in acetic acid solution with pH test paper [C(H+) = ionized C(acetic acid)].

Ionization degree of CH3COOH = ionized C(acetic acid) Total initial concentration of acetic acid = C(H+) Total initial concentration of acetic acid.

If you choose C and look down, the reading you read is larger than the actual one; If you look up, the reading is smaller than the actual one, so the subtraction of the two readings is larger than the subtraction of the actual two real volumes. Therefore, the actual specific reading is less than 40ml

Option C Because the upward view is too large, 50ml, so the actual volume of the original solution is less than 50ml, and then because it is an upward view, the actual volume of the liquid poured is greater than 10ml, so the number less than 50 minus the number greater than 10 must be less than 40

When you choose C to look down, the actual reading is less, so it starts to fall below 50; When looking up, it is actually larger than the reading, so if it is higher than 10, it will be less than 40Draw a picture.

When reading from the top of the concave level, it is less than 50

When looking up at the concave level, it is larger than 10

Between the words and the outs (10, 40).