A sophomore chemistry problem of metals... Hurry, hurry!

fe + 2h+ = fe2+ +h2↑

Fe + 2Fe3+ = 3Fe2+, K(SCN) is guaranteed to be all Fe2+, no trivalent.

2Fe2+ Cl2 = 2Fe3+ +2Cl- The divalent iron of this reaction is the sum of the first two reactions.

If you've done it so far, the basics aren't bad.

In this way, there is fe as well.

That is, when the alloy is not oxidized: Fe is, from the inside, subtract the excess mass (i.e., the part of the O element of Fe2O3) that is oxidized

That is, the total mass of the alloy that has not been oxidized, with which the mass is removed. Get results.

(1) Set x grams of sodium metal.

2na+2h20===2naoh + h2x(g) 36x 46 80x 46 to solve the equation (80x 46) :(100-36x 46)=22:100 (2) from the reaction formula 2na+2h20===2na+ h2, there will be na+, which consumes water.

So it turns out that the water should have, i.e. grams.

From the reaction formula, it is calculated that the overflow of hydrogen gas is immediate.

The mass of the solution is:

Analysis: A certain amount of iron, ferric oxide, copper oxide mixture into the sulfuric acid solution, after full reaction, the metal ions in the filtrate are only ferrous ions.

All 3 substances react with sulfuric acid, but Fe is more active than Cu and can be replaced with Cu from the solution, while Cu does not react with sulfuric acid, so the filter residue must contain copper.

Add 2 moles per liter of NaOH solution to the filtrate dropwise until 40 ml begins to precipitate, indicating that the sulfuric acid is excessive, the filter residue contains only copper, no Fe, and all Fe elements remain in the solution.

So, let the amount of sulfuric acid reacted with NaOH be x mol

h2so4 + 2naoh → na2so4 + 2h2o

x---2mol/l *

Solution: x =

The amount of ferrous sulfate is the same as the amount of sulfuric acid reacted with the mixture, let the amount of its substance be y mol

feso4 ～ h2so4

Solution: y = mol

So the amount concentration of FeSo4 substance = =

That's it].

It is better to do it according to the conservation of charge.

According to the sodium hydroxide added, after the reaction, the hydrogen ion in the solution = hydroxide ion = sulfuric acid after the full reaction The cation in the filtrate is ferrous ion and hydrogen ion Anion is sulfate ion, because the solution is acidic, so the hydroxide ion can be ignored.

According to the conservation of charge ferrous ions = sulfate ions[(hydrogen ions(0,08mol)]2

CFE=hehe, I haven't done chemistry for a long time, see if it's right.

Put a certain amount of iron, trioxane ferric oxide, and copper oxide mixture into the sulfuric acid solution, and wait until the full reaction, only ferrous ions are left in the filtrate.

All three substances react with sulfuric acid, but Fe is more active than Cu and can be displaced

Hydrogen generation is.

i.e. h2 is, h+ gets electrons is.

And these electrons come from metals.

If it's all na, it needs to be consumed, and the quality is.

All of them are mg, then they need to be consumed, and the mass is 6g

All AL, then it needs to be consumed, and the quality is.

All is zn, consumption, and the mass is 65*

Only so there must be zn to choose d

Sodium, magnesium, aluminum, zinc, write the equation, you know that the atomic weight of a molH2 is 46, 24, 18, 65

Hydrogen production i.e., using 12g of alloy produces 1molH2, and 48g of alloy is used.

Contrast: 46, 24, 18, 65

Sodium, magnesium, aluminum can produce 1molH2 in less than 48 grams, so zinc must be present.

Knowledge lies in the accumulation of bits and pieces, but also in unremitting efforts.

I wish you progress in your studies and go to the next level!

In addition, you rush the right rate, and please remember, thank you for your cooperation! (*

4--5 The amount of precipitation decreases, and the ions are the reductions of Al ions.

The Al ion --- 3OH ion becomes an Al(OH)3Al(OH)3---OH ion and becomes AL2 backwards to have 1 AL ion, so there is AL2(SO4)3, so 3 of the 4 OH ions are given to AL

1 part is to give mg ions, so there is a part mgSO4

So 1:1

The amount of OH- consumed is 5-4=1

al(oh)3+oh-=alo2- +2h2o1 1

al3+ +3oh-=al(oh)3↓

The amount of OH- consumed by Mg2+ is 4-3=1

mg2+ +2oh-=mg(oh)2

11 Al2(SO4)3 contain 2 Al3+, so the amount of Al2(SO4)3 is 1 2=

1 mgSO4 contains 1 mg2+, so the amount of MgSO4 is the ratio of Al2(SO4)3 to the amount of MgSO4 is::1

This question is a bit difficult to solve

The first step is to list the known conditions, Al2(SO4)3 and MGso4 are dissolved in water to produce three ions, AL3+, MG2+ and (SO4)2-, where AL3+ will be hydrolyzed AL3+ +3H2O = AL(OH)3+3H+ (you may not have learned this, because under normal conditions, AL(OH)3 is easier to exist in solution than AL3+, but the amount of AL(OH)3 is very low, It is not enough to form a precipitate to form a hydrolytic equilibrium, and strictly speaking, the equal sign should also be reversible).

The second step is to clarify the conditions, at this time, the KOH added dropwise to the solution will first react with the H+ produced by hydrolysis, thereby promoting the balance of Al3+ +3H2O=AL(OH)3+3H+ to the right to produce a large amount of AL(OH)3 to form a visible precipitate, and finally make AL3+ all converted to AL(OH)3, continue to add KOH dropwise will react with the remaining Mg2+ to form MG(OH)2 precipitate, and then add KOH dropwise will dissolve AL(OH)3. Needless to say, here.

Finally, according to the equation and chart you listed at the end, you will get that Al(OH)3+ consumes 1mol- from 4 to 5 in the abscissa, that is, the amount of Al3+ in the solution is 1mol, that is, the amount of Al2(SO4)3 is, so the 3molOh- of the abscissa 1 to 3 is the process of producing Al(OH)3, 3 to 4 is the process of producing Mg(OH)2, and according to Mg2+ +2OH-=mg(OH)2, we can know that the Mg2+ in the solution is, and finally the answer C can be obtained.

Since 1 volume OH dissolves all Al(OH)3, so the precipitate formed by 4 volume OH has three volumes and AL and one volume and MG, so the Al ion and Mg ion are 1 ratios, and the chemical formula of the solute is 1 to 1 Understand?

mg and oh are one to two and al is one to three. You should know this.

1、n(co2)=

m(c)=c%=2, the average molar mass of the mixture m(average)=

If m(m2CO3)=, then m(m)=

If m(mhCO3)=, then m(m)=

So the alkali metal element is potassium element, and the chemical formula of two carbonates: K2CO3, KhCO33HCO3-+H+=H2O+CO2

1 1xx=

n(hcl)=

c(hcl)=

%=(5*10^3*12*10)/

2.Let the salts be XHC3 and X2CO3

then the molecular weights are (61+x) and (60+2x).

As can be seen from the figure, there are 12 units of x2CO3, 8 units of xHCO3 3*10 -3mol 2*10 -3mol, 3*10 -3*(60+2x)+ 2*10 -3*(61+x)=x=39

are two salts of the K element.

c=8*10^-3/(32/1000)=

If you teachers don't understand, let me tell you. I was also a middle school chemistry teacher.

The energy emitted by x is greater than y

x is easier to get electrons.

Non-metallic: x y

b, d correct.

klm represents the electron configuration outside the nucleus, k is the innermost shell, the maximum is two, l is the second outer shell, m is the third layer, and the electrons in each layer are 2, the number of layers to the power of 2 n, but elements 1-20 are only 8 electrons in the third shell.

In this problem, one electron becomes the structure of a noble gas, so it is the seventh main group.

2) The arrangement of electrons outside the nucleus.

Electron number (n) 1 2 3 4 5 6 7

The electron layer symbol k l m n o p q

Summary: Extranuclear electron configuration law.

1.Layered arrangement.

2.The principle of lowest energy: Electrons are always arranged first in orbitals with lower energy.

3 Other Rules:

1) The maximum number of electrons that can be accommodated in each layer is 2n2.

2) The number of electrons in the outermost shell does not exceed 8 (when the K layer is the outermost shell, the maximum number does not exceed 2), the number of electrons in the subouter shell does not exceed 18, and the number of electrons in the penultimate shell does not exceed 32.

Example: Determine the electronic configuration of the atom of the ca element

Once we know the number of nuclear charges and the arrangement of the electron shell of an atom, we can draw a schematic diagram of the atomic structure.

2. Periodic changes in the configuration of electrons outside the nucleus.

1) The electron configuration outside the nucleus changes periodically.

The second period: the atomic number increases from 3 to 10, the outermost electron outside the nucleus increases from 1 to 8, and finally reaches the stable structure of 8 electrons.

The third period: the atomic number increases from 11 to 18, the outermost electron outside the nucleus increases from 1 to 8, and finally reaches the stable structure of 8 electrons.

Fourth Cycle: The fifth cycle and so on occur in a cyclical manner, which we call cyclical changes.

Conclusion: With the increase of atomic number, the outermost electron configuration of element atoms changes periodically.

b) Periodic changes in the radius of atoms.

2.The law of judging the size of the particle radius (one looks at the number of electron layers, the second looks at the number of nuclear charges, and the third looks at the number of electrons).

1) Whoever has more electron layers will have a larger radius.

2) The number of electron layers is the same, whoever has the more nuclear charge will have a smaller radius.

3) The number of electron layers and the number of nuclear charges are the same, whoever has more electrons will have a larger radius.