Question 32 in one chemistry, second question in chemistry question 19, and question 3 in question 2

The trivalent aluminum ion has reached its maximum ** and can no longer lose electrons, so it will not react with hydrogen ions. Metaaluminate reacts with hydrogen ions and, depending on whether there is an excess of hydrogen ions, it can produce aluminum hydroxide (not excess) or aluminum ions (excess). But metaaluminate does not react with hydroxide ions!

The aqueous solution of sodium metaaluminate is alkaline, which means that the solution contains hydroxide, so ah, they will not react As for the amphoterides, they are elemental aluminum, aluminum hydroxide and alumina, which can react with both acid and alkali.

Do trivalent aluminum ions react with hydrogen ions?

No, they are all the highest valence, and they are only oxidizing, although not strong;

Does metaaluminate react with hydrogen ions and hydroxide ions?

Metaaluminate and hydrogen ions are combined and then combined with a molecule of water to form aluminum hydroxide, which is an amphoteric compound that reacts with both hydrogen ions and hydroxide ions, both of which belong to neutralization reactions.

The previous ones are not reactive, because they are all electron-deficient ions, and the H ion cannot take electrons from the Al ion, because the trivalent Al ion is the most **...

The latter metaaluminate is neutral, and the characteristic of AL is that it can react with acids and bases, and metaaluminate can react with H ions and OH ions! As for the product of the reaction, hehe, I don't remember much...

①n1(hcl) = 2 * / 1000 =

n2(hcl) = 2 * 45 - / 1000 =

Let the Daxi contain n(oh-) = xmol and n(CO32-) = ymol

Because: oh- +h+ = h2o

co32- +h+ = hco3-

There is an equation as follows: x + y =

Because: HCO3- +H+ = CO2 +H2O

There is an equation as follows: y =

Solve x = y =

n(oh-) n(co32-) = 16 : 1

The solution contains Mg2+ and Al3+, and after adding excess NaOH, Mg2+ precipitates into Mg(OH)2, and Al3+ precipitates first and then dissolves: Al3+ Al(OH)3 AlO2-, so the precipitate is pure Mg(OH)2

n(mg(oh)2) = / 58 =

n(h) = * / 1 =

Analyze the elements contained in the salts: Al, Mg, C, H, O

Let n(c) = x mol n(al) = y mol n(o) = z mol

Conservation of valence charge according to atomic:

3n(al) +2n(mg) +4n(c) +n(h) = 2n(o)

3y + 2 * 4x + = 2z ①

Conservation of charge according to atomic clusters:

3n(al3+) 2n(mg) = n(oh-) 2n(co32-)

3y + 2 * = 16x + 2x ②

Calculated according to salt quality:

27y + 24 * 12x + 16z = ③

Simultaneous three-formula solution:

x ≈ mol

y ≈ mol

z ≈ mol

then c(al) : c(mg) : c(c) : c(h) : c(o) = 2 : 6 : 1 : 24 : 23

That is, Al2mg6ch24O23

I'm dizzy, you've hit too much.

How much less did you hit?。。 How?

a The ratio of the relative atomic masses of an element is equal to the ratio of the actual mass of the atom. Since the relative atomic mass of the C-12 atom is 12, and the mass of the atom is twice that of the C-12 atom, its relative atomic mass is 24

In order of atomic masses C12, H1, O16, N14, then the stupid mass fraction of nitrogen is 14 (12*3+1*5+16+14)*100%=

In order of atomic mass C12, H1, O16, N14, then the mass fraction of nitrogen is 14 (12*3+1*5+16+14)*100%=

There are so many questions in these three questions, only 20 rewards will be given, who will help you do it.

1.Explain that the ratio of the amount of matter before and after the reaction is 3:1

Then the methane is completely burned up and the water produced is liquid CH4(G)+2O2(G)=CO2(G)+2H2O(L).

Reaction heat: auspicious crack noisy mol2The methane combustion in the second question produces the source clear heat released by water vapor, but now under standard conditions, the generated water is liquid and cannot be calculated.