Simple chemistry in the second year of high school, chemistry questions in the second year of high s

Both oh- and h+ come from water.

The amount of the two is consistent with the ionic product constant of water. One increases, and the other naturally decreases.

CH3COONA and H2O together, CH3COONA will ionize into CH3COO- and NA+ in water, at this time there is less ionization in water H+ and OH-, you see, now there are only these four ions in the solution, so OH- ions are **ionized in water.

Now these ions are in conflict together, and CH3COO- and H+ will hydrolyze together:

The weak acid group combines with H+ in water to form a weak acid, and the solution is alkaline, CH3COO- +H2O CH3COOH + OH-

The number of oh- ions increases, but why do you ask why does the concentration increase? Because the amount of OH- per mol of water has increased, the concentration of OH- has increased.

And the above equation shows that after the water goes in, only oh- comes out, and h+ does not come out, so h+ is eaten and wiped clean by ch3coo-, then the amount of h+ contained in each mol of water is reduced, so the concentration of h+ is reduced.

Hahahaha: Clear enough.

Both oh- and h+ come from water.

Acetic acid is a weak electrolyte, so acetic acid is hydrolyzed.

Hydrolysis equation: CH3COO- +H20===CHAoOH+OH- So there is more OH in the solution, and its concentration increases.

So much hard work on the third floor!

Hard 4th floor detailed answer = =!!

Hydrogen is produced, so that Mg has N1mol, Al has N2mol, N1molmg produces hydrogen Nimol, and N2molal produces hydrogen.

24n1+27n2= n1+ solution ni= n2= i.e.

When the precipitate is all Mg(OH)2, all Al in the solution exists in the form of metaaluminate, and Mg is all Mg(OH)2, so one part Al reacts 4 parts OH-, one part MG2 parts OH-, that is, the total consumption is to add 5mol L of NaOH 100 ml.

Let the amount of magnesium be xmol and the amount of aluminum be ymol n(h2)=mg+2HCl=mgCl2+H2

xmol xmol xmol

2al+6hcl=2alcl3+3h2↑

ymol ymol ymol

24x+27y=

x+3/2y=

x=y=mass of magnesium=

Mass of aluminium = AlCl3 + 4NaOH = Naalo2 + 3NaCl + 2H2Oymol 4ymol

mgcl2+2naoh=mg(oh)2+2naclxmol xmol

n(naoh)=4ymol+xmol=1=v(naoh)=

Let's say mgxmol, alymol

The reaction was mg+2HCl=mgCl2+H2

2al+6hcl=2alcl3+3h2

So there is 24x+27y=

x+ solution gives x=, y=

Then sodium hydroxide is added for the reaction.

MgCl2+2NaOH = Mg(OH)2+2NaClalCl3+4NaOH=Naalo2+3NaCl+2H2O, so the amount of sodium hydroxide required can be calculated according to the equation.

Magnesium chloride is needed, aluminum chloride is needed, so in total.

Solution: n(h2).

Let the amount of magnesium be xmol and the amount of aluminum be ymolmg+2HCl=mgCl2+H2

xmol xmol

2al+6hcl=2alcl3+3h2↑

ymol 3/2ymol

24x+27y=

x+3/2y=

x=y=mass of magnesium=

Mass of aluminium MgCl2+2NaOH=mg(OH)2+2NaClalCl3+4NaOH=Naalo2+ 3NaCl+2H2ON(NaOH)=

v(naoh)=

I am a high school chemistry teacher.

In NaHCO3 solution, there is.

Conservation of material: Na+ H2CO3 CO32 HCO3 Conservation of charge: Na+ H+ OH2 CO32 HCO3

It can be obtained from the above two formulas.

Oh CO32 H+ H2CO3 pH is greater than 8, so oh H+, hence CO32 H2CO3

In NaHCO3 solution, the solution must be alkaline because of the hydrolysis of HCO3-.

In this solution, there is:

1) Conservation of materials: C(Na+) = C(CO32-) C(HCO3-) C(H2CO3).

2) Conservation of charge: C(Na+) C(H+) = C(HCO3-) C(OH-) 2C(CO32-).

c(h2co3) +c(h+) = c(oh-) c(co32-) a

Because the solution is alkaline, c(oh-) c(h+)-b

Combining A and B, we obtain: C(H2CO3) >C(CO32-).

is greater than, first of all, the pH is greater than 8 The solution is alkaline, then the concentration of hydroxide should be greater than the concentration of hydrogen ions. And theirs are as follows.

1. HCO3- +H2O === H2CO3 + OH- This is the ** of OH-.

2. HCO3- === H+ +CO32- This is H+** So the alkalinity means that 1 reaction is greater than 2 reactions, then you can see that the generated H2CO3 and CO32- who has more balls to adopt, hehe.

Methane and H2 are total, where methane, H2, so the heat released by the combustion of methane is the only wheel, and the amount of heat released by H2 is the total heat.

In the gas mixture, N(CH4) = XMOL N(H2) = YMOLX + Y =

x / y = 4

Solve n(ch4) =

n(h2) =

h2（g）＋1/2 o2（g）＝1h2o（l） △h＝-q 3/2 kj•mol－1

x solves x =

CH4(G) 2O2(G) CO2(G) 2H2O(L) H-Q1KJ mol Rent 1

x solves x =

So, the amount of heat emitted is + = kj

Under the same conditions, the volume ratio of wild oranges is equal to the ratio of the quantity of matter. The volume ratio is 4: Huaizhi 1, and the quantity ratio of matter is 4:1

The total volume is, and the amount of the substance is.

The amount of methane is: 4 5 =

The amount of hydrogen is; 1/5 =。

To room temperature lead trapped, water is liquid.

The total heat is: Q1 + Q32 = Because 2 moles of hydrogen emits heat from Q3, each mole is Q3 2

1.The amount of substances that consume hydrochloric acid is:

Therefore, the amount of sodium hydroxide in 10ml of solution is, then 250ml of solution contains sodium hydroxide, and the corresponding mass of sodium hydroxide is.

So the quality fraction is:

2.On the high side. 3.No, if rinsed, the amount of sodium hydroxide actually involved in the reaction is excess in 10 ml of solution.

Hello according to your title.

Then there is a -cooh

A-OH is directly attached to the benzene ring.

A-oh is on the top of the branch.

It is also known that there are only two monochlorine substitutes on the benzene ring in the molecule of this compound.

then its structural formula can be determined.

In the middle is a benzene ring.

Branched chain -chohcooh and one oh

The shape is the same as p-2 toluene (sorry, I can't upload it**).