A problem with the high 2 chemical rate, please enter a strong person

According to C Yes.

Then d is because cd is decomposed by b, and b is decomposed by pushing, so the total amount of b is there, and a is reversed, and the rest is added to be multiplied by 2l to be equal.

After careful study, the chemical reaction rate is still a bit difficult, but it is very simple to do it once you are familiar with it, good luck!

The concentration of C is 2 times that of D (as can be seen from the equation), so the concentration of D is the concentration of B that reacts, so the total concentration of B is, the total concentration of A that reacts is, the total concentration of A is, and then multiplied by the volume of 2L, that is, (all the places without units are mol L).

Either b or c or d are all from a. According to the chemical equation, the quantities of substances b and c can be found separately by using the quantities and concentrations of substances b and c given in the problem and the volume of closed containers.

n(b)=n(c)= then using the chemical equation 2b(g)===c(g)+2d(g), the amount of the substance that can be inverted into c used b is .

Then if only the equilibrium a(g)===2b occurs, n(b)+n(b) is used') = then the amount of a substance that can be reversed and reacted with is .

And in the question stem, it is mentioned that c(a)= at equilibrium, and the amount of matter at equilibrium can be calculated as.

Therefore, the amount of a substance plus the amount of a substance that reacts at equilibrium is exactly the amount of a substance that is added at the beginning.

（1）8 mol

If the second question follows the first condition, it is very simple, the amount of the total substance is reduced by the same amount as the amount of the substance that produces ammonia, that is, 4mol of N2 and H2 are consumed, 2mol NH3 is generated, and 2mol of gas is reduced, and the equilibrium gas mixture in the first question condition is 32mol, and the amount of the original gaseous substance is 32+8=40mol, and the quantity ratio of the substance is 5:4.

If you don't accept the first condition, there is no equilibrium constant, and it can't be calculated, at least I won't ......

(1)8mol

2) (a+b)/32

Because the total amount of matter of the original gas mixture is (a+b)mol; The volume of the gas mixture when the reaction reaches equilibrium is under the standard condition), i.e., it is equivalent to, so n(start) n(flat) = a+b) 32

2.Because the N2 reaction is 4mol and the H2 reaction is 12mol, and when the reaction reaches equilibrium, the volume of the gas mixture is L (under standard conditions), where the content (volume fraction) of NH3 is 25%. Therefore a = 16 and b = 48.

Therefore, the ratio is 64:58=32:29.

Probably not. Support upstairs.

1.Answer: a

means that the reactant is carbon and the state is solid (S is not the abbreviation for solid in English, G is a gas, L is a liquid, for example, H2O(G) means that the state of water is a gas).

3.As the pressure increases, the reaction rate increases, and vice versa. Increasing the volume is decreasing the pressure, and decreasing the volume is increasing the pressure.

4.When the amount of reactants increases, the rate increases, and vice versa.

5.Increasing what is not related to the reaction is increasing the pressure.

a. After coal gasification or liquefaction, it will be fully combusted, which can improve fuel efficiency.

b. The liquid fuel is sprayed out in the form of a mist, which is conducive to the full combustion of fuel and improves fuel efficiency.

c. If there is too much air, there is no need to widen it, and it cannot improve the efficiency of the fuel.

D. Crushing pulverized coal is conducive to full combustion and can improve fuel efficiency.

(1) According to 2SO2 + O2 = 2SO3 (the middle is not an equal sign, it is a symbol of the reversible reaction), the SO2 = participating in the reaction, the generated SO3 = , and the SO3 concentration =.

2) SO2 average reaction rate = (.

3) According to the reaction equation, O2 = to participate in the reaction, then the conversion rate =

3x2+y2=2x3y (reversible). The specific calculation process can be analyzed according to the "three-stage" method.

Do you want a process? Or the answer.

【Answer】The question is not clear, but I will help you to say it:

ph： nahso4 < nahso3 < na2so4 < na2so3

Explanation] HSO4- can almost completely ionize H+, which can be used as a strong acid;

The ionization of HSO3- is greater than that of hydrolysis, and the solution is weakly acidic;

Na2SO4 neutral;

SO32 - hydrolyzed, alkaline.

Yes Na2SO4 NaHSO4 NH4HSO4 [A] Na2SO4 – neutral.

NaHSO4 – acidic.

NH4HSO4 – acidic, and more acidic than , because NH4+ also hydrolyzes to produce some H+ (very little).

Conclusion: PH <

Sodium bisulfite is neutral, pH = 7, sodium bisulfite is a strong acid, pH 7, ammonium bisulfite is acidic, pH 7 but ammonium bisulfite pH is minimal due to the hydrolysis of ammonium ions to produce a part of hydrogen ions.

The first thing to do is to identify what substances (molecules or ions) are in solution

The molecules are: H2O, H2CO3 (HCO3- hydrolysis, HCO3- +H2O == H2CO3 + OH-).

The ions are: Na+ (, HCO3-, CO32- (produced by HCO3- ionization, HCO3- == H+ +CO32-), OH-, H+

There is the following equation for the concentration relationship:

C is conserved: C(HCO3-) C(H2CO3) +C(CO32-) =

Electrically neutral: positively charged Na+, H+, negatively charged HCO3-, CO32-, OH-, where CO32- is negatively charged with 2 units, so there is.

c(na+) c(h+) = c(hco3-) 2c(co32-) c(oh-)

A: False. Na+ is, while C(HCO3-) C(CO32-) = - C(H2CO3).

B: False. Carbonate with 2 units of negative electricity, which should be multiplied by 2

C: Correct. d: Because Na+=, C(HCO3-) C(H2CO3) +C(CO32-) = i.e. C(HCO3-) C(H2CO3) +C(CO32-) = Na+, and Na+ concentration = C(HCO3-) 2C(CO32-) C(OH-) C(H+) gives the relation: C(HCO3-) C(H2CO3) +C(CO32-) = C(HCO3-) 2C(CO32-) C(OH-) C(H+) Solution:

c(oh-) = c(h+) c(h2co3) -c(co32-) so it is also wrong (c(co32-) should be subtracted, not added).

The answer is C

1. Conservation of electric charge.

In sodium bicarbonate solution, because the solution is electrically neutral, the total number of positive charges of the two cations should be equal to the total number of negative charges of the three anions, from which it can be obtained:

c〔na＋〕＋c〔h＋〕＝c〔hco3－〕＋c〔oh－〕＋2c〔co32－〕

It's clear that C is not right on B.

2. Conservation of materials.

The number of sodium ions produced by the ionization of sodium bicarbonate is equal to the number of bicarbonate ions, in the solution, the sodium ions do not change, the bicarbonate ions are ionized and hydrolyzed, and the carbonate ions and carbonic acid are generated, according to the conservation of carbon atoms, it can be obtained:

c〔na＋〕＝c〔hco3－〕＋c〔co32－〕＋c〔h2co3〕= mol/l

So a is incorrect.

3 is obtained by subtracting c(na+) from the first two formulas of 1 and 2.

c（oh-）+c（co32-）=c（h+)+c(h2co3）

So d is not right.

In summary, it can be concluded that option C

Option C is the balance of positive and negative ions. Option A wants to express the conservation of elements, i.e. the total amount of Na and C is equal. B wants to express the same thing as C, but without trimming. Option d is incomprehensible.

When other conditions are certain (such as temperature, catalyst, etc.), the reaction rate is only related to the concentration, and the reaction rate increases with the concentration, and the reaction rate does not change with the concentration.

That is, V positive = K positive * [SO2] 2 * [O2], V inverse = k inverse * [SO3] 2 increases the oxygen concentration, the positive reaction rate increases, and the reverse reaction rate does not change because the SO3 concentration does not change.

If the concentration of SO3 is increased (or decreased), the reverse reaction rate increases (or decreases), while the concentration of SO2 and O2 does not change, so the positive reaction rate remains unchanged.

Rate refers to instantaneous, and unlike velocity, it is a scalar rather than a vector.

During this instantaneous lift, the concentration of the components increases, and the forward or reverse reaction increases, but the reverse instantaneous reaction rate remains the same (because it is instantaneous).

Therefore, the data in the ** of lz is right, but if it is not the reaction rate but something else, this table is buried wrong.