Easy High School Chemistry Questions, Ask for Help !!!!!! 100

If it is the first case, H2S is overdosed.

The last remaining gas is only H2S (water is in liquid form), and the content in the original gas is H2S = 70ml O2 = 30ml

If the second case is O2 excess, then the last remaining gas is SO2 or SO2 and O2, if all SO2 is SO2, then H2S = 10ml O2 = 15ml in the original gas, and the total volume is only 25ml and the title is not in harmony (the original gas volume calculated when all SO2 is the largest).

In summary, the residual gas is H2S

Please understand the meaning of excess, which does not mean that the volume of H2S is more than O2, but that the amount of gas left after the reaction is what gas is excessive.

The gas product after the reaction of H2S and O2 may be H2S; ②so2;SO2 and O2

If this is the case, it indicates an excess of H2S.

Then the volume is reduced by: 2H2S+O2=2H2O+2S.

The total volume of H2S and O2 participating in the reaction is 60 + 30 = 90, then the remaining 10 ml of gas is H2S

If it is a case or , it means that there is an excess of O2.

Then the volume is reduced by: 2H2S + 3O2 = 2H2O + 2SO2.

Then the total volume of H2S and O2 participating in the reaction is 60 + 90 = 150>100, so these two cases do not exist.

So only H2S

Can't calculate. The product in the first equation is not gas, and in the second equation, the SO2 and H2O generated react to form H2SO3 (liquid), and the solubility of SO2 at that time is not stated, and H2S can also be soluble in water. So what this 10 ml gas is is is not at all certain, it could be pure O2 or a mixture of SO2 and O2 but it can't be a mixture of H2S and SO2 or a mixture of H2S and O2.

According to the ratio of the amount of H2S and O2 substances, the reaction can be subdivided into five cases:

-2:1, exactly the reaction (1) occurs, and no gas remains.

-2:3, exactly the reaction (2) occurs, and the remaining SO2

-2:1, H2S excess, only reaction (1) occurs, H2S remains

-2:3, O2 excess, only reaction (2) occurs, O2 and SO2 remain

-2:1 and 2:3, reactions (1) and (2) occur, and the remainder so2

Be sure to give me extra points!!

A and B are constant volume sealed containers.

The amount of substances in A and B is 2:1, which is consistent with the coefficient of AB in the reaction equation, that is, the concentration ratio of A and B is constant at 2:1

The volume fraction of c in the container of equilibrium A and B is the sum of the volume fractions of both a and b.

1.The quantity and concentration of the substances in container A and B are equal, that is, the amount of substances in A is equal, and the amounts of substances b and c in A and B are also equal.

That is, 4-1=3 mol a reaction can produce 3-0=3 mol c. That is, in the equation, the coefficients of a and c should be the same, that is, x=. That is, the equation is 2a(g)+b(g)=2c(g)+d(s).

At equilibrium, the volume fraction of b is [, and the volume fraction of 1mol a at the beginning of the reaction in b is 1 9 < that is, the reaction needs to be carried out [in reverse] to increase the amount of substance b in order to reach equilibrium.

Whether the excess does not affect the progress of the reaction and the direction in which it is carried out. But when there is no d, there is no way to reverse the reaction.

At the beginning of B 1mol a,,3molc and d, the volume fraction and the volume fraction of a are equal when B equilibrium c, and the amount of substances participating in d from the beginning of the reaction to the reaction equilibrium in B is y

That is, the amount of matter whose equilibrium is a is 1+2y

The amount of matter of c is 3-2y = 1+2y

That is, y =

That is, as long as the amount of d is greater than that, B can achieve the same balance as A.

This reaction is the one that does not change in volume. The direction of equilibrium movement at constant temperature after equilibrium is independent of the volume pressure.

2mola [Note, your title is 32mola, the title should not be so nonsensical, I changed it to 2mola], 1molb is flushed into the constant volume of v container A, and 6molc and 2 mold are filled into the constant volume of 2v container B.

That is, the average concentration of gas in A is 3 v, and the average concentration of gas in B is 6 2v, which is equal to A.

That is: the pressure in A [=] the pressure in B.

Remove 3molc from B and re-equilibrate, that is, equal to 2mola, 1molb flushed into the constant volume v container A, 3molc , 2 mold filled into the volume 2 v constant volume container B. Note that the equation at this time is 2a(g)+b(g)=3c(g)+d(s).

Equilibrium B is equivalent to equilibrium after doubling the volume of equilibrium A. But the balance doesn't move. That is, the amount of matter remains the same, but the volume increases. That is, the concentration of B decreases. That is, the concentration of c in A [> the concentration of C in B.

When equilibrium is reached, the heat emitted by A is q1, that is, the energy absorbed by B (B after 3mol reduction) is Q2=q-q1 when equilibrium

That is, q2=q-4q2, that is, q2=, q1=

That is, the conversion rate is [

The solubility of Na2CO3 is small, and the distilled water required when the sample is all Na2CO3 is the most.

Sorry, the diagram is not shown in my case and cannot be explained;

However, it is estimated that this figure is first added with hydrochloric acid, no CO2 is produced, and NaCl and NaHCO3 are generated by the reaction of HCl and NaOH and Na2CO3 respectively; Then hydrochloric acid is added to produce CO2, which reacts HCl with NaHCO3 to produce NaCl and CO2; Finally, the addition of hydrochloric acid does not produce CO2 because there is no reaction anymore. Without a graph, no data is not easy to calculate. FYI.

Alas, you can't do it without that diagram... It's true!!

(2) When 20 is known, the solubility of Na2CO3 is and the solubility of NaOH is .

then the mass of the undeteriorated NaOH in the sample is (8g); The degree of deterioration of NaOH is ( ) is expressed by mass fraction]; The volume of hydrochloric acid consumed in the reaction with NaOH is (50) ml.

Solution: The second question is that I don't know what to do right. I think so, he needs to find the maximum, I assume it's all sodium hydroxide, or it's all sodium carbonate, and see how much water is needed at most. Calculate yes.

The third question is that since the gas is so, sodium carbonate is and sodium hydroxide is 8g. Therefore, it is 8g that has not deteriorated. Metamorphic sodium hydroxide can be calculated according to the conservation of atoms, and sodium carbonate is sodium, so there is sodium hydroxide metamorphosis. That's 4G. So the proportion of metamorphosis is 4 12=

The last question should be noted that when the gas is produced, the reactant is not sodium carbonate but sodium bicarbonate, and since carbon dioxide consumes hydrochloric acid, it is also consumed when sodium bicarbonate is generated, for a total of 25 ml. So the reaction with sodium hydroxide is 50ml.

Question 1:

1. Mass fraction = solute mass Total mass of solution.

Soluble mass = (v.)

Total mass of solution = solute mass + water mass = (v.)

Note that the unit should be unified here as g

Mass fraction = (v.)

Quantity fraction of a substance = Quantity of solute substance volume of solution.

Amount of solute material = v

Solution volume = total mass of solution Density = [(v.)

The quantity fraction of a substance is =v

VP Question 2:

Let the quantity of the CO substance be x the amount of the CO2 substance is y

From the question Co burned to form CO2 carbon element conservation.

Knowing x+y=28x+44y=18 The simultaneous solution yields x= y=

So 1. The CO mass is m=

2. The volume of CO2 is v = under standard conditions).

3. Mixed gas density = total mass total volume = 18 (

The mass of the liter gas should be: vm g, then the mass fraction should be: vm v The amount of the substance of the gas is:

V mol, the total mass of water and gas is: 100+vm, because its density is , so the volume of the solution is: (100+vm, then the amount concentration of its substance is:

The amount of a gaseous substance divided by the volume of the solution.

2.It can be assumed that Co and Co2 are x mol and y mol respectively, then Co in the mixed gas is 28x g and Co2 is 44y g, and the equation is 28x+44y=18;And because the volume after combustion is , i.e., the equation x+y= is obtained. Two equations solve the equation to get x=y=.

Then in the gas mixture, the mass of CO is 7g

CO2 in the gas mixture is, and the volume under standard conditions is.

The relative molecular mass is 72, and the mass ratio is 5:1,12*5:12*1, so the molecular formula is C5H12,3, and there is only one monochlorolate is 2,2 methylpropane!

2: Butyraldehyde, 2-methyl-propionaldehyde!

, the quantity fraction of the substance is, ammonia water mass = ammonia mass + water mass =, volume =, the amount concentration of the substance = 2The ratio of the amount of matter is 3:2, and the same volume of hydrogen is generated, so the metal ion A is +2 valence, and B is +3 valence, and the relative molecular mass of the two is 8x, 9x, and the amount of matter is 3N, and 2N respectively.

52nx=,3n+2n*3 2=, the solution is x=3, so the molar masses are 24g mol and 27g mol respectively

1/n）=n：m

n:n, at the same temperature and pressure, the number of molecules is the same, then the amount of matter is the same, and the volume is the same.

1. In fact, the corresponding ions undergo metathesis reaction.

The conditions of metathesis reaction: (1) strong acid to weak acid, (2) volatile acid to non-volatile acid, and (3) weak electrolyte to form.

2、h+ +hc03- = h20 + c02↑al(0h)3 + 3h+ = al3+ +3h20

The reaction between the ions produces a weak electrolyte gas precipitation.

NaHCO3+HCl=NaCl+H2O+CO2NaHCO3 will produce CO2 gas in the acidic environment of the stomach, which will cause gastric distension and may aggravate gastric perforation.

al(oh)3+3hcl==alcl3+3h2o

1 Ions or molecules react to form weak electrolytes.

Weak electrolytes, precipitates, gases are generated.

2 （h+）+hco3-）=h2o+co2al(0h)3 +3（h+）=（al3+）+3h20

The essence is an ionic reaction that produces a precipitate, gas, or water (weak electrolyte).

The conditions for this to occur are precipitation, gas or water formation, and one of them can be satisfied.

hc03- = h20 + c02↑

al(0h)3 + 3h+ = al3+ +3h20

Anions (SO4) in sodium sulfate, cations (Fe) in ferrous chloride are used in papermaking, salt (NaCO3) in detergents, sulfate ions (SO4).

C6H12O6+6O2=6CO2+6H2O (note that arrows are generally used instead of equal signs, and "enzymes" should be written above the arrows).

The first thing to look at is the appearance, the gray is the phosphate fertilizer.

Upstairs is used for papermaking, and the salt of the detergent is sodium carbonate, but sodium carbonate is a detergent and not for papermaking.

It should be anhydrous sodium sulfate, also known as anhydrous miscanthus nitrate, also known as Yuanming powder, mainly used for making water glass, glass, porcelain enamel, pulp, detergent, refrigeration mixture, desiccant, dye diluent, analytical chemical reagents, pharmaceuticals, etc.