If you ask for a chemistry question, you need to detail the process

According to the conservation of mass (conservation of elements), the raw material is Y2O3 + BaCO3 + Cuo after the reaction to generate Y2Ba4Cu6O13, the valency in the reaction is unchanged, then it can be known that BA is +2, Cu is +2, O is -2, then the algebraic sum of the valency of the compound is 0 to obtain 2Y + 4 * 2 + 6 * 2 + (13 * -2) = 0 then the Y valency is +3, and the Ymon is Y2O3 by using (valence cross method), and then the reactants and products are written out and balanced, (Using the observation method, or the 1 method, that is, the Y2Ba4Cu6O13 coefficient is set to 1) Y2O3+4BaCO3+6CuO=Y2Ba4Cu6O13+ It is found that there are 4 Cs and 8 O's missing, that is, 4CO2 Or it is inferred that the insoluble carbonate will decompose at high temperature to generate CO2, such as CaCO3Y O

Composed of 4 elements.

y is +3 valency.

y2o3 last one doesn't know.

According to the change of mass before and after the reaction, the reaction equation can be deduced: M A + N D = P B + Q C.

And because the molecular mass of the phase of A is 2 times that of D, so if it is 2mx nx, it is 2mx 64=nx 16 m n=2 1, so the answer is c

Since CH4 consumes the most oxygen per unit volume of the three gases, and 150 MLO2 is consumed by 75ml of gas when all CH4 is CH4, there is an excess of oxygen.

Let xmlCH4, YMOLH2, ZMOLCO, X+Y+Z=75CO2: X+Y=175-125=50 Z=25, that is, there are 25ml of CO and the remaining 125ml of gas is all O2 (dry to remove H2O, lye to remove CO2) 2x+ 4x+y+z=150 3x+50+25=150 x=25, that is, there is CH4 25ml

y=75-x-z=25, that is, there is H2 25ml

Since the gas mixture is relatively small, it can be speculated that there should be an excess of O2, and finally 125mLo2CH4+2O2=CO2+2H2O

x 2x x

2h2+o2=2h2o

y y/22co+o2=2co2

z z/2 z

x+y+z=75 ①

200-2x-y/2-z/2+x+y=175 ②200-2x-y/2-z/2=125 ③

Directly substituting x+y=50

Substituting Z=25

In the same way, x=y=z=25ml can be obtained

k2cr2o7+14hcl---2kcl+2crcl3+3cl2+7h2o

A Cl2 is the oxidation product of this reaction – right.

b The ratio of the amount of the oxidant to the reducing agent is 1:14 - false, is 1:6c When the electrons are transferred, the amount of the substance of the oxidized reducing agent is the opposite d From this reaction we can know the reducing hcl crcl3 - right.

This can be solved quickly with the difference method.

The CO2 and CO obtained after the combustion of CH4 in the title have a total gram, and the water has a gram, and the carbon of CH4 is completely converted into CO2 and matter and large CO

Yes, it means that the water has moles, and it is concluded that the original ch4 has moles.

Molars of CH4 are completely burned to produce CO2 in grams.

The gram cover is vertical and the gram is different, so that is to say, the less gram is O2, then the O2 of the Kepeng Yu is.

First of all, you have to understand that the degree of primary ionization of H2S is much greater than its secondary ionization, but at the same time, it is important to remember that H2S is a weak electrolyte, and its degree of ionization is very small, probably between 1-2%, which can be calculated. In other words, most of H2S exists in solution in the form of molecules, and the concentration of H2S is the largest.

Ionization of weak electrolytes while ionization of water is taken into account:

1.How to simplify the problem, because there is no actual mass, so it will be difficult to consider the problem, so you can first assume that the mass of copper is convenient to calculate), after fully reacting with dilute sulfuric acid, only copper remains in the solid, that is to say, the substance that fully reacts with oxygen is copper, according to the equation 2cu +o2=2cuo

x calculates that the mass of the product is 8g, and this mass is half of the mass of the original mixture, so the mass of the original mixture is 16g, which contains copper, then the mass of magnesium is (, and then calculate the mass ratio: 3

The upstairs process is a little more complicated, and there is no need to make the multiple-choice questions so complicated, and it will be a waste of time during the exam. It can be assumed that the mass of Cuo is 80, then the mass of the original mixture is 160, and the mass ratio of copper to magnesium is 64 to 96.

c can find that the condition of 1 is the slowest.

So d corresponds to 1

Note 3,4

It is impossible to judge under the existing conditions, only on the basis of 1,2.

C is chosen because 1 has the smallest concentration, the lowest temperature, and is still lumpy, so its reaction rate is the smallest, that is, the maximum volume value is reached at the latest.

The initial concentration of SO2 is, after 80% of the SO2 is converted, the SO2 equilibrium concentration is, and the SO3 is generated, at this time the O2 should be transformed, so the O2 equilibrium concentration is, so when the equilibrium.

The SO2 equilibrium concentration is:

The SO3 equilibrium concentration is:

The O2 equilibrium concentration is:

k= ·cso3).

The equilibrium constant k= --CO2) times (CSO2) squared.

The first question was solved upstairs.

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