High school chemistry, chemistry masters, high school chemistry masters

The equation in your book must be like this: 2h2(g) +o2(g) = 2h2o(l), right?

The unit of heat of reaction is kj mol, and here each mole refers to each mole reaction. What is a reaction per friction? It's what you write the equation like, and what is a one-friction reaction.

For example, as written above, the reaction of 2mol of hydrogen and 1mol of oxygen to produce 2mol of water is a friction reaction, and now there is 1g of hydrogen, which is hydrogen, and what happens is the reaction, so it is counted like that- = (.

v(x) v(y) = (x) (y), in the same reaction, the ratio of the reaction rate of each species is equal to the ratio of the stoichiometric numbers of each substance in the equation.

Because it is a reversible reaction, the limit of the reaction is to reach chemical equilibrium, not to react completely. So you just calculate the extreme value exactly by the reaction.

Reaction: x2(g) +y2(g) = 2z(g).

Start: Perfect Forward: 0

Full reverse: 0

Only B is reasonable.

First of all, the positive and negative electrodes are judged, and the electrode itself or the electrode attachment is reducible, and the negative electrode is easy to lose electrons.

Then write out the negative electrode reducing agent and oxidation products, reducing agent: zn, oxidation product: zn2+.

Mark the valence change, zn: 0 +2, liter 2 valence. Increasing the n valence means that one atom of the element loses n electrons.

Write the number of lost electrons: Zn - 2E- = Zn2+, and then balance according to the conservation of mass or charge if necessary.

The same goes for the positive pole. To give you another example: lead-acid batteries, for example.

The negative electrode is Pb, the negative electrode is PBO2, the electrolyte is H2SO4, and PBSO4 is insoluble in sulfuric acid.

Negative: Write PB — PBSO4 first

Valency: PB0 +2, so PB-2E- — PBSO4

According to the conservation of mass: Pb-2E- +SO42- = PBSO4

Positive: Write PBO2 — PBSO4 first

Valency: Pb +4 +2, so PbO2 + 2E- —PbSO4

According to the conservation of mass and charge: PBO2 + 2E- +4H+ +SO42- = PBSO4 + 2H2O

After writing, be sure to check whether the number of electrons gained and lost at the positive and negative electrodes is equal, if not, modify it to its least common multiple.

The first question is what you are asking about Shenma, orz

I see the answer you gave, guessing that it is δh that is completely burned by 1mol of hydrogen = =

Hydrogen is h2, 1mol is 2 grams = = 1g is given in the question, so multiply it directly by 2 ah......

And this problem is known as co(g) 1 2o2(g) = co2(g), δh= becomes.

1／（1／4mol）=283／q

What are you doing with this, = = ask questions, and make the questions clear, okay......Is there really no lack of content, not even hydrogen elements, how to calculate the heat of combustion of hydrogen according to this, what kind of thermal reaction equation does orz give to alkanes......You've got it all ......

Or is it the burning fever that you are asking, I can only guess this, I really don't know what you want to ask......

The second question is the same as ...... unitIf you don't get the same, you need to make ...... of the same unit first

Then assume that the units are the same.

3x＋y＝2z

This kind of question is very simple, don't be lazy to copy the equation on scratch paper, write out the corresponding speed according to the ratio, the speed of the same substance will come out in a comparison, the test is not to calculate and will not give strange numbers, just a few seconds of effort, I didn't use 30s to type, maybe faster than you meditate for a long time.

Question 3: Extreme Hypothesis:

x2 + y2 = 2z

Start by assuming that the x2 reaction is exactly 0

Suppose the z-reaction is exactly 0

Because it is a reversible reaction, the reaction cannot be carried out to the end, so x2 y2 2z cannot take the endpoint value. Just look for one of these ranges.

Answer: b = The gain and loss of electrons must be conserved, and if it is not conserved, the system will be charged. The consequence of non-conservation is to exaggerate, you have to throw away a used battery, and before you throw it away, you will be electrocuted.

Does this exist? It doesn't exist, because the gains and losses are conserved, so you don't have an electrocution......It's just a redox reaction, and you've ever seen redox electrons that aren't conserved?

Don't you understand how hydrogen can only gain one electron, how to lose the conservation of zinc with two electrons?

Yes, a hydrogen element can only get one electron, because it has to be conserved, so there is a trim. Don't you see there's a 2 in front of the hydrogen? That is the level that is matched for the conservation of electrons for gain and loss.

Question 1: I didn't understand it too well.

The second problem is v(x) v(y)=3 1, which can be calculated according to the formula of dividing the quantity of a substance by time and volume.

Because the time and volume are the same, the ratio of the reaction rate is the ratio of the amount of matter reacted.

The third question is to choose B.

Use the method of elimination. When a is reached, the concentration of the substance of z is 0, which is obviously impossible.

In the case of b, x2=, y2=, z=

In the case of c, the concentration of x2 is 0, which is also impossible.

When D is reached, it is the same as C.

So the answer is b

The fourth question is not easy to say, and there are a lot of knowledge points to understand the topic of galvanic batteries, so I want to talk about it in detail.

Generally, it is in the order of oxidation and reduction to see who will gain and lose electrons.

Dude, what you wrote is a bit messy, I don't understand, but the last one I understand, -e is the loss of electrons, the valency of things that lose electrons will increase, oxidation reaction, how many electrons are lost, how much the valency will change, this is the increase in oxygen loss (valency increases, loss of electrons, oxidation reaction), in the same way, it will be reduced, you know.

zn is 0 valence, and electron is negative valence, which means that 0-2 (-1) is equal to +2, so zn becomes +2 valence.

The conditions under which the metathesis reaction takes place: the reaction of precipitation, gas, weak electrolyte (water, weak acid, etc.) is generated, Na2FeO4 + 2 Koh = K2FeO4 + 2 NaOH

If there is no gas, weak electrolyte is produced, then there must be precipitation.

NaOH is not a precipitate, it is only possible that K2FeO4 is a precipitate.

Solubility of the precipitate Solubility of the soluble substance.

So the solubility of K2FeO4 is the solubility of Na2FeO4.

Generalization: Metathesis reactions are always carried out in the direction of decreasing solubility or decreasing ion concentration in solution.

Potassium ferrate can be prepared by adding a certain amount of KOH to the sodium iron rate solution at low temperature, and the reason why it can be prepared is because the solubility of potassium iron rate is smaller than that of sodium iron rate.

3no2+h2o=2hno3+no

The volume of NO2 is reduced to 1 to 3

Let the volume of no2 be x

then (1-1 3) x = 30-16

x = 21 ml.

No = 9 ml.

We used to do this kind of problem a lot, and you could draw a diagram and make an equation.

The first question is C

The greater the concentration of sulfuric acid, the greater the density, so the volume of 7mol l of sulfuric acid is greater than half the volume of 14mol l, so the mass fraction will be less than 40%.

Question B for the second question is because there is no 950ml volumetric flask in the laboratory, so you can only use a 1000ml volumetric flask to prepare a 1000ml solution, and then use a graduated cylinder to accurately measure 950ml

Hence the original b

Question 1 A is conserved by the amount of sulfuric acid substance.

14*v (concentrated) = 7*v (dilute) It can be seen that v (dilute) = 2v (concentrated) is conserved by the mass of sulfuric acid.

m(concentrated)*80%=m(rare)*x and m= v (concentrated)v(concentrated)*80%== (rare)v(rare)*x (concentrated)*80%=2* (rare)*x

Concentrate) * 40% = (dilute) * x

The greater the concentration of sulfuric acid, the greater the density; The smaller the concentration, the smaller the density: (concentrated)> diluted) and the answer is obtained: x= (concentrated)*40% (diluted)>40%Question 2 b can be judged directly:

950ml can only be used with a larger and closest volumetric flask of 1000 ml

The first question is C, because the greater the concentration of sulfuric acid, the greater the density: the second question will know by yourself.