200 points will be added for Senior 1 Chemistry and 200 points for Senior 1 Chemistry

Here's the idea of this question:

The composition of the sample should be figured out, it should contain KOH, K2CO3(, H2O(, obviously, KOH accounts for 90%, so that the mass of KOH and K2CO3 in 10g can be calculated respectively, and the molar number of K+ can be found.

Use the neutralization to calculate the number of moles of k+.

The mass of the solid can be calculated by adding up the moles of k+ in 1 and 2 to correspond to the molar number of kcl in the final solid.

This problem is to focus on the mole number of k+, figure out what solid is at the beginning, and what solid is at the end, and it is easy to do.

If you don't understand something, you can ask!

A: My opinion is A.

The main reactions in this question are: potassium hydroxide and hydrochloric acid, potassium carbonate and hydrochloric acid, and finally hydrochloric acid and potassium hydroxide.

After writing the above reaction equation, it was found that there was only one solute in the final neutralized solution, which was potassium chloride. The potassium ions in potassium chloride are both in the original solid and later neutralized, which is very messy; The chloride ions in potassium chloride** are very simple, they are the chloride ions in 250 ml of hydrochloric acid, and this is easy to do. n(hcl)=n(cl-)=n(k+)=n(kcl), you see, isn't it very simple to calculate chloride ions?

Solution: n(kcl)=n(hcl)=,m(kcl)=.

Hope it helps. Thank you.

A small amount) = 2Al(OH)3 +3K2SO4

Al(OH)3+KOH=KALO2+2H2O produces a precipitate of AL(OH)3:N(AL(OH)3)=m N=There are two possibilities:

A small amount of the first type of KOH directly yields Al(OH)3 precipitate, N(KOH)=3 N(AL(OH)3)=, C(KOH)=

The second kind of KOH is excessive, and after obtaining all Al(OH)3 precipitates, the remaining Al(OH)3 precipitates are dissolved, N1(KOH)=6N(Al2(SO4)3)=6, N2(KOH)=2N(Al2(SO4)3)-N(Al(OH)3 precipitates)=2, N total (KOH)=N1(KOH)+N2(KOH)=, C(KOH)=

Answer: Fe+2HCl=FeCl2+H2

mg+2hcl=mgcl2+h2↑

2al+6hcl=2alcl3+3h2↑

If a single metal is put in, it takes 9g to produce each, so if you want to produce a mixture of less than 10g, Al is essential, and the answer is C

You have to ask the first question, how do I choose C.

Divide by 14 and 2 is 12 (2 is a multiple of 14), and 6 is 8 (8 is a multiple of 14).

Look at it clearly, option c is not a double bond here, there are 2 h atoms on each point the red dot in the middle of option d is already full of c atom electrons, there can be no more h only methanol is 1:4, only glycerol is 1:3 8=1:3 is between 1:4 and .

Therefore, as long as methanol and glycerol are mixed in a certain proportion, n(c):n(h)=1:3 can be prepared.

This difference is the remainder of the molecular weight of this ch organic matter divided by 14, because for monoolefins, it is exactly removed, and the difference is that if you add 2, it will be exactly eliminated.

Two sets of peaks appeared on NMR, indicating the existence of two different equivalent hydrogen; The area is 3:2, which means that the ratio of the number of these two types of hydrogen is 3:2 (don't you understand the concept of equivalent hydrogen).

Methanol is 1:4 and glycerol is 1:8 3, then the two are mixed, and the hydrocarbon ratio can be feasible at any value between these two values.

The first: the difference of four means that the hydrocarbon is four less hydrogen than the alkane homolog (the alkane homolog should be two larger), then there will be two carbon-carbon double bonds or one carbon-carbon triple bond, and in the same way, if there are six less, then it will be 8 hydrogen less than the alkane homologue, so it can be benzene or its homologue. The second:

In a: two types of hydrogen refer to two allele hydrogens, one on the benzene ring and the other on the methyl group; B: There are four alleles on the ring, and there are six alleles on the methyl group, and you mark the position of the carbon, and each position should be able to connect two hydrogens, and the allele hydrogen of the methyl group goes without saying.

D: The situation is almost the same, the carbon on both sides of the position of your target carbon can be connected with two hydrogens, and I don't need to say the hydrogen of the methyl group, the so-called four equivalent hydrogen on the methyl group means that after any one of the four hydrogens is replaced, the structure is the same. They are allisotopic hydrogen to each other.

Third: This is not averaged out, because methanol is c:h=1:

4, Glycerol c:h=1:8 3, because 1:

4 1:3 1:8 3 So the mixture after mixing them in a certain proportion can reach c:

h=1:3.Of course, you can also think of it as a weighted average.

Difference 2 is the opposite of balance 2, e.g. 10 divided by 6 equals 2....Difference 2

The gas mixture is moles, where CO is moles and CH4 or C2H4 is moles. Since methane and ethylene are both 4 hydrogen atoms in 1 molecule, the molar hydrogen atom before the reaction must be formed after the reaction, that is, the first empty fill. The ionic equation in b is CO2 + Ca2+ +2OH -=CaCO3+ gram is the mass of CO2, which is moles, and X is CH4 from the conservation of carbon

The amount of total gas substances in liters is, because the amount of CO and X substances is equal, so X is, CO combustion produces CO2, X combustion produces CO2 and H2O, concentrated sulfuric acid only absorbs H2O, and because there are 4 hydrogen in methane and ethylene molecules, methane or ethylene combustion is produced, so concentrated sulfuric acid is heavier.

b gains, CO2 is, i.e., =, produces, x should be a C inside, so it is methane.

1. The mass of a weight gain is the mass of H2O.

ch4+2o2=co2+2h2o

c2h4+3o2=2co2+2h2o

The water produced by both is the same.

n=n(h2o)=2n=

m(h2o)=

2. The mass of Ca2+ +2OH-+ CO2=CaCO3+H2OB is the mass of CO2.

n(co2)=

Methane can only produce CO2

And ethylene can produce CO2

Therefore, the structure of x is simply ch2=ch2

co2 +ca 2+ +2oh- ===caco3 ↓+h2o

X is methane, and the structure is not easy to write on knowing, you read chemistry textbooks.

The calculation in the question is 1 8mol, so 1mol is 8 times this time.

n(c)= *

n(h)=The following is whether you have made a mistake in the socks.

I didn't understand what you said.

I think there's a simpler way:

Let the liquid hydrocarbon be cxhy.

Dioxide chain code wheel shed letter carbon--c

Water – hc: mold h=3:5

So x , y is 3n 5n

When n takes 2, it satisfies 1mol cxhy for 82g

One-third mole of aluminum corresponds to one-half mole of oxygen, which corresponds to one mole of electrons.

Iron is greater than 18, manganese is less than 18, vanadium is less than 18, and chromium is less than 18Answer a

Can the chemical equation of thermite reaction be written?

Hydrogen does not react with water, because even if it does, the product is still water and hydrogen, which is the same as the reactant. So in reality the reaction is just a reaction of carbon monoxide and water. Equation: CO+H2O*****=CO2+H2

Therefore, there is a formula for the transfer rate: conversion rate v = end - n (h2) 1 - n (h2).

Your answer is actually that the anions are completely hydrolyzed, at which point the anions are converted into hydroxide ions, and the cations are not hydrolyzed, so actually according to your answer, the solution is an alkaline solution. But this is the opposite of what the stem describes.

The information in the first question is not clear, only know that CO has been transformed, and the second one is a simple example, CO32-+H2O=HCO3-+OH-

Here m cannot be determined whether it is a number of 2 or more, the hydrolysis of salt is step-by-step, and one step can only hydrolyze to produce one oh-

From the question, the ratio of ammonium ions to aluminum ions to sulfate ions is 1:1:2, and the amount of barium sulfate precipitated from 2mol alum is 3:

4. If all the sulfate ions are to be precipitated, the hydroxide will be excessive and cause the dissolution of aluminum hydroxide, but there is also an ammonium ion in this question, and the extra hydroxide ions in the precipitation of sulfate are just combined with ammonium to form ammonia monohydrate.

2. kal(SO4)+2BA(OH)2=KALO2+2BASO4+2H2O (note the precipitation symbol).

Just change the above chemical equation to ions.

The reason is that when the sulfate is precipitated, aluminum ions are also precipitated, but the ratio of the amount of barium sulfate required for the precipitation of these two ions is 4:3, that is, when the sulfate is completely precipitated, the hydroxide is excessive, and the remaining hydroxide is reacted with aluminum hydroxide to form potassium metaaluminate.

1. Consider the reaction between sulfate ions and barium ions first, then consider the reaction between aluminum ions and hydroxide ions, and consider the reaction between ammonium and hydroxide ions after excess.

2ba2++2so42-+al3++4oh-==2baso4↓+alo2-+2h2o

First of all, sulfate ions and aluminum ions should be precipitated, sulfate ions should find barium ions, aluminum ions should find hydroxide ions to combine into precipitates, and ammonium ions should find hydroxide ions to combine into ammonia monohydrate, and then think about leveling like this.

Alum is potassium aluminum sulfate docahydrate, and barium hydroxide is added to precipitate sulfate ions and aluminum ions, so barium sulfate, aluminum hydroxide and potassium hydroxide are generated.