2011 Chemistry Review Material !!!!!!!!!! Urgently

n=6 option d The mass of the matter will increase when the pressure increases, so n+2>4+3

The pressure in the container is increased by 5%, indicating that the total amount of substance is multiplied, i.e., 5*

A decrease of 1 3 in the concentration of x indicates that the amount of x participating in the reaction is: 3*1 3=1mol

4x（g）+3y（g）==2q（g）+nr（g）

4 3 2 n

1 3/4 2/4 n/4

So the y participating in the reaction is: 3 4=, and the reaction produces q of 2 4= and r of n 4= mol

The total amount of substance after the reaction is:

Then column the equation: (3-1) + (=

The solution is n=6 By the way, the upstairs solution is simple and correct, like one.

Na2CO3 + CaCl2 = CaCO3 + 2NaCl The mass of hydrochloric acid precipitation remains unchanged because Na2CO3 is excessive, and hydrochloric acid reacts with Na2CO3 first.

na2co3 + 2hcl = 2nacl + co2 + h2o106 73xx=

After 50 150 g of hydrochloric acid, here 100 g of hydrochloric acid is consumed to react with CaCO3 + 2HCl = CaCl2 + CO2 + H2O100 73

yy=10g so the precipitated mass is 10g

Generating CaCO3 consumes Na2Co3, which is set to Zgna2CO3 + CaCl2 = CaCO3 + 2NaCl106 100Z 10Z=

So the total mass of Na2CO3 x+z=

caco3 + 2hcl = cacl2 + co2 + h2o100 73

xx = 5 g of precipitate mass of 5 g

na2co3 + 2hcl = 2nacl + co2 + h2o106 73yy=

The mass of Na2CO3 in this packet of white powder is:

Precipitation occurs when mixing.

2NaCO3 + CaCl2 = 2CaCO3 + 2NaCl, white precipitate is calcium carbonate When 50g of hydrochloric acid is added, there is no precipitation at this time, and the chemical balance is as follows:

2na2co3+cacl2+2hcl=ca(hco3)2+4nacl3.When 150 g of hydrochloric acid is given, the precipitate disappears, and the equation is as follows:

na2co3+cacl2+2hcl=2nacl+cacl2+co2+h2o

Hydrochloric acid quality 50*150*

Sediment mass HCl relative mass = CaCO3 = 40 + 12 + 16 * 3 = 100 Na2CO3 = 23 * 2 + 12 + 16 * 3 = 106

According to formula 2 hydrochloric acid is a 1:1 relationship with calcium ions, so m 100 = m = 10g according to formula 1 Na ions and calcium ions are 1:1 relationship 10 100 = m 106, so the mass of Na2CO3 m =

1. fe + h2so4 = feso4 + h22. mg + h2so4 = mgso4 + h23.

al + 3/2 h2so4 = 1/2 al2(so4)3 + 3/2 h2

4.Zn + H2SO4 = ZnSO4 + H2 and dilute sulfuric acid do not react.

In terms of relative molecular weight, Zn>Fe>Al>mg

9g of iron is a mixture of a certain metal, assuming that the 9g is all that metal.

Because the relative molecular mass of Zn is the largest, the molar mass of 9g metal, Zn is the smallest is 9 65=

Zn reacts with dilute sulfuric acid to form H2 (refer to the above equation).Under the standard condition, it means that even if it is full of impurities, the amount of raw material is minimized (to ensure that the least amount of H2 generated) cannot generate such a small amount of hydrogen, so it means that impurities are substances that cannot react with dilute sulfuric acid, and the substances that can actually participate in the formation of H2 reflection are less than 9g.

Select D cu. according to the title

Wish: Happy New Year! Learn and progress!

Ideas for solving such problems:

Assuming that these 9g are all iron, then a complete reaction with a sufficient amount of dilute sulfuric acid can give an amount of H2 (I won't calculate it specifically).

Compare with the amount of H2 provided by the question.

If the amount of H2 obtained from 9g of iron is greater than the amount provided in the question, then there must be a substance in the mixture with the same mass of iron that gives a lower amount of H2 than the given amount, and you can make a judgment to exclude the answer.

If the H2 obtained by 9g of iron is less than the amount provided by the question, then there must be a substance in the mixture with the same mass of H2 that is higher than the amount of iron, and the answer is excluded.

You look back at the metal activity order table cu< H Cu cannot react with dilute sulfuric acid. Then 9g of iron just reacts completely to form H2< p>

In the case of pure iron, 9g of iron can reflect 9 56 = hydrogen i.e. already greater than.

No data is required.

Copper and carbon before burning become copper oxide after burning, and carbon directly becomes carbon dioxide in the air.

According to the equality of the mass before and after the reaction, the mass of carbon is the mass of oxygen in copper oxide, and the mass fraction of carbon becomes the mass fraction of oxygen in copper oxide, 16 (16+64)=1 5=20%.

Assuming that NAI has X G, NaCl has 10-X G

2nai+cl2=2nacl+i2 A sufficient amount of chlorine reacts with sodium iodide, so sodium iodide is fully reacted completely; And because the final weight is weighed after heating, the iodine in the product will be volatilized, so the final solid is the sodium chloride that existed before the reaction and the sodium chloride product of the reaction.

According to the equation, the amount of Nai and NaCl is the same as X 150 (150 is the molar mass of NaCl), so the mass of the product NaCl is X 150* So 10-x + X 150* = Solve to get X= So the mass of NaCl is, the mass fraction is.

I hope you can read it.

The original solid mixture was Nai and NaCl, and when a sufficient amount of chlorine was introduced, all of them were converted to NaCl, so the final solid mass reduction was the amount of iodine more than chlorine. Let the mass of NaCl in the mixture be X.

2nai+cl2=2nacl+i2

300 117 available, (300-117) (

x = mass fraction of NaCl in the mixture =x m mixture=

Let the mass of sodium iodide be x, then (x m(nai))*m(naci)+(10-x)=,

The electrons obtained by the gas are 2*(b, CV1000 is the amount of the substance used in the OH root, due to just the precipitation. Composed.

Mg mg2+ 2oh- ALal3 3oh- can give 2* (b simplification yields a

It is the amount of OH root material used, which is burned and precipitated until the mass no longer changes, and the solid P g is obtained, and the solid is an oxide, and the increased oxygen element of the oxide is CV 1000mol, and the increased mass (CV 1000) is 16g

So p m (cv 1000) 16 p = m + vc 125

3。As with B, N M+(CV 1000) 17

In the two extreme cases where the two values before and after are all mg and al, they should be in this range because they are mixtures.

Select C and let the mixture of mg mg magnesium and aluminum contain al x mol, mg y mol then 27x+24y=m (Equation 1).

2al + h2so4 =al2(so4)3 +3h2

xmg + h2so4 = mgso4 + h2

y y yal2(so4)3 + 6koh = 2al(oh)3 + 3k2so4

3x xmgso4 + 2koh = mg(oh)2 + k2so4

y 2y y

2al(oh)3 = al2o3 + 3h2o

xmg(oh)2 = mgo + h2o

y y according to the above reaction equation has.

amount of hydrogen) (Equation 2).

3x+2y=cv 1000 (amount of potassium hydroxide) (Equation 3).

51x +40y=p (mass of oxide) (Equation 4).

78x+58y=n (mass of hydroxide precipitate) (Equation 5).

Simultaneous 2,3 solution gives c=1000b so a pair.

Vertical 1, 3, 4

Equation 1 - Equation 4 3 gives 6y=m-9cv 1000

Equation 2 - Equation 3 17 gives 6y = p-17cv 1000

P=m+vc 125 so b pair.

Eq. 3 12-Eq. 1 gives 9x=12cv 1000

Equation 3 29 - Equation 5 gives 9x=29cv 1000

Get n=m+17vc 1000 so c is wrong.

If there is m g mg metal, then its oxide mass is 5m 3

mg --mgo

m 5m/3

If there is a m g al metal, its oxide mass is 17 m 9

Then 2al --al2o3

m 17m/9

Since the sample is a mixture of mg and al.

Therefore, the mass p of the oxide is between 5 m 3 and 17 m 9. Therefore d is right.

In the past few days, I have been reviewing the college entrance examination, and I haven't touched the ...... of inorganic chemistry for almost half a yearI don't have a holiday, I don't have energy, I'm sorry.

Choose C n=m+17vc 1000 units are not uniform.

1, The number of h is the same as the number of oh.

2. The quality of precipitation is the quality of extra OH.

3, The mass of the final solid, which is m plus the mass of oxygen obtained by hydrogen combustion4, it is not difficult to grasp the problem of electron transfer.

Hope it helps, a few years without doing it should be right!

The gas mixture composed of a, b (relative molecular mass: a is greater than b) two common gases contains only n h two elements, regardless of the ratio of a and b mixed in, the ratio of the masses of n h two atoms in the mixed gas is always greater than 14:3, then a is nh3 b is n2 if the mass ratio of n h two atoms in the mixed gas is 14:

1, then the ratio of the amount of substances in a b in the gas mixture is 1:1

Because the gas composed of n h in the range of high school is only h2 n2 nh3, and the ratio of n and h in these three gases is: 0:2 24:0 14:3 respectively

In the title, it is said that the mass ratio is always greater than 14:3, so it is composed of two gas components with a ratio of n and h greater than or equal to 14:3, that is, n2 and nh3, and in the last air, you can let n2 contain xmol and nh3 contain ymol, then (28x+14y) 3y=14:

1 Simplification is x:y=1:1

1. Soda lime will react with chlorine and consume chlorine, so it cannot be selected.

2. Although silica gel can absorb water, chlorine will react with silica gel and corrode away silica gel (which is a mixture of sodium silicate and other organic colloids), and may produce other aromatic hydrocarbon chloride impurities, so it cannot be selected;

3. Concentrated sulfuric acid is actually very viscous, and it is difficult for gas to pass through, and chlorine will directly push the liquid in the U-shaped tube up and spray it out from the right end, so it is not good;

4. The function of anhydrous calcium chloride is to absorb water vapor and only let chlorine pass through, which is used to show that only hypochlorous acid has a bleaching effect, not chlorine itself.

Chlorine gas is bleaching, because hypochlorous acid produced by the reaction of water is bleaching, so it cannot be a wet cloth strip in three places.

A soda lime is composed of sodium hydroxide and calcium hydroxide that react with chlorine.

If you are not able to hold liquid, the liquid desiccant will be pressed out from the other end due to the pressure of the gas.