Classic Chemistry Questions !!!!!

This is done with the limit.

The amount of substances that produce H2 is n=

All are Al2Al+H2SO4=Al2(SO4)3+3H2 X

x = all mg or zn

mg(zn)+h2so4=mg(zn)so4+h2↑y

y = Since it is an alloy, the limit value cannot be taken.

Hence the amount of matter< n "Choosing C< P>

Use the extreme value method. All aluminum is required.

That is, the minimum amount required.

All zinc is required.

Zinc is the same as magnesium H2 so.

The amount of possible substances is between c

The mol ratio of the hydrogen produced by mg and zn is 1:1

al is the amount of hydrogen is.

Because of the presence of aluminum, the total amount of alloy is less.

And if it's all alloys, it's all aluminum.

The amount of alloy is.

Therefore, the value range is between to. c

It reacts with the acid to form liters of H2 which is H2, and the electrons of are transferred together.

Use the extreme assumption method (also called the limit method).

Let all be Al and provide 3mol of electrons per mole of Al, so if all Al then the amount of Al matter is >

Let all of them be zn (or mg is the same) and donate 2 mol of electrons per mole, so the amount of matter is and because it is impossible not to be all one metal, and the mixture is three metals, the sum of the total amount of matter is between the two, so only c is the same.

mg, al, and zn are x, y, z mol, x+x+, respectively

x+y+z=

When y=0, x+y+z=mol

When y>0, x+y+z< mol

When x=z=0, x+y+z=mol

Therefore, the answer should be C

si + 2naoh + h2o ==na2sio3 + 2h2↑

2al+2oh－+2h2o=2alo2－+3h2↑

The molar number of equimass silicon and aluminum is m 28 and m 27 respectively, 1mol of silicon corresponds to 2mol of hydrogen, 1mol of aluminum corresponds to 3 2mol of hydrogen, and the number of moles of silicon and aluminum corresponding to the generation of hydrogen is m 14 and m 18, so the volume ratio of hydrogen released is 9:7

Si+2NaOH+H2O==Na2SiO3+2H2 28G Si yields 4G H2

2al+2naoh+2h2o==2naalo2+3h2 54g al to generate 6g h2

Therefore, the ratio of the volume of hydrogen released by the same mass of silicon and aluminum into a sufficient amount of NaOH solution is = 4 28) :(6 54) = 9:7 after sufficient reaction

M average = 25, with the extreme method, if it is a mixture of CO and H2, H2 accounts for 3 26;If it is a mixture of H2 and O2, H2 accounts for 7 30So the answer is A

Consider (NH4)2SO4 first, because SO42- is only present in (NH4)2SO4 and is not disturbed by NH4Cl and KCL.

Now 24mol K2SO4 is 24mol SO4 2-, so the amount of (NH4)2SO4 species required = 24mol

24mol (NH4)2SO4 can provide 48mol NH4+, and the amount of NH4+ in the nutrient solution = 50mol, so the amount of NH4Cl = 50 - 48 = 2mol

The amount of substances containing K+ in the nutrient solution = 16 + 24*2 = 64mol, so the amount of KCl substances required = 64mol

X, Y, and Z mol are required respectively

k+ equal: x=16*1+24*2

NH4+ equal: y+2z=50*1

cl-equal: x+y=50*1+16*1

SO42-equal: z=21*1

Using any three of the above equations, we can solve:

x=64y=2

z=24

(1) The hydrochloric acid happens to be completely reacted, indicating that all the hydrogen elements in the hydrochloric acid have generated hydrogen, that is:

2hcl---h2

2 moln mol to obtain n = 1 mol (2) let Alcl3 be a mol and MgCl2 be b mol, according to the conservation of chlorine and the mass of metal elements, it can be obtained:

3a + 2b = 1 (1 mol is the amount of hydrochloric acid).

27a + 24b= ( magnesium-aluminum alloy) is solved to obtain a = b=, so the mass of magnesium in the alloy is 24b, i.e. grams.

Under standard conditions, hydrogen is, according to the equation, every 2mol of hydrochloric acid consumed, 1mol of hydrogen is generated, so the corresponding relationship between hydrogen consumption of 1mol of hydrochloric acid mg and al reaction with HCL to generate H2 is 2mg-2H2, 2Al-3H2, let mg have xmol, Al has ymol. Column equation 1: 24x+27y= 2:

x+(3 2)y= Solve x and multiply it by 24.

Solution: (1) Let the original mass of the solution be the mass of.

By equation: m = 175g

mkcl=nkcl=

C Prima = 2) Solution: The ratio of the amount of water to NaOH is 10:1, and the amount of NNA=10G 40G mol=

nh2o=mh20=18g/mol*

v=45ml

CNA+=3, (1) is given by solubility x

x/（100+x）=w

x=100w/(1-w)

2) The mass concentration of the device is c (must be formed into a standard unit, and the volume is l).

c=n/v=(m/ / 1000m/ (mol/l)

3) You may wish to have 1g of solution, and the volume of the solution is 1 d solute mass = 1*w The amount of solute material = 1*w

c= 1*w/ / 1/d* =1000wd/ (mol/l)

Be sure to pay attention to the answer, and be sure to turn it into a standard form).

Solution 4: Due to the complete reaction, take b as a reference, a=3b, c=2b

The reaction is written as (3b)x2+(b)y2 =(2b)?

Put b out, followed by 2b x3 y and select c (the reaction can also refer to n2+3h2=2nh3).

5 Solution: From the standard condition, the temperature and pressure are consistent, and it can be seen from the Kraberon equation that the volume ratio is the ratio of the mass of the matter.

Let the amount of the substance be 4n, 3n, 2n respectively by 28 * 4n + 32 * 3n + 44 * 2n = 100g

The amount of N=N2 is O2 is CO2 and mol

Multiply separately. N2 volume = , O2 is; CO2 is.

Be sure to pay attention to the units, and pay attention to the conversion of non-standard units.

1. Let the mass of the original solution be X, the mass of the solute in the solution before and after dilution is the same, that is, X=175g, the mass of the solute is, the amount of the substance of KCL is, and the amount of the substance of the original solution is the concentration.

2. The ratio of water molecules to NaOH is 10:1, let water xg, x 180 = 40 10, x = 720g, and the amount concentration of the substance is c =

3、（1）100w

2）1000m/

3)1000wd/

c=2c, the reaction formula is 3x2+y2=2z, and z is x3y5The average molecular mass of the gas mixture is (28*4+32*3+44*2) 9=volume=100

is the same as the end of the world, one in one.