Senior 1 Chemistry Fill in the Blank Questions, Senior 1 Chemistry Fill in the Blank Questions

There are 10 electrons outside the nucleus of ion B+ and B is known to be Na

Then the sum of the atomic numbers of a and b is 24Again, they are short-period elements. Let the atomic number of a be x, then x+x+8=24

x=8, x+8=16, so abc are o, na, s, na2o2, na2s, respectively

One point) s (six points) na (one point) na+[s (six points)]2-na

It's all the same ... Direct copying. Excuse me.

al p (meet the conditions of the question, but not the taste of the question, the question is good and want the answer later); o na s

2.Not; na2o2,na2o,na2s

3.The comrade above wrote very well, and the "point" could not be played, so I didn't write it again.

Yes O (oxygen) B is Na (sodium) and C is S (sulphur).

I should be able to understand it, it's sodium oxide).

Na2O2 (sodium peroxide).

Na2S (sodium sulfide).

Na2S2 (sodium persulfide).

Na2 (SO3) (sodium sulfite).

Na2 (SO4) (sodium sulfate).

3.[na+]2s (8 points around s, you know what I mean) na+]ss[na+] (two sulfurs are surrounded by eight points, and there is a common electron pair between them).

The first time such a problem,**If you don't make it clear, please point it out.。

Mass in 1 l hydrochloric acid = = 1190 g

Solute HCl mass = 1190 * g (not counted first) Amount of HCl species = 1190 * =

So the amount of concentration of the substance=

The amount concentration of the 500ml substance is prepared as the amount of HCl substance contained = the volume of concentrated hydrochloric acid mentioned above = = =

200ml of 98% concentrated sulfuric acid (density =, the amount concentration of its substance is (.

Pure sulfuric acid weight: 200* grams.

Amount of Concentration of Substance:

Take 50mol l of the solution, then the concentration of the solution is (98%), take 50mol l of the solution, then the concentration of the amount of the substance of the solution is (, which contains H2SO4 (;

g) Add 100 ml of sulfuric acid solution to the original solution, then the concentration of the resulting solution is (, which contains H2SO4 (.

Same as the original concentration, so still.

After adding 100ml, there is a total of 200+100=300ml

200ml of 98% concentrated sulfuric acid (density =, the amount concentration of its substance is (. take 50mol l of the solution, then the concentration of the solution is , which contains; Add 100 ml of sulfuric acid solution to the original solution, and the concentration of the resulting solution is, which contains H2SO4.

=(1000* )/98

98% (take 50ml?) )

Take 50ml? ）

g = 50* take 50ml, right? ）

The value of the sulphuric acid taken by the owner of the mol l building is incorrect. I guess it's the volume.

Feel free to ask.

1. When formulating, the instruments that must be used are 1 4 5 6 The instrument that is also missing is the rubber head dropper In this experiment, the glass rod is used twice, and its functions are stirring and draining

2. To be weighed = 10 g (please attach the calculation process) caustic soda solid, the solid should be weighed in a dry beaker.

3. One step that must be carried out before using the volumetric flask is to check whether the volumetric flask is leaking The specific operation method is to add tap water, after capping the cork, press the stopper with the index finger of one hand, hold the part above the bottleneck mark with the remaining fingers, and hold the bottom edge of the bottle with the fingertip of the other hand, and stand upside down for two minutes. If it doesn't leak, hold the bottle upright, rotate the stopper 180°, and try it upside down again

Fourth, the preparation, can generally be divided into the following steps: weighing, calculation, dissolution, shaking, transfer, washing, volume and cooling. The correct sequence of operations is 2 1 3 8 5 6 7 4

5. In the process of determining the volume of the solution in the volumetric flask, the method of completing the small amount of water added in the later stage is to add water to the volumetric flask with a rubber head dropper to the scale mark

A: CH3CH2C00CH2CH2CH3 Propyl propionate B: CH3CH2C00NA Sodium propionate C:

CH3CH2C00H Propionic Acid D: CH3CH2CH2OH n-Propanol A B+D: CH3CH2C00CH2CH2CH3 + H2O-sodium hydroxide CH3CH2C00H+ CH3CH2CH2OH

b e ch3ch2ch2oh-copper oxide the d after ch3ch2cho copper sulfate should be changed to e.

(1) Guess that A is an ester, D (should be E) can generate C, indicating that the number of carbon atoms is the same, there are three each, and E is propionaldehyde, C is propionic acid, CH3CH2COOH, then A is propyl propionate CH3CH2COOCH2CH2CH3

b is sodium propionate ch3ch2coona

2)ch3ch2cooch2ch2ch3+naoh---ch3ch2coona+hoch2ch2ch3

2ch3ch2cho+o2---2ch3ch2cooh(3) distillation, propionaldehyde evaporated, sodium propionate obtained by cooling through a condenser tube is left in the solution, and acidification can be obtained.

b is sodium propionate, c is propanol, and a is propyl propionate.

The rest is hard to type.

A will decompose into sodium salts and alcohols under alkaline.

d is the oxidation of propanol to acetone fractionation.

a: c-c-c-o-c (there is a carbon-oxygen double chain here)-c-c b: c-c-c (carbon and oxygen double chain here)-o-na c:

c-c-c (carbon and oxygen double chain here)-o-h Equation 1: c-c-c-o-c (there is a carbon and oxygen double chain here)-c-c - > c-c-c (carbon and oxygen double chain here)-o-na + c-c-c-o-h Equation 2: c-c-c-o-h - > c-c-c (with carbon and oxygen double health).

**: Fractionation (hydrogen on the carbon can be filled).

1.There are three solutions, sodium chloride, calcium chloride and Alcl3, which react with a sufficient amount of silver nitrate solution: if the amount of the substances containing the solute is equal, the ratio of the amount of substances that form the precipitate is (1:2:3).

NaCl, MgCl2, and AlCl3 contain Cl ions, respectively, so the ratio of the amount of precipitated substances produced by the amount of equal substances is 1:2:3.

If the mass of the precipitate is equal, the ratio of the amount of solute contained in the three solutions (6:3:2).

Let all generate 6mol of AGCL, NaCl, MgCl2 and AlCl3 contain Cl ions respectively, NaCl is 6 1=6mol, MgCl2 is 6 2=3mol, and AlCl3 is 6 3=2mol

2.(6) The number of nitrogen atoms contained in gram urea is the same as that contained in ammonium chloride.

The amount of the substance of NH4Cl is n=m m=

1 NH4Cl contains 1 n atom, so the amount of matter of n atom is.

The chemical formula of urea is CO(NH2)2 and contains 2 n atoms, so the amount of substance requiring urea is.

The mass of urea required is m=nm=

1, 220 grams 112 liters to the 22nd power and 22 powers.

2, 98*2 g 4 moles per liter 2 moles per liter.

3, Same CO2

4, liters. 5.5 grams without moles.

6.95 grams per mole 24

Moles per liter.

In that order, they are:

g (c:12,o:16 12+16*2=44,44*5=200g) liters (standard gas molar volume constant: liters).

23rd power*5 (23rd power is Afogadro's constant) and 23rd power*2 (numerator number multiplied by 2).

g (（32+16*4+2）*2=208g)10mol/l (4/

5mol/l (2/

3. Same. co2

l ((cl:17 17*3?.)

6. Because it is a divalent metal, the general formula is: MCL2, and the Cl- contained in it, then the salt is, the molar mass is, the appropriate amount is 95, and the molar mass of the metal is 95g mol-71g mol=24g mol

24mol

Avogadro's Law Law of Conservation of Mass The gas is CO or H2