Forehead. Chemistry topic in the first year of high school. Some chemistry topics in the first year

a) (1) Na2O2 covalent bonds, ionic bonds.

2）nahso4,nahso3

b) (1) Carbon-carbon double bond, carboxyl group.

2) C2H6O CH3CH2CH2OH (propanol), CH3CH(CH3)OH

3) CH2 = CH-COOH + C2H5OH = = = = CH2 = CH-COOC2H5 + H2O (condition is concentrated sulfuric acid, heated) substitution or esterification.

A is H, B is O, C is Na, and D is S

1, Na2O2, covalent bond (O-O), ionic bond (Na-O), NaHSO3

1。Carboxyl group (can be combined with NaHCO3 to CO2), carbon-carbon double bond (addition reaction with carbon tetrachloride solution of bromine).

ch3ch2ch2oh

CH3CH2OH== concentrated sulfuric acid CH2==CHCOOCH2CH3+H2O, esterification reaction.

a] a: h, b: o, c: na, d: s

1) Na2O2, ionic bond, covalent bond.

2）nahso4、nahso3

b) (1) Carbon-carbon double bond, carboxyl group.

2) C2H6O, CH3CH2CH2OH CH3CH(OH)CH33)CH2=CHCOOH+CH3CH2OH --CH2=CHoCH2CH3+H2O, which belongs to esterification reaction.

Na2O2 ionic bond Covalent bond.

nahso4 nahso3

Carboxyl carbon-carbon double bond.

C2H6O ch3-ch2-ch2-oh ch3-ch0h-ch3ch2=ch-coOH + CH3CH2OH = ch2=ch-Cooch2ch3 +H2O lipidation reaction.

Question 1: A is H, B is O, C is Na, D is S

The answer to the first question is Na2O2 and the second answer is non-polar covalent and ionic bonds.

nahso4,nahso3

Question 2: 1) Carboxyl double bonds.

2）c3h6o

3) CH2=CH-COOH+C2H5OH---CH2=CH-COOC2H5+H2O esterification reaction.

Wow dizzy. I was stunned.

2) It should be written as a. Answer:

It can be seen relatively quickly, and it may be a little overlooked. If you have any questions, you can contact me again.

b b c d c 56 na ratio of (17*:(176

The number of molecules is only the ratio of the quantity of matter = 3;4

Ratio of atomic number =; :1

The outermost electron number of an element r is one-third of its total number of electrons.

1. If the number of electrons in the outermost shell is 1 and the total number of electrons is 3, then the element is Li and the oxide is Li2O

2. If the number of electrons in the outermost shell is 2 and the total number of electrons is 6, then the element is C, and the oxide is Co, Co2

3. If the number of electrons in the outermost shell is 3 and the total number of electrons is 9, then the element does not exist.

4. If the number of electrons in the outermost shell is 4 and the total number of electrons is 12, then the element does not exist.

5. If the number of electrons in the outermost shell is 5 and the total number of electrons is 15, then the element is P, and the oxide is P2O5 and P2O3

6. If the number of electrons in the outermost shell is 6 and the total number of electrons is 18, then the element does not exist.

7. If the number of outermost electrons is 7 and the total number of electrons is 21, then the element does not exist.

8. If the number of electrons in the outermost shell is 8 and the total number of electrons is 24, then the element does not exist.

Visible: The oxide of this element cannot be dro3

For the main group of elements, the valency has a large relationship with the number of outermost electrons: the highest positive valence = the number of outermost electrons of the atom (except o and f) |Negative price|=8—n n is the number of electrons in the outermost shell of the atom.

If you choose d d, the electronic configuration will either be 286 or 26 and the outermost shell will not be one-third of the total.

In the case of the higher one, the number of electrons in the outermost shell is generally positive valency 8 minus the outermost shell, which is negative valency.

A is 21 B is 285 C is 285 There are multiple valencies of phosphorus and nitrogen to remember.

The maximum number of electrons in the outermost shell is 8.

When the number of electron shells is 2.

2 +x ==3x

x==1 is li r2o, i.e. li2o

When the number of electron layers is 3.

2+8+x==3x

x==5 i.e. p i.e. p2o5 or p2o3 when the number of electron layers is 4.

x is already greater than 8, impossible.

Therefore it is impossible to be d