The Easiest Way to Understand Chemistry. Chemistry Concepts, Problems and Solutions. Sergey D Skudaev
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Iron: 1s 2 2s 2 2p 63s 2 3p 64s 2 3d 6
Cobalt: 1s 2 2s 2 2p 63s 2 3p 64s 2 3d 7
Nickel: 1s 2 2s 2 2p 63s 2 3p 64s 2 3d 8
Copper: 1s 2 2s 2 2p 63s 2 3p 64s 1 3d 10
Zinc: 1s 2 2s 2 2p 63s 2 3p 64s 2 3d 10
Gallium: 1s 2 2s 2 2p 63s 2 3p 64s 2 3d 10 4p 1
What is interesting is that Vanadium has 3d3 4S2 electrons and the next element, Chromium (Cr), should have 3d4 4S2 electrons, but actually it has 3d54S1 electrons. One electron passed from the 4S orbital to the 3d orbital.
Nickel has 8 electrons on the d orbital. The next element, Copper (Cu), should have 9 electrons on the 3d orbital and 2 electrons on the 4S orbital, but actually Copper has 10 electrons on the 3d orbital and 1 on the 4S orbital. One electron passed from the 4S orbital to the 3d orbital. The electronic configuration of Copper is [Ar] 3d104S1.
The complete list of electronic configurations of all the chemical
elements cab be found in Wikipedia:
http://en.wikipedia.org/wiki/Electron_configurations_of_the_elements_%28data_page%29
All elements can be divided into metals and nonmetals.
In the periodic table, metals are on the left, nonmetals are on the right.
Group number shows how many electrons are in the outermost orbital. These electrons are called valence electrons. For example, Na (sodium) is in the first group. It has one electron on the outermost orbital Na can easily give this electron to Cl (chlorine). Cl is in the seventh group. It has 7 electrons and it takes one electron from Na. As a result Na becomes a positive ion Na+ and Cl becomes negative ion Cl -. Ions with opposite charge form ionic bonds.
Ionic bonds usually form crystal structures. That is why salt is made of crystals.
Carbon C, is located in the 4th group. It has four valence electrons. As a result C forms four covalent bonds with four atoms of Cl. C does not give its electrons to Cl. Carbon and chlorine share electrons. When atoms share electrons, they form covalent bonds.
Oxides
Oxides are produced when metals or nonmetals react with oxygen.
Oxygen is located in the 6th group and has 6 valence electrons. It tends to gain 2 more electrons to become a complete octet and its valence is 2.
2Ca + O2 = 2CaO
In nature, metal oxides exist in clay. Clay is a mixture of the oxides
SiO2
Al2O3
K2O
Na2O
MgO
CaO
Fe2O3
TiO2
Bases
In reaction with water metals or metal oxides produce a base:
2Na +2H2O = 2NaOH + H2
CaO + H2O = Ca (OH) 2
Bases dissociate in water and produce a negative hydroxide OH – ion.
Acids
Non metal oxides are NO2, SO3, P2O5
In reaction with water non metal oxides produce acids:
H2O + SO3 = H2SO4 – sulfuric acid
H2O + NO2 = HNO3 – nitric acid
H2O + CO2 = H2CO3 – Carbonic acid
Acids dissociate in water and produce proton of hydrogen H+.
Salts
When an acid reacts with a base, a salt is produced.
NaOH + HNO3 = NaNO3 + H2O
To calculate the percentage composition of NaNO3, find the molecular mass of NaNO3
23 (Na) +12 (N) +16*3 (O3) = 83
The molecular weight of NaNO3 = 83
83 – 100%
23 (Na) – X%
X = 23 * 100 /83 = 27.7% of Na
83 – 100%
12 (N) – X %
X = 12 * 100/83 = 14.5% of N
83 – 100%
48 (O) – X %
X = 48 * 100/83 = 57.8% of O
Some salts are more soluble in water; some are less soluble or not soluble at all. When a non-soluble salt is produced as a result of an acid and base reaction, a precipitate is formed.
Ca (OH) 2 + H2CO3 = CaCO3 +2H2O
CaCO3 is not soluble in water. A white precipitate is formed.
Salts may react with each other and new salts are produced:
Equivalent proportions
All chemical reactions occur in equivalent proportions.
1. How many grams of Ca Cl2 are spent in the following reaction:
10g?g
Na2CO3 + Ca Cl2 = CaCO3 +2NaCl
All compounds react with each other in certain proportions. In a given reaction one mole of Na2CO3 reacts with one mole of CaCl2
A mole is MW (Molecular mass) in grams.
For Na2CO3 MW is 23 *2 +12 +48 = 106 g. = 1 mole
For CaCl2 MW is 40 +35*2 = 110g. = 1 mole.
106 g Na2CO3 react with 110g CaCl2
10g Na2CO3 react with X g CaCl2
X= 10 * 110 / 106 = 10.38 g CaCl2
2. How many grams of CaCO3 are produced if 100 ml of 0.5 M solution of Na2CO3 reacts with an unlimited volume of solution CaCl2?
0.5 M
2. Na2CO3 + Ca Cl2 = CaCO3 +2 NaCL
100 ml
1 liter of I M solution of Na2CO3 contains 106 g
How many grams of Na2CO3 in 1 liter of 0.5 M solution?
1 M – 106 g
0.5 M – X g
X