Answer:
Please answer my question
18.The octet rule tells us that in every chemical
reactions, elements will either gain or lose electrons to attain the noble gas electron
configuration. This stable<span> electron configuration is known as the octet configuration
since it is composed of 8 valence. Oxygen’s electron configuration is 1s2 2s2
2p4. So when</span> oxygen reacts with
other elements to form compounds, it completes the octet configuration by
taking 2 electrons from the element
it reacts with
19. Actually pure metals are made up not of
metal atoms but rather of closely packed cations (positively charge particles).
These cations are then surrounded by a pack of mobile valence electrons which
drift from one part of the metal<span> to
another. This is called metallic bond.</span>
20. This is the
energy which is needed to break a single bond. When the dissociation energy is
large, this means that the compound is more stable. Since carbon to carbon
bonds have high dissociation energy, therefore they are not very reactive.
21. Network solids are type of solids
in which the atoms are covalently bonded to one another, so they are very
stable. It takes higher temperature to melt them because breaking these
covalent bonds required greater energy. Some examples are:
- Diamond
<span>-Silicon Carbide</span>
Answer:
The Relative Formula Mass of Fe(NO₃)₂ is 179.8524 grams
Explanation:
The Relative Formula Mass is the mass of one mole of a compound expressed in grams, obtained by adding together the Relative Atomic Masses, RAM, of the elements which makes the compound
The Relative Formula Mass of a compound is the same as its Relative Molecular Mass
The relative formula mass of Fe(NO₃)₂ is given as follows;
The relative atomic mass of Fe = 55.845 amu
The relative atomic mass of nitrogen, N = 14.0067 amu
The relative atomic mass of oxygen, O = 15.999 amu
Therefore, we have;
The formula mass of Fe(NO₃)₂ = (55.845 + 2×(14.0067 + 3×15.999)) amu = 179.8524 amu
The Relative Formula Mass of Fe(NO₃)₂ = 179.8524 grams.
Missing question: What is the rate constant for the reaction?
<span>[RS2](mol L-1) Rate (mol/(L·s))
0.150 0.0394
0.250 0.109
0.350 0.214
0.500 0.438</span>
Chemical reaction: 3RS₂ → 3R + 6S.
Compare second and fourth experiment, when concentration is doubled, rate of concentration is increaced by four. So rate is:
rate = k·[RS₂]².
k = 0,438 ÷ (0,500)².
k = 1,75 L/mol·s.
