Answer:
a
Explanation:
The formation of ion occurs when an atom that is said to be neutral gains or losses electrons.
At the time it gains electrons, it is regarded that a negative ion (anion) is formed.
When it loses electron, it is regarded that a positive ion (cation) is formed.
Atomic number = No of protons and electrons occurring in a neutral atom.
Given that:
Protons = 14
electron = 18
Net Charge = no of proton - no of electron
= 14 - 18
= -4
Mass number = 14 + 15 = 29
So, the chemical symbol = 
For ion with
27 proton, 32 neutrons and 25 electrons
Net charge = 27 - 25
= +2
Mass number = 27 + 32 = 59
Thus, the chemical symbol = 
It is indeed true. Molecular Polarity occur in covalent bonds. Like the water molecule h20
Answer:
The three methods we can use to separate a mixture are:
1.sublimation
example: sand and camphor
2.centrifugation
example:cream from milk
3.evaporation
example:salt from sea water
Answer: True.
Explanation:
Molar mass is the mass of a given substance divided by the amount of that substance, measured in g/mol. For example, the atomic mass of titanium is 47.88 amu or 47.88 g/mol. In 47.88 grams of titanium, there is one mole, or 6.022 x 1023 titanium atoms.
The characteristic molar mass of an element is simply the atomic mass in g/mol. However, molar mass can also be calculated by multiplying the atomic mass in amu by the molar mass constant (1 g/mol). To calculate the molar mass of a compound with multiple atoms, sum all the atomic mass of the constituent atoms.
For example, the molar mass of NaCl can be calculated for finding the atomic mass of sodium (22.99 g/mol) and the atomic mass of chlorine (35.45 g/mol) and combining them. The molar mass of NaCl is 58.44 g/mol.
Beryllium Oxide has a molecular formula Be-O
% covalent character in Be-O = 100 - % ionic character in Be-O
Now,
% ionic character = {1 - exp[(-0.25)(E(Be)-E(O))²]}*100
where,
E(Be) and E(O) are the electronegativity values of Be and O
Based on the Paulings scale:-
E(Be) = 1.57
E(O) = 3.44
% ionic = {1 - exp[(-0.25)(1.57-3.44)²]}*100 = 58.28%
Therefore, % covalent = 100 -58.28 = 41.72%
Ans: b) 41 percent