Noble gases have complete valence electron shells
Answer:
Along period electronegativity and ionization energy increases.
Along group electronegativity and ionization energy decreases.
Explanation:
Along period:
As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. Thus the attraction of the atoms for valance electrons increases. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required, and electronegativity also increases.
Along group:
As we move from top to bottom in periodic table the atomic sizes increases.The electrons are added in next energy level in every next element. Thus the valance electrons farther away from the nucleus and hold of nucleus becomes weaker, because of weak nuclear attraction atomic radii increases and electronegativity and ionization energy decreases.
Answer:
b. 6.02 x 1023 molecules
Explanation:
The formula mass of ammonia is 14 + 1 × 3 = 17.
The number of moles in 27.6g ammonia is 27.6 ÷ 17 = 1.62 mol.
A mole is 6.02 × 10²³, so the number of hydrogen atoms in a 1.62 moles of ammonia is 1.62 × 6.02 × 10²³ × 3 = 2.93 × 10² atoms.
Answer:
-514 kJ/mol
Explanation:
The bond enthalpy which is also known as bond energy can be defined as the amount of energy needed to split one mole of the stated bond. The change in enthalpy of a given reaction can be estimated by subtracting the sum of the bond energies of the reactants from the sum of the bond energies of the products.
For the given chemical reaction, the change in enthalpy of the reaction is:
Δ
[2(409) + 4(388) + 3(496) - 4(630) - 4(463)] kJ/mol = 818 + 1552 + 1488 - 2520 - 1852 = -514 kJ/mol
