Answer:
Elements having same valence electrons are placed in <u>same group.</u>
Explanation:
First, let's start with some basic concepts of modern periodic table:
1. Modern Periodic table : It is the arrangement of element in the increasing order of their atomic numbers
The Modern periodic table is divided into Periods and groups .
Periods : These are the horizontal rows. There are seven periods in the periodic table . Period 1 has 2 element. Period two and three has 8 elements , period 4 and 5 have 18 elements and the period 6 and 7 have 32 elements.
Same period have same number of atomic orbital(Shell)
Group : The group is the vertical columns . There are 18 groups in the modern periodic table.Those element which have same group number will also have same number of electron in their outermost shell. The number of electron in the outermost shell determines the valency of the element.
So, elements showing same valency are placed in same group.
All alkali are place in group 1 and have 1 valance electron in the outermost shell
Answer: 10.9 mol.
Explanation:
- To understand how to solve this problem, we must mention the reaction equation where water produced from PbO₂.
Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
- Now, it is a stichiometric oriented problem, that 1 mole of PbO₂ produces 2 moles of H₂O.
Using cross multiplication:
1.0 mole of PbO₂ → 2.0 moles of H₂O
5.43 moles of PbO₂ → ??? moles of water
The moles of water produced = (5.43 x 2.0) = 10.86 moles ≅ 10.9 moles.
Explanation:
By losing or gaining electrons from its outermost orbit
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
