This element is found in group 3A, period 3
<h3>Further explanation
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The maximum number of electrons that can be filled in the nth electron shell is 2n²(n=shell)
-
K shell (n = 1) maximum 2 x 1² = 2 electrons
- L shell (n = 2) maximum 2 x 2² = 8 electrons
- M shell (n = 3) maximum 2 x 3² = 18 electrons
- N shell (n = 4) maximum 2 x 4² = 32 electrons
Electron configuration of element X : 2.8.3 , so :
K shell = 2 ⇒1s²
L shell = 8⇒2s²2p⁶
M shell = 3⇒ 3s²3p¹
Block p: group 13-18 (has a 2p-6p configuration), also called a representative element because it includes metals, non-metals and metalloids
The outer shell 3s²3p¹ : located in group 3A and period 3
group⇒valence electron ⇒3
period⇒the greatest value of the quantum number n⇒3
A: 2H₂ + O₂ → 2H₂O
H: 4 H:4
O: 2 O: 2
The equation is balanced.
B. 2S + 3O₂ → 2SO₃
S: 2 S: 2
O: 6 O: 6
The equation is balanced.
C. Li + Cl₂ → LiCl
Li: 1 Li: 1
Cl: 2 Cl: 1
The equation is not balanced.
2Li + Cl₂ → 2LiCl
Li: 2 Li: 2
Cl: 2 Cl: 2
D: 2K + 2H₂O → H₂ + 2KOH
K: 2 K: 2
H: 4 H: 4
O: 2 O: 2
The equation is balanced.
E: 2Fe + Cu(NO₃)₂ → 2Cu + Fe(NO₃)₂
Fe: 2 Fe: 1
Cu: 1 Cu: 2
N: 2 N: 2
O: 6 O: 6
The equation is not balanced.
2Fe + 2Cu(NO₃)₂ → 2Cu + 2Fe(NO₃)₂
Fe: 2 Fe: 2
Cu: 2 Cu: 2
N: 4 N: 4
O: 12 O: 12
The following equations that are balanced are A, B, and D.
The temperature will be 1200K if the volume remained constant
calculation
This is calculated using gay lussac law formula, that is P1/ T1=P2/T2 since the volume is constant
P1 = 100 Kpa
T1= 300 K
P2= 400 Kpa
T2=?
by making T2 the subject of the formula T2 =( P2 xT1)/P1
T2 is therefore = (400 KPa x 300 K) / 100 Kpa = 1200 K
