453 divided by 224
density is roughly 2.02 g per ml
as a ml is 1 cm3 density is 2.02 grams per centimeter cubed
<span>Begin by classifying which energy level, and indirectly principal quantum number, n, resembles to the N shell.
no. of orbitals =n2
In your case, the fourth energy level will contain
n=4⇒no. of orbitals= 4^2=16
The number of subshells is given by the principal quantum number.
no. of subshells=n
In your case, the fourth energy level will have
no. of subshells = 4 this is the answer
to check:
the fourth energy shell will can hold a thoroughgoing of no. of electrons=2⋅42=32 e−</span>
This is the Lewis Dot structure for oxygen
Answer:
The answer is 375.54 g of AgBr
Explanation:
Mass (g) = Concentration (mol/L) x volume (L) x Molecular Weight of AgBr (g/mol)
Mass = 2M x 1L x 187.77 g/mol
Mass = 375.54g
Answer:
At Equilibrium
[COCl₂] = 0.226 M
[CO] = 0.054 M
[Cl₂] = 0.054 M
Explanation:
Given that;
equilibrium constant Kc = 1.29 × 10⁻² at 600k
the equilibrium concentrations of reactant and products = ?
when 0.280 moles of COCl2(g) are introduced into a 1.00 L vessel at 600 K. [COCl²]
Concentration of COCl₂ = 0.280 / 1.00 = 0.280 M
COCl₂(g) ----------> CO(g) + Cl₂(g)
0.280 0 0 ------------ Initial
-x x x
(0.280 - x) x x ----------- equilibrium
we know that; solid does not take part in equilibrium constant expression
so
KC = [CO][Cl₂] / COCl₂
we substitute
1.29 × 10⁻² = x² / (0.280 - x)
0.0129 (0.280 - x) = x²
x² = 0.003612 - 0.0129x
x² + 0.0129x - 0.003612 = 0
x = -b±√(b² - 4ac) / 2a
we substitute
x = [-(0.0129)±√((0.0129)² - 4×1×(-0.003612))] / [2 × 1 ]
x = [-0.0129 ± √( 0.00017 + 0.01445)] / 2
x = [-0.0129 ± 0.1209] / 2
Acceptable value of x =[ -0.0129 + 0.1209] / 2
x = 0.108 / 2
x = 0.054
At equilibrium
[COCl₂] = (0.280 - x) = 0.280 - 0.054 = 0.226 M
[CO] = 0.054 M
[Cl₂] = 0.054 M
