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Papessa [141]
3 years ago
13

The maximum number of energy sublevels found in any energy level at ground state is:

Chemistry
2 answers:
algol [13]3 years ago
6 0

Answer:

The maximum number of energy sublevels found in any energy level at ground state is one

Explanation:

The energy state in general can be specified by four quantum numbers:

1) n = principal quantum number = 1,2,3...

2) l = angular momentum where l = 0...n-1

where: l = 0 corresponds the 's' sublevel

l = 1, corresponds to the 'p'

l = 2, corresponds to the 'd' sublevel

3) m = magnetic quantum number which extends from -l....+l

4) s = spin = +1/2 or -1/2

For the ground state n = 1 and l = 0 which corresponds to a single 's' sublevel.

Jlenok [28]3 years ago
4 0

the sublevels of the first four principal energy levels and the maximum number of electrons that the sublevels can contain are summarized in Table 5.1.

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Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1
Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

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SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of SbCl_5 is  increasing .So, the equilibrium will shift in the right direction.

b)

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Concentration of SbCl_5  = 0.195 M

Concentration of SbCl_3  = 6.98\times 10^{-2} M

Concentration of Cl_2  = 6.98\times 10^{-2} M

On adding more [SbCl_5 to 0.370 M at equilibrium :

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Initially

0.370 M         6.98\times 10^{-2}M    

At equilibrium:

(0.370-x)M   (6.98\times 10^{-2}+x)M  

The equilibrium constant of the reaction  = K_c

K_c=2.50\times 10^{-2}

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K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}

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x = 0.0233 M

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[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

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