Missing in your question Ka2 =6.3x10^-8
From this reaction:
H2SO3 + H2O ↔ H3O+ + HSO3-
by using the ICE table :
H2SO3 ↔ H3O + HSO3-
intial 0.6 0 0
change -X +X +X
Equ (0.6-X) X X
when Ka1 = [H3O+][HSO3-]/[H2SO3]
So by substitution:
1.5X10^-2 = (X*X) / (0.6-X) by solving this equation for X
∴ X = 0.088
∴[H2SO3] = 0.6 - 0.088 = 0.512
[HSO3-] = [H3O+] = 0.088
by using the ICE table 2:
HSO3- ↔ H3O + SO3-
initial 0.088 0.088 0
change -X +X +X
Equ (0.088-X) (0.088+X) X
Ka2= [H3O+] [SO3-] / [HSO3-]
we can assume [HSO3-] = 0.088 as the value of Ka2 is very small
6.3x10^-8 = (0.088+X)*X / 0.088
X^2 +0.088 X - 5.5x10^-9= 0 by solving this equation for X
∴X= 6.3x10^-8
∴[H3O+] = 0.088 + 6.3x10^-8
= 0.088 m ( because X is so small)
∴PH= -㏒[H3O+]
= -㏒ 0.088 = 1.06
Answer:
Jewelry
Explanation:
Gold and silver are metals which are malleable, ductile, and lustrous. Also they are noble metals. since they are noble metals they are not affected by acid, oxygen, atmosphere and etc. hence gold and silver are used or suitable for making jewlery.
Answer: 1. CaO + H2O => Ca(OH)2
2. P4 + 5O2 => 2P2O5
3. 2 Ca + O2 => 2 CaO
4. 8 Cu + S8 => 8 CuS
5. CaO + H2O => Ca(OH)2
6. S8 + 8 O2 => 8 SO2
7. 3 H2 + N2 => 2 NH3
8. H2 + Cl2 =>2 HCl
9. 16 Ag + S8 => 8 Ag2S
10. Cr + O2 => 2Cr2O3
11. 2Al + 3Br2 => 2AlBr3
12. 2Na + I2 => 2NaI
13. 2H2 + O2 =>2 H2O
14. 4 Al + 3O2 => 2 Al2O3
Explanation:
Answer: Option (5) is the correct answer.
Explanation:
It is known that the ground state electronic configuration of silicon is
.
And, we know that when an atom tends to gain an electron then it acquires a negative charge and when an atom tends to lose an electron then it acquires a positive charge.
As
has a +4 charge which means that it has lost 4 electrons. Hence, the electronic configuration of
is
.
According to the Aufbau principle, in the ground state of an atom or ion the electrons fill atomic orbitals of the lowest energy levels first, before filling the higher energy levels.
As 2p orbital is filled after the filling of 2s orbital.
Therefore, we can conclude that 2p orbital will be occupied by the electrons of highest energy for the
ground-state ion.
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