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PIT_PIT [208]
3 years ago
7

What information can be learned about the element chlorine (Cl) by looking at the periodic table? Select all that apply.

Chemistry
2 answers:
olasank [31]3 years ago
5 0

That it is number 17   35.45  it's period is 3.

Hope that helped ^^

pantera1 [17]3 years ago
3 0
It's a halogen. It's a non metal. It's valency is 1. It is highly electronegative and non metallic in nature.
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Calculate [h3o+] and [s2−] in a 0.10 m solution of the diprotic acid hydrosulfuric acid. (for hydrosulfuric acid ka1 = 9.0 × 10−
Rudiy27
The equation for the first dissociation is:
H₂S(aq) + H₂O(l) ⇄ HS⁻(aq) + H₃O⁺(aq)
The acid dissociation constant, Ka1 is 9.0 x 10⁻⁸ 
Construct ICE table and obtain their equilibrium concentrations:
                  H₂S(aq) + H₂O(l) ⇄ HS⁻(aq) + H₃O⁺(aq)
I (M):             0.1                            0              0
C (M):            -x                            +x            +x
E (M):        0.1 -x                            x              x
So:
9.0 x 10⁻⁸ = \frac{X^{2} }{0.1-x}
x = 9.4 x 10⁻⁵ 
From the equilibrium table:
[H₃O⁺] = x = 9.4 X 10⁻⁵ M
[HS⁻] = x = 9.4 X 10⁻⁵ M
The equation for the second dissociation of the acid is:
HS⁻(aq) + H₂O(l) ⇄ S⁻²(aq) + H₃O⁺(aq)
The acid dissociation constant Ka2 is 1.0 x 10⁻¹⁷
Construct ICE table and obtain their equilibrium concentrations:
                    HS⁻(aq) + H₂O(l) ⇄ S²⁻(aq) + H₃O⁺(aq)
I (M):          9.4 x 10⁻⁵                   0              9.4 x 10⁻⁵
C (M):            -x                            +x            +x
E (M):     9.4 x 10⁻⁵ -x                    x              9.4 x 10⁻⁵ + x
So:
1.0 x 10⁻¹⁷ = \frac{(x)(9.4 x 10^{-5} + x) }{(9.4 x 10^{-5} - x) }
x = 1.0 x 10⁻¹⁷ M
Therefore the equilibrium concentrations are as follows:
[S²⁻] = x = 1.0 x 10⁻¹⁷ M
[H₃O⁺] = 9.4 x 10⁻⁵ + 1.0 x 10⁻¹⁷ M = 9.4 x 10⁻⁵ M
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