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dolphi86 [110]
3 years ago
6

What is the most active non metallic element in group 16

Chemistry
1 answer:
Umnica [9.8K]3 years ago
7 0
Metals=Reactivity increases as we move through the elements in the period table from top to bottom, and left to right. Nonmetals=Reactivity increases as we move through the elements in the periodic table, as we move from bottom to the top, and right to left.  Since reactivity of nonmetals increases going up the periodic table, oxygen is therefore the most reactive nonmetal in the group. So, the answer is oxygen
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The pH of a water is measured to be 7.5. The system is open to atmosphere and the temperature is 25 oC. Assume that the system i
Dafna1 [17]

Explanation:

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pK_{a1}=-\log[K_{a1}]

6.35=-\log[K_{a1}]

K_{a1}=4.467\times 10^{-7}

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pK_{a2}=-\log[K_{a2}]

10.33=-\log[K_{a2}]

K_{a2}=4.467\times 10^{-11}

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7.5=\log[H^+]

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H_2CO_3\rightleftharpoons HCO_3^{-}+H^+

 C                        0          0

At equilibrium

  (C-x)                     x        x

HCO_3^{-}\rightleftharpoons CO_3^{2-}+H^+

x                      0              0

At equilibrium

(x -y)                   y             y

Expression of an second dissociation constant will be given as:

K_{a2}=\frac{y\times y}{(x-y)}

4.677\times 10^{-11}=\frac{y^2}{(x-y)}..[1]

x+y=[H^+]

x+y=3.162\times 10^{-8}...[2]

Solving [1] and [2]:

x = 3.045\times 10^{-8} M

y = 1.1702\times 10^{-9} M

Expression of an first dissociation constant will be given as:

K_{a1}=\frac{x\times x}{(C-x)}

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4.467\times 10^{-7}=\frac{(3.045\times 10^{-8} M)^2}{(C-(3.045\times 10^{-8} M))}

Solving for C:

C = 3.253\times 10^{-8} M

At equilibrium , concentration of species:

Carbonic acid :

[H_2CO_3]=(C-x)=3.253\times 10^{-8} M-3.045\times 10^{-8} M

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Carbonate ion :

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[HCO_3^{-}]=(x-y)=3.045\times 10^{-8} M-1.1702\times 10^{-9} M=2.928\times 10^{-8} MTotal carbonates:[TC]

[TC]=[H_2CO_3]+[HCO_3^{-}]+[CO_3^{2-}]=C

= [TC} = 3.253\times 10^{-8} M

8 0
3 years ago
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