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Vesna [10]
3 years ago
13

What is the total number of atoms in a 75 g sample of zinc?

Chemistry
1 answer:
velikii [3]3 years ago
8 0
Jinx kxknxbxbxbjxj dbdbcj
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Which option contains a correct statement about the size of
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a molecule always has two or more atoms

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2 years ago
In a 1.0x10^-4 M solution of HClO(aq), identify the relative molar amounts of these species:HClO, OH-, H3O+, OCl-, H2O
yarga [219]
HClO is a weak acid, which means the ions do not fully dissociate. The hydrolysis reaction for the hypochlorous acid is:

HClO + H2O ⇄ H3O+ +OCl-

Then the equilibrium constant, Ka, of dilute HClO would be:

K_{a} = \frac{[ H_{3}  O^{+} ][O Cl^{-} ]}{HClO}

Then we do the ICE table. I is for the initial concentration, C for the change and E for the excess.
      
          HClO       + H2O   ⇄   H3O+ +  OCl-
I     1.0x10^-4                          0             0
C        -x                                 +x           +x 
E  (1.0x10^-4 - x)                     x             x

Substituting the excess (E) concentration to the Ka equation:

K_{a} = \frac{[x ][x]}{1.0 \ x \  10^{-4} - x }

Simplifying the equation would yield a quadratic equation:

x^{2} + K_{a}x-(1.0 \ x \ 10^{-4}) K_{a}=0

The Ka for HClO is an experimental data which was determined to be 2.9 x 10^-8. Substitute this to the equation, determine the roots, then you get the value for x, which is the concentration of H3O+ and ClO-. Just use your calculator feature Shift-Solve.

x = 1.688 x 10^-6 M = [H3O+] = [ClO-]

Then, you can determine the conc of [OH-] through pH.

pH = -log {H3O+] = -log [1.688 x 10^-6] = 5.77
pOH = 14 - pH = 14 - 5.77 = 8.23
pOH = 8.23 = -log [OH-]
[OH-] = 5.89 x 10^-9 M

Also, since HClO is (1.0x10^-4 - x), then it's concentration would be:
[HClO] = 1.0x10^-4 - 1.688 x 10^-6 = 9.83 x10^-5 M

Let's summarize all concentrations:
[HClO] = 9.83 x10^-5 M
[OH-] = 5.89 x 10^-9 M
[H3O+] = [ClO-] = 1.688 x 10^-6 M
Since the solution is dilute, H2O is relatively higher in concentration.

Thus in relative amounts, the order would be

H2O >>> HClO > H3O+ = ClO- > OH-


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