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bagirrra123 [75]
3 years ago
9

carol accidentally dropped a container of salt and sodium chloride, into a container of a dark grey metal. carol is pretty sure

that the gray metal is iron fillings. using the table above, help carol seperate the two substances
Chemistry
1 answer:
o-na [289]3 years ago
4 0

Explanation:

The best way to separate the mixture is by using a bar magnet. We explore the magnetic property of the iron fillings in the mixture to separate it out.

  • The mixture is spread out thinly on a flat surface.
  • A bar magnet is hovered above the spread of the mixture
  • The magnet attracts the iron fillings to itself
  • This will leave the salt behind in the mixture.
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Which solution has a higher percent ionization of the acid, a 0.10M solution of HC2H3O2(aq) or a 0.010M solution of HC2H3O2(aq)
Svetach [21]

Answer:

The solution 0.010 M has a higher percent ionization of the acid.

Explanation:

The percent ionization can be found using the following equation:

\% I = \frac{[H_{3}O^{+}]}{[CH_{3}COOH]} \times 100    

Since we know the acid concentration in the two cases, we need to find [H₃O⁺].          

By using the dissociation of acetic acid in the water we can calculate the concentration of H₃O⁺ in the two cases:

1. Case 1 (0.1 M):

CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺(aq)   (1)

0.1 - x                                         x                     x

Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}  (2)

Where:

Ka: is the dissociation constant of acetic acid = 1.7x10⁻⁵.

1.7 \cdot 10^{-5} = \frac{x^{2}}{0.1 - x}  

1.7 \cdot 10^{-5}*(0.1 - x) - x^{2} = 0

By solving the above equation for x we have:

x = 1.29x10⁻³ M = [CH₃COO⁻] = [H₃O⁺]

Hence, the percent ionization is:      

\% I = \frac{1.29 \cdot 10^{-3} M}{0.1 M} \times 100 = 1.29 \%                      

   

2. Case 2 (0.01 M):

The dissociation constant from reaction (1) is:

Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}

With [CH₃COOH] = 0.01 M

1.7 \cdot 10^{-5} = \frac{x^{2}}{0.01 - x}  

1.7 \cdot 10^{-5}*(0.01 - x) - x^{2} = 0

By solving the above equation for x:

x = 4.04x10⁻⁴ M = [CH₃COO⁻] = [H₃O⁺]    

Then, the percent ionization for this case is:

\% I = \frac{4.04 \cdot 10^{-4} M}{0.01 M} \times 100 = 4.04 \%

As we can see, the solution 0.010 M has a higher percent ionization of the acetic acid.

Therefore, the solution 0.010 M has a higher percent ionization of the acid.

I hope it helps you!    

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