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patriot [66]
3 years ago
10

What's the difference between distillation and fractional distillation?

Chemistry
1 answer:
Elan Coil [88]3 years ago
6 0

Answer:

The main difference between distillation and fractional distillation are:Simple distillation separate liquid with boiling point gaps of at least 50 degrees whereas fractional distillation separates liquid with closer boiling point.

Explanation:

<h2>hope it helps</h2>
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CO2 i think if you understand so mark me as brainliest please.

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At a given temperature, the elementary reaction A − ⇀ ↽ − B , in the forward direction, is first order in A with a rate constant
ZanzabumX [31]

Answer:

0.24

Explanation:

We are given that

Rate constant for A=k=0.0180/s

Rate constant for  B,k'=0.0750/s

We have to find the value of equilibrium constant for the reaction

A\rightleftharpoons B

Equilibrium constant, for k=\frac{k}{k'}

Using the formula

k=\frac{0.0180}{0.0750}=0.24

Hence, the value of the equilibrium constant for the reaction

A\rightleftharpoons B at this temperature=0.24

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3 years ago
Electrolysis can be used for all of the following, except A. separating metals from ores. B. plating objects with metal coatings
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3 years ago
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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