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vivado [14]
3 years ago
14

What is the average isotopic mass of an element that has the following abundances in nature? Isotopic mass: X20 X22 X23 Abundanc

es in natures: 90.5 % 8.0 % 1.5 %
Chemistry
1 answer:
Paladinen [302]3 years ago
8 0

Answer: 20.2

Explanation:

Mass of isotope 1 = 20

% abundance of isotope 1 = 90.5% = \frac{90.5}{100}=0.905

Mass of isotope 2 = 22

% abundance of isotope 2 = 8.0% = \frac{8}{100}=0.08

Mass of isotope 3 = 23

% abundance of isotope 3 = 1.5% = \frac{1.5}{100}=0.015

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(20\times 0.905+(22\times 0.08)+(23\times 0.015]

A=20.2

Therefore, the average atomic mass of the element is 20.2.

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A student prepares a aqueous solution of butanoic acid . Calculate the fraction of butanoic acid that is in the dissociated form
ella [17]

Answer:

15.4%

Explanation:

If Ka = 0.54 mM = 1.51x10⁻⁵

Then;

C₄H₈O₂               -------->            C₄H₇O₂⁻          +           H⁺

I                    0.54x10⁻³                             0                                0

E                   0.54x10⁻³(1-x)                      0.54x10⁻³x                0.54x10⁻³x

Recall that x is the percentage degree of dissociation

From the ICE table;

Ka = [C₄H₇O₂⁻] [ H⁺]/[C₄H₈O₂]

1.51x10⁻⁵=(0.54x10⁻³x) (0.54x10⁻³x)/ 0.54x10⁻³(1-x)  

1.51x10⁻⁵ = 0.54x10⁻³x^2/1-x

1.51x10⁻⁵(1-x) = 0.54x10⁻³x^2

1.51x10⁻⁵ - 1.51x10⁻⁵x = 0.54x10⁻³x^2

Hence;

0.54x10⁻³x^2 + 1.51x10⁻⁵x - 1.51x10⁻⁵=0

x^2 + 0.028x - 0.028 = 0

Solving the quadratic equation here;

x = 0.154 or −0.182

Ignoring the negative result, x = 0.154

Hence, fraction of butanoic acid that is in the dissociated form in this solution = 15.4%

3 0
3 years ago
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