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kobusy [5.1K]
3 years ago
11

Zinc has an average atomic mass of 65.37 amu. Jack is trying to figure out what the most abundant isotope of zinc is, but he doe

sn't have access to the internet, so the average atomic mass is the only information he has. Jack decides to made an educated guess as to the most abundant isotope of zinc by assuming there are only two isotopes of zinc. If Jack assumes there are only two isotopes of Zn, what he be most likely to decide is zinc's most abundant isotope?
Question 5 options:

Zinc-64


Zinc-65


Zinc-68


Zinc-66
Chemistry
1 answer:
Mekhanik [1.2K]3 years ago
5 0

Zinc's most abundant isotope : Zinc-65

<h3>Further explanation </h3>

Isotopes are atoms whose no-atom has the same number of protons while still having a different number of neutrons.

So Isotopes are elements that have the same Atomic Number (Proton)

Atomic mass is the average atomic mass of all its isotopes

In determining the mass of an atom, as a standard is the mass of 1 carbon-12 atom whose mass is 12 amu

Mass atom X = mass isotope 1 . % + mass isotope 2.%

To decide zinc's most abundant isotope, then choose the closest mass number

or we can check the difference with the average mass number, if the value is the smallest, then that isotope has the largest abundant

  • Zinc-64

\tt \dfrac{65.37-64}{65.37}=0.021

  • Zinc-65

\tt \dfrac{65.37-65}{65.37}=0.005

  • Zinc-68

\tt \dfrac{68-65.37}{65.37}=0.04

  • Zinc-66

\tt \dfrac{66-65.37}{65.37}=0.009

The closest = Zinc-65(the smallest)

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HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)

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(B) The dissociation reaction of Co(H_2O)_6^{3+} will be:

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The equilibrium expression :

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HC_2H_3O_2(aq)\rightleftharpoons H^+(aq)+C_2H_3O_2^-(aq)

The equilibrium expression of HC_2H_3O_2 will be:

K_a=\frac{[H^+][C_2H_3O_2^-]}{[HC_2H_3O_2]}

(B) The dissociation reaction of Co(H_2O)_6^{3+} will be:

Co(H_2O)_6^{3+}(aq)\rightleftharpoons H^+(aq)+Co(H_2O)_5(OH)^{2+}(aq)

The equilibrium expression of Co(H_2O)_6^{3+} will be:

K_a=\frac{[H^+][Co(H_2O)_5(OH)^{2+}]}{[Co(H_2O)_6^{3+}]}

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The equilibrium expression of CH_3NH_3^+ will be:

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------------------  x  -------------------  x  ----------------------  x  -------------------  =
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=  31.5 g CO₂

3.8 g O₂             1 mole               2 moles CO₂          44.0095 g
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=  10.6 g CO₂

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