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kow [346]
4 years ago
11

Vanadium has two naturally occurring isotopes, 50V with an atomic mass of 49.9472 amu and 51V with an atomic mass of 50.9440. Th

e atomic weight of vanadium is 50.9415. The percent abundances of the vanadium isotopes are ________% 50V and ________% 51V.
Chemistry
1 answer:
stira [4]4 years ago
8 0

Answer:

The percent abundances of the vanadium isotopes are 0.25% 50V and 99.75% 51V.

Explanation:

Let 50V be isotope A

Let 51V be isotope B

Mass number of isotope A (50V) = 49.9472

Mass number of isotope B (51V) = 50.9440

Abundance of isotope A (50V) = A%

Abundance of isotope B (51V) = B% = 100 — A

The atomic weight of vanadium

= 50.9415

Atomic weight = [(Mass of A x Abundance of A)/100] + [(Mass of B x Abundance of B) /100]

50.9415 = [(49.9472xA%)/100] + [(50.9440x(100 — A))/100]

50.9415 = [49.9472A%/100] + [(5094.40 — 50.9440A%)/100]

Multiply through by 100

5094.15 = 49.9472A% + 5094.40 — 50.9440A%

Collect like terms

5094.15 — 5094.40 = 49.9472A% — 50.9440A%

—0.25 = — 0.9968A%

Divide both side by — 0.9968

A% = —0.25/ — 0.9968

A% = 0.25%

B% = 100 — 0.25 = 99.75%

The percent abundances of the vanadium isotopes are 0.25% 50V and 99.75% 51V.

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Answer:

a) The reaction is first order, that is, order 1. Option C is correct.

b) The half life of the reaction is 23 minutes. Option B is correct

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Explanation:

First of, we try to obtain the order of the reaction from the data provided.

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40 0.37x10-2

50 0.28x10-2

70 0.15x10-2

Using a trial and error mode, we try to obtain the order of the reaction. But let's define some terms.

C₀ = Initial concentration of the reactant

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t = time since the reaction started

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We Start from the first guess of zero order.

For a zero order reaction, the general equation is

C₀ - C = kt

k = (C₀ - C)/t

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We try first order next, for first order reaction

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k = [In (C₀/C)]/t

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