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Novosadov [1.4K]
3 years ago
5

A first-order reaction has a half-life of 16.7 s . How long does it take for the concentration of the reactant in the reaction t

o fall to one-fourth of its initial value?
Chemistry
1 answer:
andrew11 [14]3 years ago
5 0

It takes 33.4 s for the concentration of A to fall to one-fourth of its original value.

A <em>half-life</em> is the time it takes for the concentration to fall to half its original value.

Assume the initial concentration is 1.00 mol/L. Then,

\text{1.00 mol/L }\stackrel{\text{1st half-life }}{\longrightarrow}\text{ 0.50 mol/L } \stackrel{\text{ 2nd half-life }}{\longrightarrow}\text{0.25 mol/L}\\

The concentration drops to one-fourth of its initial value in two half-lives.

∴ Time = 2 × 16.7 s = 33.4 s

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Ans: Moles of Fe(OH)2 produced is 5.35 moles.

Given reaction:

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3 years ago
Vanadium has an atomic mass of 50.9415 amu. It has two common isotopes. One isotope has a mass of 50.9440 amu and a relative abu
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Answer:

Average atomic mass of the  vanadium = 50.9415 amu

Isotope (I) of vanadium' s abundance = 99.75 %= 0.9975

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Atomic mass of Isotope (II) of vanadium ,m' = ?

Average atomic mass of vanadium =

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m'= 49.944 amu

Explanation:

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