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Jet001 [13]
3 years ago
9

A radioisotope forms a stable isotope after it undergoes radioactive decay. Which is the best conclusion to draw from this state

ment?
A. It undergoes alpha decay. B. It undergoes beta decay. C. It does not participate in a decay series. D. It does not have a long half-life.
Chemistry
2 answers:
noname [10]3 years ago
8 0

Answer:

C

Explanation:

because that's just the answer:)

Black_prince [1.1K]3 years ago
3 0

Answer:

C. It does not participate in a decay series.

Explanation:

From this statement, we can deduce that a radioisotope that forms a stable isotope after it undergoes radioactive decay suggests that it does not participate in a decay series.

  • It could have emitted any form of radioactive particles which can be alpha or beta.
  • We do not know if it has a long or short half life because the value is not given.
  • But since the radioactive decay in one step produces a stable isotope, we can conclude that it did not participate in a decay series.
  • A decay series involves a radioactive decay in multiple steps.
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At 400 K oxalic acid decomposes according to the reaction:H2C2O4(g)→CO2(g)+HCOOH(g)In three separate experiments, the intial pre
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Answer:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

Explanation:

For the reaction:

H₂C₂O₄(g) → CO₂(g) + HCOOH(g)

At t = 0, the initial pressure is just of H₂C₂O₄(g). At t= 20000 s, pressures will be:

H₂C₂O₄(g) = P₀ - x

CO₂(g) = x

HCOOH(g) = x

P at t=20000 is:

P₀ - x + x + x = P₀+x. That means P at t=20000s - P₀ = x

For 1st point:

x = 92,8-65,8 = 27

Pressure of H₂C₂O₄(g) at t=20000s: 65,8-27 = 38,8

2nd point:

x = 130-92,1 = 37,9

H₂C₂O₄(g): 92,1 - 37,9 = 54,2

3rd point:

x = 157-111 = 46

H₂C₂O₄(g): 111-46 = 65

Now, as the rate law is :

v = k P[H₂C₂O₄]

Based on integrated rate law, k is:

(- ln P[H₂C₂O₄] + ln P[H₂C₂O₄]₀) / t = k

1st point:

k = 2,64x10⁻⁵

2nd point:

k = 2,65x10⁻⁵

3rd point:

k = 2,68x10⁻⁵

The averrage of this values is:

k = 2,66x10⁻⁵

That means law is:

v = 2,66x10⁻⁵ P[H₂C₂O₄]

I hope it helps!

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