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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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In atmospheric chemistry, the following chemical reaction converts SO2, the predominant oxide of sulfur that comes from combusti
Misha Larkins [42]

Answer:

Explanation:

From the given information;

The chemical reaction can be well presented as follows:

\mathtt{SO_{2(g)} + \dfrac{1}{2}O_{2(g)} }  ⇄ \mathtt{3SO_{2(l)}}

Now, K is known to be the equilibrium constant and it can be represented in terms of each constituent activity:

i.e

K = \dfrac{a_{so_3}}{a_{so_2} a_{o_2}^{\frac{1}{2}}}

However, since we are dealing with liquids solutions;

K = \dfrac{1}{\dfrac{Pso_2}{P^0}\Big ( \dfrac{Po_2}{P^0} \Big)^{1/2}}   since the activity of a_{so_3} is equivalent to 1

Hence, under standard conditions(i.e at a pressure of 1 bar)

K = \dfrac{1}{Pso_2Po_2^{1/2}}

(b)

From the CRC Handbook, we are meant to determine the value of the Gibb free energy by applying the formula:

\Delta _{rxn} G^o = \sum \Delta_f \ G^o (products) - \sum \Delta_fG^o (reactants) \\ \\ = (1) (-368 \ kJ/mol) - (\dfrac{1}{2}) (0) - ((1) (-300.13 \ kJ/mol)) \\ \\ = -368 \ kJ/mol + 300.13 \ kJ/mol \\ \\  \simeq -68 \ kJ/mol

Thus, for this reaction; the Gibbs frree energy = -68 kJ/mol

(c)

Le's recall that:

At equilibrium, the instantaneous free energy is usually zero &

Q(reaction quotient) is equivalent to K(equilibrium constant)

So;

\mathtt{\Delta _{rxn} G = \Delta _{rxn} G^o + RT In Q}

\mathtt{0- \Delta _{rxn} G^o = RTIn K } \\ \\ \mathtt{ \Delta _{rxn} G^o = -RTIn K }  \\ \\  K = e^{\dfrac{\Delta_{rxn} G^o}{RT}} \\ \\  K = e^{^{\dfrac{67900 \ J/mol}{8.314 \ J/mol \times 298 \ K}} }

K =7.98390356\times 10^{11} \\ \\  \mathbf{K = 7.98 \times 10^{11}}

(d)

The direction by which the reaction will proceed can be determined if we can know the value of Q(reaction quotient).

This is because;

If  Q < K, then the reaction will proceed in the right direction towards the products.

However, if Q > K , then the reaction goes to the left direction. i.e to the reactants.

So;

Q= \dfrac{1}{Pso_2Po_2^{1/2}}

Since we are dealing with liquids;

Q= \dfrac{1}{1 \times 1^{1/2}}

Q = 1

Since Q < K; Then, the reaction proceeds in the right direction.

Hence, SO2 as well O2 will combine to yield SO3, then condensation will take place to form liquid.

8 0
3 years ago
Whats the correct order of these?
Igoryamba
1109, 5.9 * 10 squared, -0.041, -4.2 times 10 to the -3, and -7.6 times 10 to the -5
Hope this helps
3 0
3 years ago
How does maximum boiling azeotropic mixture is separated using fractional distillation?
Evgesh-ka [11]

Answer:

By heating the mixture to maximum boiling point and then the solution is distilled at a constant temperature without having a change in composition.

Explanation:

An azeotropic mixture is also called a constant boiling mixture and it is a mixture of two or more liquids whose proportions cannot be altered by simple distillation due to the fact that when an azeotropic mixture is boiled, the vapor has the same proportions of constituents as the unboiled mixture.

Now, maximum boiling azeotropic mixture are the solutions with negative deviations that have an intermediate composition  for which the vapor pressure of the solution is minimum and as a result, the boiling point is maximum. At that point, the solution will distill at a constant temperature without having a change in composition.

4 0
2 years ago
Of the following forms of carbon dioxide, which has the lowest entropy?
saw5 [17]
Those atoms which have same atomic numbers but different atomic misses are lowest entropy.
4 0
3 years ago
Will give brainliest
Allisa [31]

Answer:

12.5

1.95x10^-10

1.5

0.5

in that order

Explanation:

4 0
3 years ago
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