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shusha [124]
3 years ago
5

22. Radon has a half-life of 3.83 days. How long will it take a 225 g sample to decay to 14.06 g? (3pts.)

Chemistry
1 answer:
prohojiy [21]3 years ago
8 0

Answer:

15.32 days

Explanation:

From the question given above, the following data were obtained:

Half-life (t½) = 3.83 days

Original amount (N₀) = 225 g

Amount remaining (N) = 14.06 g

Time (t) =.?

Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:

Original amount (N₀) = 225 g

Amount remaining (N) = 14.06 g

Number of half-lives (n) =?

N = N₀ / 2ⁿ

14.06 = 225 / 2ⁿ

Cross multiply

14.06 × 2ⁿ = 225

Divide both side by 14.06

2ⁿ = 225 / 14.06

2ⁿ = 16

Express 16 in index form with 2 as the base

2ⁿ = 2⁴

n = 4

Thus, 4 half-lives has elapsed.

Finally, we shall determine the time. This can be obtained as follow:

Half-life (t½) = 3.83 days

Number of half-lives (n) = 4

Time (t) =.?

n = t / t½

4 = t / 3.83

Cross multiply

t = 4 × 3.83

t = 15.32 days

Therefore the time for 225 g sample of Radon to decay to 14.06 g is 15.32 days

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<u>Answer:</u>

<u>For a:</u> The standard Gibbs free energy of the reaction is -347.4 kJ

<u>For b:</u> The standard Gibbs free energy of the reaction is 746.91 kJ

<u>Explanation:</u>

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}           ............(1)

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The given chemical equation follows:

2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)

<u>Oxidation half reaction:</u>   Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-       ( × 2)

<u>Reduction half reaction:</u>   3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)

We are given:

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Putting values in equation 1, we get:

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Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

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The given chemical equation follows:

6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)

<u>Oxidation half reaction:</u>   Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-       ( × 6)

<u>Reduction half reaction:</u>   2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)

We are given:

n=6\\E^o_{cell}=-1.29V\\F=96500

Putting values in equation 1, we get:

\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

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