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ella [17]
3 years ago
8

The half life of an isotope X is 3 days how long will it take for 1/2 of a 10 G sample to decay

Chemistry
1 answer:
lorasvet [3.4K]3 years ago
3 0

Answer:  It will take 3 days for half of a 10 g sample to decay.

Explanation:

Half-life of sample of an isotope X = 3 days

\lambda=\frac{0.693}{t_{\frac{1}{2}}}=\frac{0.693}{3 days}=0.231 day^{-1}

N=N_o\times e^{-\lambda t}

\lnN=-\lambda t\lnN_o

Sample decayed = \frac{N_o}{2}=\frac{10 g}{2 g}=5 g

N=N^o-\text{sample decayed}=10 g-5 g=5 g, time = t

\ln[5 g]=-0.231 day^{-1}\times t\ln[10]

\ln[\frac{5}{10}]=-0.231\times t

\ln\frac{1}{2}=-0.693=-0.231\times t

t = 3 days

It will take 3 days for half of a 10 g sample to decay.

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nexus9112 [7]

Answer:

pH = 3.70

Explanation:

Moles of aniline in solution are:

0.0500L × (0.0350mol / 1L) = <em>1.750x10⁻³ mol aniline</em>

Aniline is in equilibrium with water, thus:

C₆H₅NH₂ + H₂O ⇄ C₆H₅NH₃⁺ + OH⁻ Kb = 3.8x10⁻¹⁰

HCl reacts with aniline thus:

HCl + C₆H₅NH₂ → C₆H₅NH₃⁺ + Cl⁻

At equivalence point, all aniline reacts producing  C₆H₅NH₃⁺.  C₆H₅NH₃⁺ has its own equilibrium with water thus:

C₆H₅NH₃⁺(aq) + H₂O(l) ⇄ C₆H₅NH₂(aq) + H₃O⁺(aq)

Where Ka is defined as:

Ka = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺] = Kw / Kb = 1.0x10⁻¹⁴ / 3.8x10⁻¹⁰ = <em>2.63x10⁻⁵ </em><em>(1)</em>

<em />

As all aniline reacts producing C₆H₅NH₃⁺, moles in equilibrium are:

[C₆H₅NH₃⁺] = 1.750x10⁻³ mol - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in (1):

2.63x10⁻⁵ =  [X] [X] / [1.750x10⁻³ mol - X]

4.6x10⁻⁸ - 2.63x10⁻⁵X = X²

0 = X² + 2.63x10⁻⁵X - 4.6x10⁻⁸

Solving for X:

X = -2.3x10⁻⁴ → False answer, there is no negative concentrations

X = 2.0x10⁻⁴ → Right answer

As [H₃O⁺] = X, [H₃O⁺] = 2.0x10⁻⁴.

Now, pH = -log[H₃O⁺]. Thus, pH at equivalence point is:

<em>pH = 3.70</em>

<em></em>

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