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lbvjy [14]
2 years ago
13

Plutonium-239 undergoes alpha decay and has a half-life of 24,000 years. After 72,000 years, how much of an initial 100.0 g samp

le will remain?
A. none
B. 50.0 g
C. 25.0 g
D. 12.5 g
Chemistry
1 answer:
dybincka [34]2 years ago
8 0
After 72000 years, the half life would have passed 3 times.

This means that (1/2)^3 = 1/8 of the original sample would remain.

The original sample is 100.0 g, meaning that 1/8 of the sample would be 12.5 g.

So, the answer is D.
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Explanation:

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On the basis of the information above, what is the approximate percent ionization of HNO2 in a 1.0 M HNO2 (aq) solution?
enot [183]

Answer:

The answer is "2%"

Explanation:

Equation:

HNO_2\ (aq) \leftrightharpoons  H^{+} \ (aq) + NO_2^{-}\ (aq) \\\\\  K_a = 4.0\times \ 10^{-4}

H^{+}=?

Formula:

Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}

Let

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x^2 = (4.0\times 10^{-4})\times 1.0\\\\x = ((4.0\times 10^{-4})\times 1.0)^{0.5} = 2.0 \times 10^{-2} \  M\\\\

therefore,

[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M

Calculating the % ionization:

= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\

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