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Rasek [7]
3 years ago
8

A scientist digs up sample of arctic ice that is 458,000 years old. He takes it to his lab and finds that it contains 1.675 gram

s of krypton-81.
If the half-life of krypton-81 is 229,000 years, how much krypton-81 was present when the ice first formed?

Use the formula N = N0 .
Chemistry
1 answer:
Fiesta28 [93]3 years ago
3 0

Answer:

6.70 grams of krypton-81 was present when the ice first formed

Explanation:

Let use the below formula to find the amount of sample

N= N_0(\frac{1}{2})^n

where

n = \frac{t}{t_{\frac{1}{2}}}

here

t =  458,000 years

t_{\frac{1}{2}} = 229,000

\frac{t}{t_{\frac{1}{2}}} = \\frac{ 458,000}{229,000}

n = \frac{t}{t_{\frac{1}{2}}} = 2.000

Now substituting the values

1.675 = N_0(\frac{1}{2})^{2.000}}

1.675 = N_0\times (0.2500)

N_0= \frac{1.675}{0.2500}

N_0=6.70

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\boxed{\text{25. 20 L; 26. 49 K}}

Explanation:

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\begin{array}{rcrrcl}p_{1}& =& \text{100 kPa}\qquad & V_{1} &= & \text{10.00 L} \\p_{2}& =& \text{50 kPa}\qquad & V_{2} &= & ?\\\end{array}

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\begin{array}{rcl}100 \times 10.00 & =& 50V_{2}\\1000 & = & 50V_{2}\\V_{2} & = &\textbf{20 L}\\\end{array}\\\text{The new volume will be } \boxed{\textbf{20 L}}

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