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3 years ago

How many milligrams remain of a 15.0 mg sample of radium-226 after 6396 years? The half life of radium-226 is 1599 years.

1 answer:

3 0

There are 4 half lives of radium-226 in 6396 years. 15.0 divided by half 4 times (15.0/2= 7.5, 7.5/2 = 3.75, 3.75/2 = 1.875) equals 1.875 mg of radium-226.

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Answer:

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Explanation:

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2 years ago

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vesna_86 [32]

Explanation:

Relation between pressure, latent heat of fusion, and change in volume is as follows.

$\frac{dP}{dT} = \frac{L}{T \times \Delta V}$

Also, $\frac{L}{T} = \Delta S^{fusion}_{m}$

where, $\Delta V^{fusion}_{m}$ is the difference in specific volumes.

Hence, $\frac{dP}{dT} = \frac{\Delta S^{fusion}_{m}}{\Delta V^{fusion}_{m}}$

As, $\Delta S^{fusion}_{m} = \frac{L}{T} = \frac{6010}{273.15}$ = 22.0 J/mol K

And, $\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}$ ...... (1)

where, $d_{H_{2}O}$ = density of water

$d_{ice}$ = density of ice

M = molar mass of water = $18.02 \times 10^{-3} kg$

Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.

$\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}$

= $\frac{18.02 \times 10^{-3}}{997} - \frac{18.02 \times 10^{-3}}{920}$

= $-1.51 \times 10^{-6}$

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$\frac{dP}{dT} = \frac{22}{-1.51 \times 10^{-6}}$

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or, = 145 bar/K

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= 681.5 bar

Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.

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