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3 years ago

A 0.133 pg sample of iron-59 produces 242 β-particles per second. what is the half-life of this isotope (in hours)

1 answer:

34kurt3 years ago

4 0

1070 hours.
1 mole of iron-59 would mass 59 grams, so 0.133 picograms would be 0.133x10^-12 / 59 = 2.25x10^-15 moles of iron-59. Multiplying by Avogadro's number, we can determine the number of atoms of iron-59 we have, so: 2.25x10^-15 * 6.02214x10^23 = 1.35x10^9
Since we have 242 decays over a period of 1 second, we can divide the
number of atoms left by the original number of atoms
(1350000000 - 242)/1350000000
= 1349999758/1350000000
= 0.999999820740741
And calculate the logarithm to base 2 of that quotient.
ln(0.999999820740741)/ln(2)
= -1.79259275281191x10^-7/0.693147180559945
= -2.58616467481524x10^-7
The reciprocal of this number will be the half life in seconds. So
-1/2.58616467481524x10^-7
= -3866729.79388461
And dividing by 3600 (number of seconds in an hour) will give the half-life in
hours.
-3866729.79388461 / 3600 = -1074.091609
So the half life in hours to 3 significant figures is 1070 hours.
Dividing that figure by 24 gives a half life of 44.58 days which is in pretty close agreement to the official half-life of 44.495 days for iron-59.

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2 years ago

A gas exerts a pressure of 1.8 atm at a temperature of 60 degrees celsius. What is the new temperature when the pressure of the

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The answer for the following problem is mentioned below.

Therefore the final temperature of the gas is 740 K

Explanation:

Given:

Initial pressure of the gas ( $P_{1}$ ) = 1.8 atm

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Initial temperature of the gas ( $T_{1}$ ) = 60°C = 60 + 273 = 333 K

To solve:

Final temperature of the gas ( $T_{2}$ )

We know;

From the ideal gas equation;

we know;

P × V = n × R × T

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we can tell from the above equation;

P ∝ T

(i.e.)

$\frac{P}{T}$ = constant

$\frac{P_{1} }{P_{2} }$ = $\frac{T_{1} }{T_{2} }$

Where;

$P_{1}$ = initial pressure of a gas

$P_{2}$ = final pressure of a gas

$T_{1}$ = initial temperature of a gas

$T_{2}$ = final temperature of a gas

$\frac{1.8}{4}$ = $\frac{333}{T_{2} }$

$T_{2}$ = $\frac{333*4}{1.8}$

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3 years ago

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no. of moles = no. of molecules ÷ 6.02 × 10²³ molecules

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2 years ago

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Answer:

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3 years ago