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Otrada [13]
3 years ago
5

Iron-59 is used in medicine to diagnose blood circulation disorders. The half life of iron-59 is 44.5 days. How much of a 2.000

mg sample will remain after 133.5 days? Show your work
Chemistry
1 answer:
kolezko [41]3 years ago
7 0

Answer:

0.258 mg of iron remains.

Explanation:

To solve this problem we can use the formula

M₂ = M₀ * 0.5^{\frac{t}{44.5} }

Where M₂ is the mass remaining, M₀ is the initial mass, and t is time in days.

Using the data given by the problem:

M₂ = 2.000 mg * 0.5^{\frac{133.5}{44.5} }

M₂ = 0.258 mg

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Charge q is 1 unit of distance away from the source charge S. Charge p is six times further away. The force exerted between S an
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Answer:

<u>36 times</u>

Explantaion:

The force exerted between two charged particles is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

In the case of S and q: The distance of separation is 1 unit

F(Sq) α S*q/1²

In the case of S and p: the distance is 6 units

F(Sp) α S*p/6²

therefore:

F(Sq) = 36 * F(Sp)

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3 years ago
Describe why the formation of sand is a physical change
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Calculate the change in entropy when 10.0 g of CO2 isothermally expands from a volume of 6.15 L to 11.5 L. Assume that the gas b
USPshnik [31]

Answer:

The change in entropy of the carbon dioxide is 1.183\times 10^{-3} kilojoules per Kelvin.

Explanation:

By assuming that carbon dioxide behaves ideally, the change in entropy (\Delta S), measured in kilojoules per Kelvin, is defined by the following expression:

\Delta S = m\cdot \bar c_{v}\cdot \ln \frac{T_{f}}{T_{o}}+m\cdot \frac{R_{u}}{M}\cdot \ln \frac{V_{f}}{V_{o}} (1)

Where:

m - Mass of the gas, measured in kilograms.

\bar c_{v} - Isochoric specific heat of the gas, measured in kilojoules per kilogram-Kelvin.

T_{o}, T_{f} - Initial and final temperatures of the gas, measured in Kelvin.

V_{o}, V_{f} - Initial and final volumes of the gas, measured in liters.

R_{u} - Ideal gas constant, measured in kilopascal-cubic meter per kilomole-Kelvin.

M - Molar mass, measured in kilograms per kilomole.

If we know that T_{o} = T_{f}, m = 0.010\,kg, R_{u} = 8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K}, M = 44.010\,\frac{kg}{kmol}, V_{o} = 6.15\,L and V_{f} = 11.5\,L, then the change in entropy of the carbon dioxide is:

\Delta S = \left[\frac{ (0.010\,kg)\cdot \left(8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)}{44.010\,\frac{kg}{kmol} } \right]\cdot \ln \left(\frac{11.5\,L}{6.15\,L}\right)

\Delta S = 1.183\times 10^{-3}\,\frac{kJ}{K}

The change in entropy of the carbon dioxide is 1.183\times 10^{-3} kilojoules per Kelvin.

4 0
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Also be aware that 2 molecules can have the same % composition but different formulas.

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The crust of the Earth is thickest beneath the continents. 
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