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Natasha2012 [34]
3 years ago
13

Chromium-51 is a radioisotope that is used to assess the lifetime of red blood cells The half-life of chromium-51 is 27.7 days.

If you begin with 48.0 mg of this isotope, what mass remains after 47.9 days have passed?
Chemistry
1 answer:
oee [108]3 years ago
5 0

Answer:

After 47.9 days, will remain 14.5mg of the isotope

Explanation:

The radioactive decay follows always first-order kinetics where its general law is:  

Ln[A] = -Kt + ln[A]₀

<em>Where [A] is actual concentration of the atom, k is rate constant, t is time and [A]₀ is initial concentration. </em>

We can find rate constant from half-life as follows:

t_{1/2} = \frac{ln2}{K}

K = ln 2 / 27.7 days

K = 0.025 days⁻¹

Replacing, initial amount of isotope is 48.0mg = [A]₀ , K is 0.025 days⁻¹ and t = 47.9 days:

Ln[A] = -Kt + ln[A]₀

Ln[A] = -0.025 days⁻¹*47.9 days + ln (48.0mg)

ln [A] = 2.6726

[A] = e^ (2.6726)

[A] = 14.5mg

<h3>After 47.9 days, will remain 14.5mg of the isotope</h3>

<em />

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Answer:

The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.

Explanation:

Cu (s) + 4 HNO_3 (aq) \rightarrow Cu(NO_3)_2 (aq) + 2 H_2O (l) + 2 NO_2 (g)

Moles of copper = \frac{2.01 g}{63.55 g/mol}=0.03163 mol

According to reaction, 1 mol of copper gives 2 moles of nitrogen dioxide gas.

Then 0.03613 moles of copper will give:

\frac{2}{1}\times 0.03163 mol=0.06326 mol of nitrogen dioxide gas

Moles of nitrogen dioxide gas = n = 0.06326 mol

Pressure of the gas = P

P = Total pressure - vapor pressure of water

P = 726 mmHg - 23.8 mmHg = 702.2 mmHg

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Volume of the gas = V

PV=nRT

V=\frac{0.06326 mol\times 0.0821 atm L/mol K\times 298.15 K}{0.924 atm}

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The volume of NO₂ gas collected over water at 25.0 °C is 1.68 Liters.

3 0
3 years ago
Nitrogen can ionically bond with an unknown element X from group 2. How many ions of element X are required to create a stable i
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I can tell you there certainly is enough information, so we can eliminate the fourth option right away.
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Answer:

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Explanation:

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4.95cm^3\frac{8.95g}{1cm^3} \frac{1mol}{63.55g} = 0.697 moles

Moles of nitric acid are:

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As 1 mol of Cu reacts with 4 moles of HNO₃:

0.697 mol Cu × (4mol HNO₃ / 1mol Cu) = 2.79 moles of HNO₃ will react. That means Cu is limiting reactant.

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<em>Where P is pressure (735torr / 760 = 0.967atm); n are moles (1.394mol); R is gas constant (0.082atmL/molK); T is temperature (28.2°C + 273.15 = 301.35K). </em>

Thus, volume is:

V = nRT / P

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V = 35.6L

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Answer:

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