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4vir4ik [10]
4 years ago
14

Suppose you have 100 grams of a radioisotope with a half-life of 100 years. How much of the isotope will you have after 200 year

s?
Chemistry
2 answers:
Ulleksa [173]4 years ago
8 0
        Amount Remaining     Years       #half lives 
             100g                            0                 0
             50 g                           100                1                  
             25g                            200               2                      




Brilliant_brown [7]4 years ago
8 0

Answer:

m = 25 g

Explanation:

To do this, we need to use the general expression for Half life:

<em>A = Ao e^-tλ (1)</em>

<em>Where:</em>

<em>A: concentration or mass of the substance after t time has passed</em>

<em>Ao: Initial concentration or mass of the substance</em>

<em>t: time that has passed.</em>

<em>λ: lambda that is relationed to half life time.</em>

The value of λ can be calculated with the following expression:

λ = ln2 / t(1/2)   (2)

So, let's calculate first the value of lambda, and then, we replace it in expression (1) to know the mass of the radioisotope:

λ = ln2/100

λ = 6.93x10^-3

Now, let's use (1) to calculate the mass after 200 years:

m = 100e^(-200*6.93x10^-3)

m = 100e^(-1.386)

<em>m = 25 g</em>

<em>And this is the mass of the isotope after 200 years.</em>

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Explanation:

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4 years ago
A student reacts 5.0 g of sodium with 10.0 g of chlorine and collect 5.24 g of sodium chloride. What is the percent yield of thi
Ede4ka [16]

Answer: The percent yield of this combination reaction is 41.3 %

Explanation : Given,

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First we have to calculate the moles of Na and Cl_2.

\text{Moles of }Na=\frac{\text{Given mass }Na}{\text{Molar mass }Na}

\text{Moles of }Na=\frac{5.0g}{23g/mol}=0.217mol

and,

\text{Moles of }Cl_2=\frac{\text{Given mass }Cl_2}{\text{Molar mass }Cl_2}

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Now we have to calculate the limiting and excess reagent.

The balanced chemical equation will be:

2Na+Cl_2\rightarrow 2NaCl

From the balanced reaction we conclude that

As, 2 mole of Na react with 1 mole of Cl_2

So, 0.217 moles of Na react with \frac{0.217}{2}=0.108 moles of Cl_2

From this we conclude that, Cl_2 is an excess reagent because the given moles are greater than the required moles and Na is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NaCl

From the reaction, we conclude that

As, 2 mole of Na react to give 2 mole of NaCl

So, 0.217 mole of HCl react to give 0.217 mole of NaCl

Now we have to calculate the mass of NaCl

\text{ Mass of }NaCl=\text{ Moles of }NaCl\times \text{ Molar mass of }NaCl

Molar mass of NaCl = 58.5 g/mole

\text{ Mass of }NaCl=(0.217moles)\times (58.5g/mole)=12.7g

Now we have to calculate the percent yield of this reaction.

Percent yield = \frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100

Actual yield = 5.24 g

Theoretical yield = 12.7 g

Percent yield = \frac{5.24g}{12.7g}\times 100

Percent yield = 41.3 %

Therefore, the percent yield of this combination reaction is 41.3 %

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