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IgorLugansk [536]
2 years ago
11

How many grams of NaHCO3 are needed to prepare 250 mL of 0.50 M NaHCO3?

Chemistry
1 answer:
forsale [732]2 years ago
6 0

Answer:

10.5g

Explanation:

First, let us calculate the number of mole of NaHCO3 present in the solution. This is illustrated below:

Volume = 250mL = 250/1000 = 0.25L

Molarity = 0.5M

Mole =?

Molarity = mole /Volume

Mole = Molarity x Volume

Mole = 0.5 x 0.25

Mole = 0.125 mole

Now, we shall be converting 0.125 mole of NaHCO3 to grams to obtain the desired result. This can be achieved by doing the following:

Molar Mass of NaHCO3 = 23 + 1 + 12 +(16x3) = 23 + 1 +12 +48 = 84g/mol

Number of mole of NaHCO3 = 0.125 mole

Mass of NaHCO3 =?

Mass = number of mole x molar Mass

Mass of NaHCO3 = 0.125 x 84

Mass of NaHCO3 = 10.5g

Therefore, 10.5g of NaHCO3 is needed.

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Polonium- 210 , Po210 , decays to lead- 206 , Pb206 , by alpha emission according to the equation Po84210⟶Pb82206+He24 If the ha
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Answer:

0.269 g

Explanation:

Po_{210} ^{84}  ⟶  Pb_{206} ^{82}+  + He_{4} ^{2}

Data:

Half-life of  Po_{210} ^{84}  (T(1/2)) = 138.4 days

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Time = 338.8 days  

Isotopic masses  

Po_{210} ^{84} = 209.98 g/mol  

Pb_{206} ^{82} = 205.97 g/mol  

Concepts  

Avogadro’s number: This is the number of constituent particles that are contained in a mol of any substance. These constituted particles can be atoms, molecules or ions). Its value is 6.023*10^23.  

The radioactive decay law is  

N=Noe^(-λt)

Where:  

No = number of atoms in t=0

N = number of atoms in t=t (now) in this case t=338.8 days  

λ= radioactive decay constant  

The radioactive constant is related to the half-life by the next equation  

λ= \frac{ln 2}{t(1/2)}

so  

λ= \frac{ln2}{138.4 days}  =0,005008 days^(-1)

No (Atoms of  Po_{210} ^{84}  in t=0)

To get No we need to calculate the number of atoms of  Po_{210} ^{84}   in the initial sample. We have a sample of 0,561 g of PoCl4. If we get the number of moles of PoCl4 in the sample, this will be the number of moles of  Po_{210} ^{84}  in the initial sample.  

This is:

\frac{0,561 g of PoCl4}{350. 79 g of PoCl4 /mol} = 0,001599 mol of  PoCl4

This is the number of mol of  Po_{210} ^{84} in the initial sample.

To get the number of atoms in the initial sample we use the Avogadro’s number = 6.023*10^23  

0,001599 mol of  Po_{210} ^{84} * 6.023*10^23 atoms/ mol of  Po_{210} ^{84} = 9.632 *10^20 atoms of  Po_{210} ^{84}

Atoms after 338.8 days

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N=Noe^(-λt)

N=9.632*10^20 e^(-0,005008 days^(-1) * 338.8 days) =1.765*10^20

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The number of atoms  Po_{210} ^{84} transformed is equal to the number of atoms of Pb_{206} ^{82}  produced.  

The number of atoms of Po_{210} ^{84} transformed is No - N  

9.632 *10^20 – 1.765 *10^20 = 7.866*10^20

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This number of moles have a mass of:

(0,001306 moles of Pb_{206} ^{82} )* (205.97 g of Pb_{206} ^{82} /mol of Pb_{206} ^{82} ) = 0.269 g  

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