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Mashutka [201]
3 years ago
6

How many hydrogen moles are in 2.5 moles of aspartame c14h18n2o5

Chemistry
1 answer:
STALIN [3.7K]3 years ago
3 0
Firstly, we need to convert 3g aspartame into moles aspartame. In order to do this we have to find the molecular mass of aspartame (the total weight of each atom of the molecule combined. This figure can be used to construct a conversion factor so that the grams may be converted into moles. Molecular weights for each atom can be found on any periodic table. Avagadro's number (6.022*10^23) is a constant value that expresses the number of molecules in one mole of a substance. 

The molecular weight for aspartame is 294.3 grams per mole. 

The process of finding how many atoms of H there are in 3.00g of aspartame would be like this: 

1. 3g * 1mol/294.3g = .01mol aspartame (this is converting grams to moles) 

2. .01mol * 6.022*10^23 = 6.022*10^21 (This is finding the number of molecules) 

3. (6.022*10^21) * 18 = 1.08*10^23 

This 3rd calculation is done because in part 2, you calculated the number of molecules of aspartame there were in 3g. In each molecule of aspartame there are 18 hydrogen atoms. So the final answer is: 
1.08*10^23 hydrogen atoms.
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The given question is incomplete. The complete question is as follows.

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca^{2+}(aq) and 0.0390 M Ag^{+}(aq). What will be the concentration of Ca^{2+}(aq) when Ag_{2}SO_{4}(s) begins to precipitate? What percentage of the Ca^{2+}(aq) can be separated from the Ag(aq) by selective precipitation?

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The given reaction is as follows.

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When Ag_{2}SO_{4} precipitates then expression for K_{sp} will be as follows.

         K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]

        1.20 \times 10^{-5} = (0.0390)^{2} \times [SO^{2-}_{4}]

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Now, equation for dissociation of calcium sulfate is as follows.

         CaSO_{4} \rightleftharpoons Ca^{2+} + SO^{2-}_{4}

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     4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788

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Now, we will calculate the percentage of Ca^{2+} remaining in the solution as follows.

               \frac{0.00625}{0.05} \times 100

                 = 12.5%

And, the percentage of Ca^{2+} that can be separated is as follows.

                     100 - 12.5

                     = 87.5%

Thus, we can conclude that 87.5% will be the concentration of Ca^{2+}(aq) when Ag_{2}SO_{4}(s) begins to precipitate.

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