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inn [45]
3 years ago
15

What is the mass (in grams) of 9.09x10^24 molecules of methanol (CH3OH)?

Chemistry
1 answer:
zloy xaker [14]3 years ago
8 0

The amount of a substance can be converted into grams of that substance by using the molar mass. It is a physical property of a substance which describes the mass per mole of the substance. The molar mass of CH3OH is 32.04 g/ mol. Multiplying the given amount of CH3OH with the molar mass will yield to the mass of the substance.

<span>9.09x10^24 molecules CH3OH ( 1 mol / 6.022x10^23 molecules) (32.04 g/mol) = 483.63 g CH3OH

Dividing the molar mass of N2O which is 44.08 g/mol to the given amount will yield to the number of moles.

0.183 g N2O / 44.08 g/ mol =0.0042 mol N2O</span>
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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 conce
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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.

      Ag_{2}SO_{4} \rightleftharpoons 2Ag^{+} + SO^{2-}_{4}

[Ag^{+}] = 0.0390 M

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}]

       [SO^{2-}_{4}] = 0.00788 M

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

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

      K_{sp} = [Ca^{2+}][SO^{2-}_{4}]

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

           [Ca^{2+}] = 0.00625 M

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               \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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