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Rudiy27
3 years ago
13

How many moles of water are in 1.23 x 10^18 water molecules?

Chemistry
2 answers:
ANEK [815]3 years ago
4 0
Divide the number of molecules you have by, 6.022 x 10^23. This will give you the mols of water, or the mols of anything, since there is always 6.022 x 10^23 molecules in 1 mol of substance. 
1.23x10^24 atoms/6.022x10^23 atom/mol = 2.04 mol H20 
TiliK225 [7]3 years ago
3 0

<span>1.23x10^24 atoms/6.022x10^23 atom/mol = 2.04 mol H20 </span>
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A chemist titrates 120.0 mL of a 0.4006 M hydrocyanic acid (HCN) solution with 0.6812 MNaOH solution at 25 °C. Calculate the pH
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Explanation:

The given reaction equation is as follows.

      HCN + NaOH \rightarrow NaCN + H_{2}O

Hence, initial moles of HCN will be as follows.

        Moles = Molarity × Volume

                   = 0.4006 M \times 120 ml

                   = 48.072 mol

Now, we will calculate the volume of NaOH as follows.

             M_{1}V_{1} = M_{2}V_{2}

            0.4006 \times 120 = 0.6812 \times V_{2}  

              V_{2} = 70.57 ml

At the equivalence point, moles of both HCN and NaOH will be equal. And, total volume will be as follows.

               120 ml + 70.57 ml = 190.57 ml

Initial concentration of CN^{-} is as follows.

               \frac{48.072}{190.57}

               = 0.25 M

Now, the equilibrium equation will be as follows.

             CN^{-} + H_{2}O \rightleftharpoons HCN + OH^{-}

Initial:          0.25

Equilbm:  0.25 - x           x             x

Expression to find K_{a} is as follows.

       K_{a} = \frac{x^{2}}{0.25 - x}

We know that,  pK_{a} = -log K_{a}            

          K_{a} = antilog (-9.21)

                      = 6.16 \times 10^{-10}

As,     K_{b} = \frac{K_{w}}{K_{a}}                

     \frac{10^{-14}}{6.16 \times 10^{-10}} = \frac{x^{2}}{0.25 - x}

                 x = 0.2 \times 10^{-2} = [OH^{-}]

Now, we will calculate the pH as follows.

               pOH = -log[OH^{-}]

as pOH = 2.698

And,      pH + pOH = 14

               pH = 14 - 2.698

                     = 11.30

Thus, we can conclude that pH at equivalence is 11.30.        

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