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Rasek [7]
3 years ago
8

Based on this equation: 2AL + 3CuCl2 -> 2AlCl3 + Cu ... how many grams of copper(II) chloride dihydrate would be required to

react completely with 1.20g of aluminum metal?
Chemistry
1 answer:
PolarNik [594]3 years ago
6 0
2Al + 3CuCl2 ➡ 2AlCl3 + 3Cu. Moles of Al = 1.2/27 = 0.045 moles 2 moles of Al reacts with 3 moles of CuCl2. Therefore, 1 mole of CuCl2 reacts with =2/3 x moles of Al = 2/3 x 0.045 = 0.03 moles.
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A weak monoprotic acid is titrated with 0.100 MNaOH. It requires 50.0 mL of the NaOH solution to reach the equivalence point. Af
larisa86 [58]

Explanation:

The given reaction is as follows.

        HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Hence, number of moles of NaOH are as follows.

        n = 0.05 L \times 0.1 M

           = 0.005 mol

After the addition of 25 ml of base, the pH of a solution is 3.62. Hence, moles of NaOH is 25 ml base are as follows.

             n = 0.025 L \times 0.1 M

                = 0.0025 mol

According to ICE table,

         HA(aq) + NaOH (aq) \rightleftharpoons NaA(aq) + H_{2}O(l)

Initial:     0.005 mol   0.0025 mol              0                  0

Change: -0.0025 mol  -0.0025 mol        +0.0025 mol

Equibm:   0.0025 mol    0                         0.0025 mol

Hence, concentrations of HA and NaA are calculated as follows.

          [HA] = \frac{0.0025 mol}{V}

        [NaA] = \frac{0.0025 mol}{V}

       [A^{-}] = [NaA] = \frac{0.0025 mol}{V}

Now, we will calculate the pK_{a} value as follows.

          pH = pK_{a} + log \frac{A^{-}}{HA}

       pK_{a} = pH - log \frac{[A^{-}]}{[HA]}

                  = 3.42 - log \frac{\frac{0.0025 mol}{V}}{\frac{0.0025}{V}}

                  = 3.42

Thus, we can conclude that pK_{a} of the weak acid is 3.42.

           

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Answer:

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Explanation:

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The transfer of energy that occurs when a force is applied over a distance is WORK.

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vovikov84 [41]
A. It absorbs energy.
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Alenkinab [10]

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Explanation:

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