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goldfiish [28.3K]
2 years ago
8

Calculate the amount, in moles, of each of the following. You may round atomic masses to one decimal place.

Chemistry
1 answer:
Vlada [557]2 years ago
7 0

Answer:

The answer to your question is:

a) 0.023 mol of CdS

b) 0.12 mol of MoO3

c) 7.3 mol of AlPO4

Explanation:

a) CdS

MW = 112.4 + 32

      = 144 g

                                144g ------------------ 1 mol

                                  3.28g --------------  x

                                 x = (3.28 x 1) / 144

                                 x = 0.023 mol

b) MoO3

MW = 96 + 3(16) = 144 g

                              144g -------------------  1 mol

                               17.5g -----------------   x

                               x = (17.5 x 1) / 144

                               x = 0.12 mol

c) AlPO4

MW = 27 + 31 + 4(16)

MW = 122 g

                            122 g ------------------  1mol

                            890 g ------------------   x

                            x = (890 x 1) / 122

                            x = 7.3 mol of AlPO4

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In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.
Mamont248 [21]

Answer:

Approximately 1.30 \times 10^{-2}, assuming that this acid is monoprotic.

Explanation:

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After 12.5\% of that was dissociated, the concentration of both \rm H^{+} and \rm A^{-} (conjugate base of this acid) would become:

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Concentration of \rm HA in the solution after dissociation:

(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}.

Let [{\rm HA}], [{\rm H}^{+}], and [{\rm A}^{-}] denote the concentration (in \rm mol \cdot L^{-1} or \rm M) of the corresponding species at equilibrium. Calculate the acid dissociation constant K_{\rm a} for \rm HA, under the assumption that this acid is monoprotic:

\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}.

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