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Verdich [7]
3 years ago
5

How many grams of magnesium acetate are in 8.95x10^23 formula units?

Chemistry
1 answer:
Natasha_Volkova [10]3 years ago
5 0

Answer:

211.63 g.

Explanation:

  • Particles could refer to atoms, molecules, formula units.
  • <em>Knowing that every one mole of a substance contains Avogadro's no. of molecules (NA = 6.022 x 10²³).</em>

<em><u>Using cross multiplication:  </u></em>

1.0 mole → 6.022 x 10²³ molecules.

??? mole → 8.95 x 10²³ molecules.

  • The no. of moles of magnesium acetate = (8.95 x 10²³ molecules) (1.0 mole) / (6.022 x 10²³ molecules) = 1.486 mol.

∴ The grams of magnesium acetate are in 8.95 x 10²³ formula units = n x molar mass = (1.486 mol)(142.394 g/mol) = 211.63 g.

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25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca
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Answer:

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\rm HCl is a monoprotic acid. In other words, each \rm HCl\! would release up to one proton \rm H^{+}.

On the other hand, \rm KOH is a monoprotic base. Each \rm KOH\! formula unit would react with up to one \rm H^{+}.

Hence, \rm HCl molecules and \rm KOH\! formula units would react at a one-to-one ratio.

{\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l).

Therefore, that 0.02625\; \rm mol of \rm HCl molecules would neutralize exactly the same number of \rm NaOH formula units. That is: n(\mathrm{NaOH}) = 0.02625\; \rm mol.

Calculate the concentration of a \rm NaOH solution where V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3} and n(\mathrm{NaOH}) = 0.02625\; \rm mol:

\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}.

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