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Mashcka [7]
3 years ago
11

What is the difference between a molecular formula and a molecular model?

Chemistry
1 answer:
snow_tiger [21]3 years ago
6 0
 <span>Molecular compounds, which are represented by molecules, are usually made of non-metals only (or of metalloids and non-metals). Ionic compounds, which are represented by formula units, are made of metals and non-metals. 

More detail if you're interested: Molecules and formula units are the representative particles for molecular and ionic compounds, respectively. By that I mean, one unit of a molecular compound is a molecule...a bundle of atoms covalently bonded that exists separately from all the other molecules. One unit of an ionic compound is a formula unit. A formula unit is a representation of the compound's formula. For example, the formula unit of NaCl is one Na^+1 ion and one Cl^-1 ion. The formula unit of AlCl3 is one Al^+3 ion and three Cl^-1 ions. Ionic compounds don't have separate bundles of atoms like molecular compounds do, so the formula unit is just the smallest number of ions that it takes to represent the formula. </span>
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• We obtained the above 10.00-mL solution by diluting a stock solution using a 1.00-mL aliquot and placing it into a 25.00-mL vo
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Answer:

a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b) 0.0035 mole

c)  0.166 M

Explanation:

Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.

The equation of the reaction is expressed as:

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

1 mole         3 mole

The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.

b)  if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

H_3PO_4 \ + \ 3NaOH -----> Na_3 PO_4 \ + \ 3H_2O

10 ml            17.50 ml

(x) M              0.200 M

Molarity = \frac{0.2*17.5}{1000}

= 0.0035 mole

c) What was the molar concentration of phosphoric acid in the original stock solution?

By stoichiometry, converting moles of NaOH to H₃PO₄; we have

= 0.0035 \ mole \ of NaOH* \frac{1 mole of H_3PO_4}{3 \ mole \ of \ NaOH}

= 0.00166 mole of H₃PO₄

Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:

Molar Concentration =  \frac{mole \ \ of \ soulte }{ Volume \ of \ solution }

Molar Concentration = \frac{0.00166 \ mole \ of \  H_3PO_4 }{10}*1000

Molar Concentration = 0.166 M

∴  the molar concentration of phosphoric acid in the original stock solution = 0.166 M

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