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Mashcka [7]
4 years ago
11

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

Chemistry
1 answer:
snow_tiger [21]4 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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What is the molarity of a 1.44 (m/v)% solution of an HCOOOH (Formic acid)?
AlexFokin [52]

This problem is providing the mass-volume percent of a formic acid solution so its molarity is required and found to be 0.313 M after the following calculations.

<h3>Molarity</h3>

In chemistry, units of concentration provide a measurable understanding of the relationship between the relative amounts of both solute and solvent. In the case of molarity, one must relate moles of solute and liters of solution as follows:

M=\frac{mol\ solute}{Volume\ solution \ in \ L}

In such a way, when given this mass-volume percent of 1.44% for the formic acid in the solution, one can assume there is 100 mL of solution and 1.44 g of solute (formic acid), which means one must convert the volume to liters and the mass to moles with:

mol\ solute=\frac{1.44g}{46.03g/mol} =0.0313mol\\\\Volume\ in \liters: 100mL*\frac{1L}{1000mL}=0.100L

Finally, we plug in these numbers in the equation for the calculation of molarity:

M=\frac{0.0313mol}{0.100L}\\\\M=0.313M

Learn more about molarity: brainly.com/question/10053901

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2 years ago
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Calculate the theoretical yield for the bromination of both stilbenes
podryga [215]

Answer:

cinnamic acid - 150 mg

cis-stilbene - 100 μL

trans- stilbene - 100 mg

pyridinium tribromide - 200-385 mg

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moles of cinnamic acid = 0.150 g/148.16 g/mol = 0.001 mols

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cis-stilbene (100 ul = 0.1 ml)

moles of cis-stilbene = 0.1 ml x 1.01 g/mol/180.25 g/mol = 0.00056 mols

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Theoretical mass of dibromoproduct formed = 0.00055 mol x 340.05 g/mol = 0.19 g

Explanation:

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