Solids and Liquids

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Solids and Liquids Chapter 14

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Solids and Liquids. Chapter 14. Intermolecular forces. Intramolecular forces. The forces that occur among molecules that cause them to aggregate to form a solid or a liquid. Bonding forces that hold atoms together within a molecule. Intermolecular forces. - PowerPoint PPT Presentation

Transcript of Solids and Liquids

Page 1: Solids and Liquids

Solids and Liquids

Chapter 14

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Intermolecular forces• Bonding forces that hold

atoms together within a molecule

Intramolecular forces• The forces that occur

among molecules that cause them to aggregate to form a solid or a liquid

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Intermolecular forces

• Dipole-dipole attraction- molecules with dipole moments can attract each other by lining up so that the positive and the negative ends are close to each other.

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• Hydrogen bonding- hydrogen is bound to two highly electronegative atoms, such as nitrogen, oxygen or fluorine.

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• London dispersion forces- the forces that exist among noble gas atoms and nonpolar molecules

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Solutions

Chapter 15

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Soluble vs. Insoluble

Solubility (or Soluble)• Something that is capable

of being dissolved in a solvent

Insolubility (or insoluble)• Incapable of being dissolved

within a given solution

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Solubility Rules

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Examples

• State which of the following compounds are soluble and which are not– KBr = – PbCO3 =

– BSO3 =– zinc hydroxide =

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– KBr = Soluble– PbCO3 = Insoluble

– BSO3 =Insoluble– zinc hydroxide = Insoluble

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Types of solutions• Solution-a homogeneous mixture• Solvent-the largest substance that is present within the solution• Solutes-the smallest substance within the solution• Aqueous solution-solutions with water as a solvent• Saturated-a solution that contains as much solute as will dissolve at

the temperature• Unsaturated- a solution that has not reached the limit of solute that

will dissolve• Concentrated-a large amount of solute dissolved• Dilute-a relatively small amount of solute is dissolved• Supersaturated- a solution that contains more dissolved material

than could be dissolved by the solvent under normal circumstances

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Molarity vs Molality

Molarity• (M)• Describes the amount of

solute in moles and the volume of the solution in liters.

• The number of moles of solute per volume of solution in Liters

• Moles of Solute = mol Liters of Solution L

Molality• The amount of substance of

solute, divided by the mass of solvent.

• The number of moles of solute divided by the number of kilograms of solvent

• Moles of solute = mol Kg of Solution kg

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Example of calculating Molarity

• Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50L of solution.

• Step 1: calculate the number of moles in NaOH– 11.5g x 1mol = 0.288moles

40 g• Step 2: divide number of moles by the number of

Liters– 0.288moles = 0.192M

1.50 L

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Example of calculating Molality

• What is the molality of a solution made from 2.4 moles of NaCl and 0.80 kg of water?

• Divide number of moles by the number of kilograms:– 2.4 moles = 3.0m

0.80 kg

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Net Ionic Equation

• Ionic equations are used to describe the chemical reaction while also clearly indicating which of the reactants and products exist primariy as ions in aqueous solutions.

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Rules for writing net ionic equations

1. Start with a balanced molecular equation (meaning: an equation that shows the chemical formulas of all reactants and products, but not their physical states)

2. Break all soluble strong electrolytes (compounds with (aq) beside them into their ions

* indicate the correct formula and charge of each ion * indicate the correct number of each ion* write (aq) after each ion

3. Bring down all compounds with (s), (l), or (g) uncharged

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For example

Use the following equations to answer the question:2Na3PO4(aq) + 3CaCl2(aq) 6NaCl(aq) + Ca3(PO4)2(s)

Net ionic equation:6Na+ (aq) + 2PO4

-3(aq) + 3Ca+2(aq) + 6Cl-(aq) 6Na+(aq) +6Cl-(aq) + Ca3(PO4)2(s)