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Solutions Review Play as a slideshow!

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**Calculate the molarity of a 0**

Calculate the molarity of a L sugar solution that was prepared with 0.15 moles of sugar? M = mol/L M = 0.15 mol/0.175 L M = 0.86 M

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**Solutions are ________________ mixtures made up of very small particles.**

Homogeneous

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**Determine the volume of 0**

Determine the volume of M sugar solution that can be prepared with moles of sugar. M = mol/L 0.235 M = mol/ x L ***cross-multiply 0.235 x = 0.470 X = 2.00 L

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**Explain why solutions are classified as mixtures instead of compounds.**

The two parts, solute and solvent, are only physically combined, not chemically combined. Also, you can easily separate them with a physical change, such as evaporating off the solvent. Finally, they can combine in different ratios, they do not have a fixed ratio like compounds (example, water is always 2 H and one O, but salt water can have varying percent of salt in it).

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**A salt solution is to be added to a marine aquarium**

A salt solution is to be added to a marine aquarium. Calculate the molarity of a salt solution that is prepared by adding water to g of NaCl to give a final volume of ml. M = mol/L 18.65 g x 1 mol/ g = mol M = mol/ L M = M

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**alloy air oxygen aqueous**

Pure gold is 24 carat. 14-carat gold contains 14 parts gold and 10 parts other metals. 14-carat gold is said to be a(n) ___________, which is a type of solution. An example of a gaseous solution is ________________, which is made up mostly of ______________ and nitrogen when dry. The most common solutions are ______________ solutions. alloy air oxygen aqueous

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**Calculate the volume of a 3**

Calculate the volume of a 3.15 M NaOH (aq) solution that should be used to prepare 250. ml of M NaOH (aq). M1V1 = M2V2 3.15 M x V1 = M x 250. mL V1 = 11.9 mL

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Because of the _______________ _______________, you can see the light beams from car headlights in a fog. Tyndall Effect

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**You have 3.0 L of 3.0 M HCl. What volume of 2.0 M HCl can you make?**

M1V1 = M2V2 3.0 M x 3.0 L = 2.0 M x V2 V2 = 4.5 L

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**3. 5 L of solvent was added to 2. 0 L of a 0. 88 M solution**

3.5 L of solvent was added to 2.0 L of a 0.88 M solution. What is the new molarity of the solution? M1V1 = M2V2 0.88 M x 2.0 L = M2 x 5.5 L M2 = 0.32 M

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**Multiple choice: To increase the rate of solution of a solid in water,**

a. increase the pressure over the water. b. decrease the pressure over the water. c. crush the particles of the solid. d. chill the water.

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**You need 450 mL of 0. 15 M NaOH. All you have available is a 2**

You need 450 mL of 0.15 M NaOH. All you have available is a 2.0 M stock solution of NaOH. What volume of solvent will you add to the volume required of the stock solution to make your dilution? M1V1 = M2V2 0.15 M x 450 mL = 2.0 M x V2 V2 = mL stock solution 450 – = mL solvent

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If 15 grams of iodine are dissolved in 1000 mL of alcohol, the alcohol is the (solute, solvent) and the solution is said to be a(n) _______________. A substance that dissolves other materials is a (solute, solvent). The substance being dissolved is a (solute, solvent). tincture

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**You have 6. 0 L of 5. 0 M NaCl stock solution**

You have 6.0 L of 5.0 M NaCl stock solution. You take 800 mL of that stock solution, and add 500 mL of solvent. -How many moles of NaCl would be present in the new solution? -What is the molarity of new solution? M = mol/L 5.0 M = x mol/ L X = 4.0 mol M1V1 = M2V2 5.0 M x 800 mL = M2 x 1300 mL M2 = 3.08 M

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In (solutions, suspensions) the substances separate after standing a while. The substances (can also, can not) be separated by filtration.

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**Find the volume of a 0.75 M solution if it contains 39 grams of KOH.**

39 g x 1 mol/56.1 g = mol M = mol/L 0.75 M = mol/x L ***cross-multiply*** X = L

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**Smoke is an example of a _____________ of solid dirt and dust particles in air.**

colloid

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**How many grams of hydrochloric acid (HCl) are present in 3. 0 L of a 0**

How many grams of hydrochloric acid (HCl) are present in 3.0 L of a M solution? M = mol/L 0.750 M = x mol/ 3.0 L X = 2.25 mol 2.25 mol x g/ 1 mol = 82.0 g

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**A(n) ________________’s particles are between those of a solution and a suspension.**

colloid

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**110. 0 mL of 3. 00 M sulfuric acid has 25. 0 mL of water added to it**

110.0 mL of 3.00 M sulfuric acid has 25.0 mL of water added to it. What is the resulting concentration of the solution ? M1V1 = M2V2 3.00 M x mL = M2 x mL M2 = 2.44 M

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**How does a solution behave differently from a suspension when a beam of light is shined through it?**

A beam of light will go right through a solution and you will not see it. A suspension exhibits the Tyndall Effect where you will be able to see the beam of light.

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**Acetic acid is purchased as a solution in 17. 0 M concentrations**

Acetic acid is purchased as a solution in 17.0 M concentrations. Explain how you would prepare mL of a 5.00 M solution. M1V1 = M2V2 17 M x V1 = 5.00 M x mL V1 = mL Put mL into a 500 mL volumetric flask. Then add water until the solution is 500 mL total (to the line).

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**What is the solute in a brass alloy containing 75% copper and 25% zinc?**

Zinc (it has a smaller percentage)

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**electrolytes Non-electrolytes**

Substances that conduct electricity when dissolved are said to be ______________, while substances that do NOT conduct electricity when dissolved are said to be __________________. electrolytes Non-electrolytes

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