CHM 111 Lecture Notes - Lecture 8: Microsoft Powerpoint, Reagent

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CHM 111 Lecture Notes - Lecture 7: Chemical Formula, Reagent, Chemical Equation

CHM 111 Lecture Notes - Lecture 8: Microsoft Powerpoint, Reagent

CHM 111 Lecture Notes - Fall 2018 Lecture 10 - Spectator ion, Chemical equation, Precipitation (chemistry)

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Chapter 4: chemical quantities and aqueous reactions (continued) What would result (product) if the limiting reagent reacted: actual yield. The results (product) gained from a reaction: % yield = (actual yield / the theoretical yield) x (100%) Mixture of 2 or more materials: solute. What is dissolved (smaller amounts: solvent. Can be concentrated or diluted: concentrated. Amount of solute is greater than the solvent (solute > solvent: diluted. Amount of solute is less than the solvent (solute < solvent: molarity. Equation: m= amount of solute (moles) / amount of solution (l) When a solutes dissolve : attractive forces between. The solute pieces that hold the solute together: attractive forces between. Solvent and solute particles when the solute and solvent are mixed: solute will dissolve if . Solute and solvent have a very strong attraction. Making a concentrated solution less concentrated: using molarity.

Related Questions

I need help on number 7!!!

1. Consider the unbalanced equation for the reaction of solid lead with a silver nitrate solution.

_____Pb (s) + ______AgNO3 (ac) â ______Pb(NO3)2 (ac) + _______Ag (s)

a. Balance the equation.

b. How many moles of silver nitrate are required to react completely with 9.3 mol of lead?

c. How many moles of silver, Ag, are formed by the complete reaction of 28.4 mol of Pb?

2. For the following reaction: _____CaO (s) + _____CO2 (g) â _____CaCO3 (s), Calculate how many grams of CaCO3, are formed when 2.4 g of CaO Do they react completely with CO2?

3. For the following reaction: 2 Li (s) + F2 (g) â 2 LiF (s), find the reagent limiting with the amounts of 1.0 mol of Li and 1.0 mol of F2.

4. For the next reaction, determine the limiting reagent, calculate the yield Theoretical product in grams with 5.5 g of Al and 19.8 g of Cl2.

2 Al (s) + 3 Cl2 (g) â 2 AlCl3 (s)

5. Consider the reaction between NiS2 and O2:

2 NiS2 (s) + 5 O2 (g) â 2 NiO (s) + 4 SO2 (g)

When 11.2 g of NiS2 react with 5.43 g of O2, 4.86 g of NiO are obtained. Determine the limiting reagent, the theoretical yield of NiO and the% yield for the reaction.

6. How many atoms of mercury are in 5.8 mol of mercury?

7. Complete the following table:

Element Moles Mass Number of molecules

8. How many grams of sodium are there in 1.00 g of each of the following compounds?

a. Na3PO4

b. NaC7H5O2

c. Na2C6H6O7

9. A sample of 4.78 g of aluminum reacts completely with oxygen to form 6.67 g of aluminum oxide. Use the data to calculate the percent of composition by mass.

10. Citric acid, the compound responsible for the bitter taste in lemons, has the following composition: C - 37.51%

H - 4.40%

O - 58.29%

Calculate the empirical formula.

11. A compound containing phosphorus and oxygen has a molar mass of 219.9 g / mol and an empirical formula of P2O3. Find the molecular formula of compound.

12. It is a term used in chemistry, to describe a system in which a

or more substances, are mixed or dissolved homogeneously in another

substance.

a. Mixture

b. Solution

c. Compound

d. None of the above

13. It is the component that is in a solution, in smaller quantity.

a. Solution

b. Solvent

c. Solvent

d. Soluto

14. It is the component that is in a solution in greater quantity.

a. Solution

b. Solvent

c. Solvent

d. The answers b and c are correct.

15. The quantitative expression of the dissolved amount of solute in a quantity determined solvent is known as:

a. Solution

b. Volume

c. Concentration

d. Stoichiometry

16.Express the moles of solute among the liters of solution

a. Molality

b. Molarity

c. % weight by volume

d. % weight by weight

17.Express the moles of solute between kg of solvent

a. Molarity

b. ppm

c. ppt

d. Molality

18. What is the molarity of a solution prepared by dissolving 7.50 g of nitrate of magnesium, Mg (NO3) 2 in enough water to prepare 25.0 mL of solution?

a. 10.3 M

b. 5.6 M

c. 6.5 M

d. 2.03 M

19. How many grams of sodium chloride, NaCl are needed to prepare 125 mL of 0.037 M NaCl solution

a. 1.021 M

b. 0.268 M

c. 0.201 M

d. 0.500 M

20. Calculate the molality of a solution containing 12.5 g of ethylene glycol (C2H6O2) dissolved in 135 g of water.

21. A solution of ethylene glycol contains 21.2 g of ethylene glycol (C2H6O2) in 85.4 mL of water. Calculate the freezing point and the boiling point of the solution. (Assume the water density is 1.00 g / mL)

22. At what volume should you dilute 75 mL of a 10.0 M solution of H2SO4 to obtain a 1.75 M solution of H2SO4.

Experiment 10 Colligative Properties Introduction A pinch of wa make boiling water hotter. A handful will clear the snow from your steps. Every day we the physical properties of materials around us, but often don't know why we use these particular s or what chemistry is involved. Modifying the temperature of phase changes, by adding a second chemical to the pure solvent are examples of manipulating the Colligative Properties of a solution. this lab, we will study how varying the concentration and type of a salt affects the point of a solution. Colligative properties are properties of a solution containing 2 or more components. Typically, the primary is the solvent and the other called the such as color or hardness, properties depend on the number of particles of astute in the not the identity of the In many cases, the rokute will determine the number of solute particles (degree of ionization, solubility, etc,) in the solvent, but a colligative property does not depend upon the nature of the particles. Two colligative properties commonly encountered are boiling point elevation and freezing point depression (the temperature of of a is lower than of pure solvent). We will be studying boiling point elevation in this Most of the previous work that you have done with solutions has involved units of molarity, M, or moles of solute per liter of solution. Boiling point elevation calculations (as well as those for freezing point depression) or moles of solute per kilogram of solvent. boili change of temperature. When the temperature of a solution changes, the volume of the solution also changes. Since molarity depends on a change in wil the solution's molarity. Molality depends on the mass of the solvent. and thus is independent of temperature. n't Hoff Factor salt into and the of moles inorganic salts in an aque the based on the addition of the number dissociates, the total number of particles in the solution will be of cations and anions produced from the dissociation. Thus; Nacion Na Choaa (2moles of particles) Cacbrs Ca +2ch (3moles of particles) The quantity of particles produced per mole of solute is called the van'tHoff factor and is given the symbol, i, If 1 mole of sodium chloride were completely dissociated, iwould be 2. This theory works very well for dilute solutions and non-electrolytes (i 1, no dissociation at all since nonelectrolytes not ionize), For does not always apply to more concentrated solutions. Thus, when there are a large number of particles in a solution, they have a tendency to recombine as the salt and you don't get 100% dissociation. example, at a NaCl concentration of 0.1M, iwill be 1.81 (90% dissociation) rather than 2 (100%). Boiling Point Elevation solute causes the solution to boil at a higher When a solution is heated, the presence of a non-volatile point of the pure liquid and the temperature than the pure The difference between the boiling boiling point of the solution is: 169

CHM 111 Lecture Notes - Lecture 8: Microsoft Powerpoint, Reagent

CHM 111 Lecture Notes - Lecture 7: Chemical Formula, Reagent, Chemical Equation

CHM 111 Lecture Notes - Lecture 8: Microsoft Powerpoint, Reagent

CHM 111 Lecture Notes - Fall 2018 Lecture 10 - Spectator ion, Chemical equation, Precipitation (chemistry)

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