Question

On the line at the left, write the letter of the definition that best matches each term. 1. solution a, capable of being dissolved 2. solute b. solution with water as the solvent 3. solvent C. substance that is dissolved in a solution 4. soluble d. substance that dissolves in water to form a solution that conducts an electric current 5. alloy e. solid solution containing two or more metals 6. aqueous solution | f. homogeneous mixture of two or more substances in a single 7. electrolyte g. substance that does the dissolving in a solution Question 2 Describe how you would prepare 5.00 litres of a 2.00M solution of potassium hydroxide. Equilibrium Question 3 A mixture of hydrogen, nitrogen and ammonia is at equilibrium. When the reaction is written N2(g) + 3H2(g) = 2NH3(g) the equilibrium constant K1 = 3.5x10 a) What would be the equilibrium constant Kz for the reaction 1/2N2(g) + 3/2H2(g) = NH3(g)? b) What would be the equilibrium constant Kz for the inverse reaction 2NH3(g) = N2(g) + 3H2(e) ? Question 4 For the reaction CO (g) + NO2 (g) = CO2 (g) + NO (g) How can we maximize the amount of NO produced?

Answer 1

solution 1 1 solution : substance that does Homogeneous misture of two or more substances. 2. solute substance that is dissolved in a solution. 3. Solvent nt substance that does the dissolving in a - solution. 4. Soluble: capable of being dissolved. 5. Alloy solid solution containing 2 or more metal o. Aqueous solution: Solution with with tion water water as solvent as 1 7. Electrolyte: Substance that dissolves in water to form a solution that conducts an electric current Solution: 2 Preparation of 5.00L of 2.00M solution of koH. As we know that. salute - Molarity - moles of - --- I volume of solution in L e Given, Molarity = 2.00M Volume of solution - S.ool By putting in equo 2.00 = moles of koH 5.00L moles of KOH = 10 moles Molecular 1 weight of koH in lo moles = 10457 wt of kot = 570 g of koH l = 57 g/mold By weighing 570g of koH & dissolving in 5L of to we can prepare 2.00M KOH solution.

solution - 3 N29) + 314cg) - 2NH3 K 3.5 x 10 k: [NH₂]² 23 (لہ a) Ź Ng+ 3 Cg) & NH3 (9) k: [Nts = 1k [Na] Ž [12] K2= √3.5X108 K2 = 1.878104] ZNH Nz (g) + 3th(9) KsCNx][M]° - [NH₂) = ki Kystlog = 2.8x107 solution-4. on corgo + Norg) = Coq CqJt No Cq) I ways to mascinize the amount of No production: ů By increasing the concentration of Reactants i. Colg) 4 No, (g) (i) by removing Con from the product side.