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Entry Task: May 1 st - 2 nd Block #1

Entry Task: May 1 st - 2 nd Block #1. Collect Entry Tasks Sheets!!. Agenda:. Discuss Solutions review w.s . DON’T LOSE IT!! Ice cream “lab”. 1. What would be the percent concentration of each of the following solutions?. a. 54.0 g of AgNO 3 g is dissolved in 128 g of water. .

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Entry Task: May 1 st - 2 nd Block #1

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  1. Entry Task: May 1st - 2nd Block #1 Collect Entry Tasks Sheets!!

  2. Agenda: • Discuss Solutions review w.s. • DON’T LOSE IT!! • Ice cream “lab”

  3. 1. What would be the percent concentration of each of the following solutions? a. 54.0 g of AgNO3 g is dissolved in 128 g of water. 54 g AgNO3 X 100 128g water + 54 g AgNO3 = 182 g of solution = 29.7%

  4. 1. What would be the percent concentration of each of the following solutions? b. 4.22 g of K2CO3 is dissolved in 426 mL of water. 4.22 g K2CO3 X 100 426 g water + 4.22 g K2CO3 = 430.22 g of solution = 0.981%

  5. 2. What mass of solute is needed to produce each of the indicated solutions? a. 500.0 g of a 6.40% NaCl solution. X g NaCl = 6.40% X 100 500g of solution = 32.0 g NaCl 0.064 X 500

  6. 2. What mass of solute is needed to produce each of the indicated solutions? b. 136 g of a 14.2% LiNO3 solution. X g LiNO3 = 14.2% X 100 136 g of solution = 19.3 g LiNO3 0.142 X 136

  7. 3. Calculate the grams needed of solutes to prepare the indicated volume and concentration of the solutions given? a. 340. mL of a 1.82 M aluminum nitrate solution. mm= 212.991g X g Al(NO3)3 = 1.82 M 0.340 L solution = 0.619 molAl(NO3)3 0.340 X 1.82 0.619 molAl(NO3)3 212.991g Al(NO3)3 1 molAl(NO3)3 =132 g Al(NO3)3

  8. 3. Calculate the grams needed of solutes to prepare the indicated volume and concentration of the solutions given? b. 25.0 mL of a 4.26 M potassium cyanide solution. mm= 65g X g KCN = 4.26 M 0.025 L solution 0.1065 molKCN 0.025 X 4.26 0.1065 molKCN 65 g KCN 1 molKCN =6.92 g KCN

  9. 4. What should the final volume(mL) of each solution be so that the amount of solute dissolved will produce the indicated concentration. a. 2.86 g of copper(I) carbonate for a 0.640 M solution. mm= 187.089g 1 molCu2CO3 2.86 g Cu2CO3 = 0.0153 molCu2CO3 187.089g Cu2CO3 0.0153 molCu2CO3 = 0.640 M X L solution 0.0153÷0.640 = 0.0239 L

  10. 4. What should the final volume(mL) of each solution be so that the amount of solute dissolved will produce the indicated concentration. b. 12.62 g of calcium hydrogen carbonate for a 1.28 M solution.mm= 161.994g 1 molCa(HCO3)2 12.62 g Ca(HCO3)2 = 0.0779 molCa(HCO3)2 161.994g Ca(HCO3)2 0.0779 molCa(HCO3)2 = 1.28 M X L solution 0.0779 ÷ 1.28 = 0.0608 L

  11. 5. What will be the final concentration of a solution prepared by dissolving the indicated solute in enough water to produce the indicated volume of solution? a. 15.4 g of strontium acetate filled up to 340. mL.mm= 205.62g 1 molSr(C2H3O2)2 15.4 g Sr(C2H3O2)2 = 0.0749 molSr(C2H3O2)2 205.62g Sr(C2H3O2)2 0.0749 molSr(C2H3O2)2 = 0.220 M 0.340 L solution

  12. 5. What will be the final concentration of a solution prepared by dissolving the indicated solute in enough water to produce the indicated volume of solution? b. 176.2 g of Iron(III) sulfite filled up to 1.42 liters. mm= 351.591g 1 molFe2(SO3)3 176.2 g Fe2(SO3)3 = 0.501 molFe2(SO3)3 351.591g Fe2(SO3)3 0.501 molFe2(SO3)3 = 0.359 M 1.42 L solution

  13. 6. What will be the final concentration of the solution indicated that will result from the following dilutions? M1V1=M2V2 a. 14.0 mL of a 4.20 M Na2CO3 solution is diluted to 86.0 mL. (4.20 M)(14.0ml) = ( XM)(86.0ml) 58.8 = 67.9M 86.6

  14. 6. What will be the final concentration of the solution indicated that will result from the following dilutions? M1V1=M2V2 b. 450. mL of a 1.22 M HCl solution is diluted to 1.26 liters. (1.22 M)(450ml) = ( XM)(1260 ml) 549 = 0.436M 1260

  15. 7. To what volume should the indicated solution be diluted to produce a solution of the desired concentration?M1V1=M2V2 a. 12.0 mL of a 0.64 M KCl solution for a 0.19 M solution. (0.64 M)(12.0ml) = ( 0.19M)(X ml) 7.68 = 40.4 ml 0.19

  16. 7. To what volume should the indicated solution be diluted to produce a solution of the desired concentration?M1V1=M2V2 b. 84.2 mL of a 4.60 M KMnO4 solution for a 1.42 M solution. (4.60 M)(84.2ml) = ( 1.42M)(X ml) 387.32 = 273 ml 1.42

  17. 8. What volume of the indicated solution is needed to produce the volume and concentration of a diluted solution as indicated?M1V1=M2V2 a. 2.73 M NaOH solution to prepare 142 mL of a 0.540 M solution. (2.73 M)(X ml) = ( 0.540M)(142 ml) 76.68 = 28.1 ml 2.73

  18. 8. What volume of the indicated solution is needed to produce the volume and concentration of a diluted solution as indicated?M1V1=M2V2 b. 0.0076 M SnF2 solution to prepare 25.0 mL of a 0.00027 M solution. (0.0076 M)(X ml) = ( 0.00027M)(25.0 ml) 0.00675 = 0.888 ml 0.0076

  19. Ice Cream Lab • 8 cups of whole milk • 8 cups of heavy whipping cream • 4 teaspoon vanilla • 4 cups of sugar • Then the Dry Ice • Let’s do it- and eat 

  20. DO NOT CHEW ANY ICE!!!!Get a chunk- spit it out

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