Entry Task: May 1 st - 2 nd Block #1 Collect Entry Tasks Sheets!!
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Transcript of Entry Task: May 1 st - 2 nd Block #1 Collect Entry Tasks Sheets!!
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Entry Task: May 1st - 2nd Block #1Collect Entry Tasks Sheets!!
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Agenda:
• Discuss Solutions review w.s.– DON’T LOSE IT!!
• Ice cream “lab”
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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
128g water + 54 g AgNO3 = 182 g of solution X 100
= 29.7%
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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
426 g water + 4.22 g K2CO3 = 430.22 g of solution X 100
= 0.981%
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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 NaCl500g of solution
X 100 = 6.40%
0.064 X 500 = 32.0 g NaCl
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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
136 g of solution X 100 = 14.2%
0.142 X 136 = 19.3 g LiNO3
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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
0.340 L solution = 1.82 M
0.340 X 1.82 = 0.619 mol Al(NO3)3
0.619 mol Al(NO3)3
1 mol Al(NO3)3
212.991g Al(NO3)3
=132 g Al(NO3)3
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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 KCN0.025 L solution
= 4.26 M
0.025 X 4.26 0.1065 mol KCN
0.1065 mol KCN
1 mol KCN
65 g KCN
=6.92 g KCN
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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
0.0153 mol Cu2CO3
X L solution = 0.640 M
0.0153÷0.640 = 0.0239 L
2.86 g Cu2CO3 1 mol Cu2CO3
187.089g Cu2CO3
= 0.0153 mol Cu2CO3
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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
0.0779 mol Ca(HCO3)2
X L solution = 1.28 M
0.0779 ÷ 1.28 = 0.0608 L
12.62 g Ca(HCO3)21 mol Ca(HCO3)2
161.994g Ca(HCO3)2
= 0.0779 mol Ca(HCO3)2
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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
0.0749 mol Sr(C2H3O2)2
0.340 L solution = 0.220 M
15.4 g Sr(C2H3O2)21 mol Sr(C2H3O2)2
205.62g Sr(C2H3O2)2
= 0.0749 mol Sr(C2H3O2)2
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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
0.501 mol Fe2(SO3)3
1.42 L solution = 0.359 M
176.2 g Fe2(SO3)31 mol Fe2(SO3)3
351.591g Fe2(SO3)3
= 0.501 mol Fe2(SO3)3
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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.
58.8
(4.20 M)(14.0ml) = ( XM)(86.0ml)
86.6= 67.9M
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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.
549
(1.22 M)(450ml) = ( XM)(1260 ml)
1260= 0.436M
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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.
7.68
(0.64 M)(12.0ml) = ( 0.19M)(X ml)
0.19= 40.4 ml
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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.
387.32
(4.60 M)(84.2ml) = ( 1.42M)(X ml)
1.42= 273 ml
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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.
76.68
(2.73 M)(X ml) = ( 0.540M)(142 ml)
2.73= 28.1 ml
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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.
0.00675
(0.0076 M)(X ml) = ( 0.00027M)(25.0 ml)
0.0076= 0.888 ml
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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
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DO NOT CHEW ANY ICE!!!!Get a chunk-
spit it out