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4. Draw the structure for SCIF. Place the Fatoms in the axial positions and the Cl...
Name Instructor balanted: 2A1s) 2.4DMg) 4H5t 229400-2KAI()12420 43W2ug Date EXPERIMENT REPORT SHEET Recycling Aluminum Experimental Data:...
Name Instructor balanted: 2A1s) 2.4DMg) 4H5t 229400-2KAI()12420 43W2ug Date EXPERIMENT REPORT SHEET Recycling Aluminum Experimental Data: Synthesis of KAKSO-12 H glass ler Poper 38. 2010 Mass of Al Volume of KOH Wotch 15 mL 5.5M mL Volume of H,SO, Mass of product 43.297 .201 Calculations Moles of Al 0141mol Moles of KOH Moles of H SO Moles of product (KAKSO 12HO) Limiting reagent Theoretical yield (KAISO-12 H,O) Percent yield (KA(SO-12 H,O) n d d w ct Comment on possible impurites...
Recycling Al experiment Mass of Al : 0.49g Mass of KAl(SO4)2 12H2O synthesized: 2.86g 1. Assuming...
Recycling Al experiment Mass of Al : 0.49g Mass of KAl(SO4)2 12H2O synthesized: 2.86g 1. Assuming you started with pure aluminium scrap, calculate the theoretical mass of the precipitate, alum. Include a balanced net ionic equation for the formation of the product from Al3+. 2. Calculate your percent yield. 3. If you had a large number amount of impurities in the scrap aluminium, what would be the impact in the percent yield calculation? explain. 4. In this experiment volume of...
ed chemical equation. However, the balanced chemical equation only works in units of moles. To convert...
ed chemical equation. However, the balanced chemical equation only works in units of moles. To convert grams of aluminum to moles of aluminum the atomic mass of aluminum must be used. grams of aluminum x 1 mole Al = 1 26.98 g Al To relate the moles of aluminum to KAl(SO4)2* 12H20, the mole to mole ratio must be found in the balanced chemical equation. In this case it is: 1 mole of Al3 = 1 mole KAl(SO4)2 12H20 As...
Prelab Questions: Name 1) What is the molar mass of KAl(SO4)2 12H20 ? (the 12 waters...
Prelab Questions: Name 1) What is the molar mass of KAl(SO4)2 12H20 ? (the 12 waters are part of the molecule) 2) Calculate the mass of alum that can be produced from 14.471g of Al 3) Calculate the mass of alum that can be formed from 0.724 moles of KOH 4) Calculate the mass of alum that can be formed from 1.10 mole of H2SO4 5) a) Based on the data above which reactant is the limiting reactant? b) What...
During a "Synthesis of Alum" experiment, with 24.9mL of 1.4M KOH. After cooling the mixt Given...
During a "Synthesis of Alum" experiment, with 24.9mL of 1.4M KOH. After cooling the mixt Given the following balanced equation: OF Alum" experiment, a student uses 0.500g of aluminum foil and disson NOR. After cooling the mixture, the students adds 11.9mL of 9.OM H2SO4. 2 Al(s) + 2 KOH(aq) + 4 H,SO(aq) + 22 H200) — 3 H2(g) + 2 KAl(SO4)2-12 H20 a. Calculate the theoretical yield of alum. Show the process of finding the limiting reactant. [Molar Mass (Alum)...
experimental value fld percent sulfate in alum Calculations for Part 2 1504 in KAl(SO4)2 1. Experimental...
experimental value fld percent sulfate in alum
Calculations for Part 2 1504 in KAl(SO4)2 1. Experimental value for percent sulfate in alum BA 0.15799 504) H20 12( 18.02 ): 216.24 KA(SO4), 124,00 MM (504) = 64,32+ 128.00 2 = 192.32 glmol Results and Data Analysis SON Table 1: Determination of the Water of Hydration in Alum Crystals Mass of crucible + cover + alum 43.10509_water D ana 09959 Mass of alum 1.005sga Mass of crucible + cover + alum after...
please help with the calculations and limiting reagent\percent yeild!!! molarity of FeCl3 is .40 g/mol we...
please help with the calculations and limiting reagent\percent
yeild!!!
molarity of FeCl3 is .40 g/mol
we
had 5.0 mL of FeCl3 (0.40g/mL) which equals 0.08g divided by 162.2g
FeCl3 which equals 0.000493 moles or 4.9 x 10^-4
EXPERIMENT REPORT SHEET Synthesis of KzFe(C204)3.3 H20 15.0 mL 4a5a watch glass 43.958 : paper Experimental Data Volume of FeCl, solution Mass of K C,04H2O Mass of product 47.891 Calculations Moles of FeCl3 Moles of K C2O4 H2O Moles of product (KsFe(C204)3.3 H20)...
percent error #3 Calculations for Part 2 Ou in KAl(SO4)2 1. Experimental value for percent sulfate...
percent error #3
Calculations for Part 2 Ou in KAl(SO4)2 1. Experimental value for percent sulfate in alum BA 0.15793504) H20 12 ( 18.0a)= 216.24 KAYS0y), 12H20 MM (504)₂ = 64.327 128.00 2 = 192.32 glmor 2. Theoretical value for percent sulfate in alum are monologa dobar odbud 3. Percent error Results and Data Analysis Table 1: Determination of the Water of Hydration in Alum Crystals Mass of crucible + cover + alum 43.10509 - water Mass of crucible +...
help with all qeustions(: Prelab Ouestions: Name 1) What is the molar mass of KAI(SO.), -12H,O...
help with all qeustions(:
Prelab Ouestions: Name 1) What is the molar mass of KAI(SO.), -12H,O ? (the 12 waters are part of the molecule) 2) Calculate the mass of alum that can be produced from 14.471g of Al 3) Calculate the mass of alum that can be formed from 0.724 moles of KOH 4) Calculate the mass of alum that can be formed from 1.10 mole of H2SO4 5) a) Based on the data above which reactant is the...
help with all questions(: 1) Describe what the crystals look like? 2) If your percent yield...
help with all questions(:
1) Describe what the crystals look like? 2) If your percent yield is greater that 100%, give one plausible error you may have made. 3) Calculate the limiting reactant and theoretical yield of Cu for the following reaction: First balance the reaction CuSO4 (aq) + Al(s) → Al2(SO4)(aq) + Cu (S) 3.35 g 0.321 g 4) If the actual yield for the reaction in question (3) is 1.06 g, what is the percent yield for the...
Name Instructor balanted: 2A1s) 2.4DMg) 4H5t 229400-2KAI()12420 43W2ug Date EXPERIMENT REPORT SHEET Recycling Aluminum Experimental Data: Synthesis of KAKSO-12 H glass ler Poper 38. 2010 Mass of Al Volume of KOH Wotch 15 mL 5.5M mL Volume of H,SO, Mass of product 43.297 .201 Calculations Moles of Al 0141mol Moles of KOH Moles of H SO Moles of product (KAKSO 12HO) Limiting reagent Theoretical yield (KAISO-12 H,O) Percent yield (KA(SO-12 H,O) n d d w ct Comment on possible impurites...
ed chemical equation. However, the balanced chemical equation only works in units of moles. To convert grams of aluminum to moles of aluminum the atomic mass of aluminum must be used. grams of aluminum x 1 mole Al = 1 26.98 g Al To relate the moles of aluminum to KAl(SO4)2* 12H20, the mole to mole ratio must be found in the balanced chemical equation. In this case it is: 1 mole of Al3 = 1 mole KAl(SO4)2 12H20 As...
Prelab Questions: Name 1) What is the molar mass of KAl(SO4)2 12H20 ? (the 12 waters are part of the molecule) 2) Calculate the mass of alum that can be produced from 14.471g of Al 3) Calculate the mass of alum that can be formed from 0.724 moles of KOH 4) Calculate the mass of alum that can be formed from 1.10 mole of H2SO4 5) a) Based on the data above which reactant is the limiting reactant? b) What...
During a "Synthesis of Alum" experiment, with 24.9mL of 1.4M KOH. After cooling the mixt Given the following balanced equation: OF Alum" experiment, a student uses 0.500g of aluminum foil and disson NOR. After cooling the mixture, the students adds 11.9mL of 9.OM H2SO4. 2 Al(s) + 2 KOH(aq) + 4 H,SO(aq) + 22 H200) — 3 H2(g) + 2 KAl(SO4)2-12 H20 a. Calculate the theoretical yield of alum. Show the process of finding the limiting reactant. [Molar Mass (Alum)...
experimental value fld percent sulfate in alum
Calculations for Part 2 1504 in KAl(SO4)2 1. Experimental value for percent sulfate in alum BA 0.15799 504) H20 12( 18.02 ): 216.24 KA(SO4), 124,00 MM (504) = 64,32+ 128.00 2 = 192.32 glmol Results and Data Analysis SON Table 1: Determination of the Water of Hydration in Alum Crystals Mass of crucible + cover + alum 43.10509_water D ana 09959 Mass of alum 1.005sga Mass of crucible + cover + alum after...
please help with the calculations and limiting reagent\percent
yeild!!!
molarity of FeCl3 is .40 g/mol
we
had 5.0 mL of FeCl3 (0.40g/mL) which equals 0.08g divided by 162.2g
FeCl3 which equals 0.000493 moles or 4.9 x 10^-4
EXPERIMENT REPORT SHEET Synthesis of KzFe(C204)3.3 H20 15.0 mL 4a5a watch glass 43.958 : paper Experimental Data Volume of FeCl, solution Mass of K C,04H2O Mass of product 47.891 Calculations Moles of FeCl3 Moles of K C2O4 H2O Moles of product (KsFe(C204)3.3 H20)...
percent error #3
Calculations for Part 2 Ou in KAl(SO4)2 1. Experimental value for percent sulfate in alum BA 0.15793504) H20 12 ( 18.0a)= 216.24 KAYS0y), 12H20 MM (504)₂ = 64.327 128.00 2 = 192.32 glmor 2. Theoretical value for percent sulfate in alum are monologa dobar odbud 3. Percent error Results and Data Analysis Table 1: Determination of the Water of Hydration in Alum Crystals Mass of crucible + cover + alum 43.10509 - water Mass of crucible +...
help with all qeustions(:
Prelab Ouestions: Name 1) What is the molar mass of KAI(SO.), -12H,O ? (the 12 waters are part of the molecule) 2) Calculate the mass of alum that can be produced from 14.471g of Al 3) Calculate the mass of alum that can be formed from 0.724 moles of KOH 4) Calculate the mass of alum that can be formed from 1.10 mole of H2SO4 5) a) Based on the data above which reactant is the...
help with all questions(:
1) Describe what the crystals look like? 2) If your percent yield is greater that 100%, give one plausible error you may have made. 3) Calculate the limiting reactant and theoretical yield of Cu for the following reaction: First balance the reaction CuSO4 (aq) + Al(s) → Al2(SO4)(aq) + Cu (S) 3.35 g 0.321 g 4) If the actual yield for the reaction in question (3) is 1.06 g, what is the percent yield for the...
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