Midterm Exam Exp. 1-9 Monday night
Nick and Kel reacted 3.10 g of NiCl₂·6H₂O with 11.50 mL of 4.00 M C₂H₈N₂. They recovered 3.09 g of product, [Ni(C₂H₈N₂)₃]Cl₂. At the end, the filtrate was a light pink solution. The litmus test on the filtrate solution turned the red litmus paper blue in color where the solution touched the paper. NiCl₂·6H₂O + 3 C₂H₈N₂ → [Ni(C₂H₈N₂)₃]Cl₂(s) + 6 H₂O Molar Mass of NiCl₂·6H₂O is 237.69 g/mol Molarity of C₂H₈N₂ is 4.00 M Molar Mass of product, [Ni(C₂H₈N₂)₃]Cl₂(s) is 309.90 g/mol Based on the experiment described, select the correct answers for the questions below. The chemical, NiCl₂·6H₂O, is the ... The theoretical yield is ... The actual or experimental yield is ... The chemical, C₂H₈N₂, is the ...
****The chemical, NiCl₂·6H₂O, is the limiting reagent Since the filtrate is basic and pink, this indicates that C₂H₈N₂ is in excess. Therefore, the NiCl2·6 H2O is the Limiting Reagent (the reactant which is completely consumed or "used up" during the reaction.) ****The theoretical yield is 4.04 g (g NiCl₂·6H₂O ÷ g/mol NiCl₂·6H₂O) x g/mol [Ni(C₂H₈N₂)₃]Cl₂ =(3.10 g ÷ 237.69 g/mol) x 309.90 g/mol =4.04 g ****The actual or experimental yield is 3.09 g "They recovered 3.09 g of product, [Ni(C₂H₈N₂)₃]Cl₂." ****The chemical, C₂H₈N₂, is the chemical present in excess Since the filtrate is basic and pink, this indicates that C₂H₈N₂ is in excess.
In a similar experiment, Cooper, Zach & Nate started with 2.25 g of Cu(C₂H₃O₂)₂·H₂O and 5.13 g of NaC₇H₄SO₃N·H₂O and obtained 5.18 g of Cu(C₇H₄SO₃N)₂(H₂O)₄·2H₂O. The theoretical yield of the product is ____ g. The percent yield of the product is ____ %. Molar Mass of Copper(II) acetate monohydrate = 199.65 g/mol Molar Mass of Sodium saccharinate monohydrate = 223.18 g/mol Molar Mass of product = 535.59 g/mol
****The theoretical yield of the product is 6.04 g 1st: find moles of Cu(C₂H₃O₂)₂·H₂O = g Cu(ac)₂∙H₂O x (1 mol Cu(ac)₂∙H₂O ÷ g/mol Cu(ac)₂∙H₂O) = 2.25 x (1÷199.65) = .011269 mol Cu(ac)₂∙H₂O 2nd: find moles of the product Cu(C₇H₄SO₃N)₂(H₂O)₄·2H₂O(s) = mole Cu(ac)₂∙H₂O x (1 mol Cu(sacch)₂(H₂O)₄∙2H₂O ÷ 1 mol Cu(ac)₂∙H₂O) = .011269 x (1÷1) = .011269 mol Cu(sacch)₂(H₂O)₄∙2H₂O 3rd: find mass of product Cu(C₇H₄SO₃N)₂(H₂O)₄·2H₂O(s) = mol Cu(sacch)₂(H₂O)₄∙2H₂O x (g/mol Cu(sacch)₂(H₂O)₄∙2H₂O ÷ 1 mol Cu(sacch)₂(H₂O)₄∙2H₂O) = .011269 x (535.59÷1) = 6.04 g Cu(sacch)₂(H₂O)₄∙2H₂O ****The percent yield of the product is _85.8_ % percent yield = (actual yield ÷ theoretical yield) x 100 percent yield = [5.18 g of Cu(C₇H₄SO₃N)₂(H₂O)₄·2H₂O ÷ 6.04 g Cu(sacch)₂(H₂O)₄∙2H₂O] x 100 = 85.8 %
mol (solution) = M x L M = mol ÷ L
...
M titrant × equivalent point volume titrant × (mol analyte / mol titrant) = moles analyte Manalyte = molesanalyte / Lanalyte
....
Sid Trick titrated an 8.50 mL aliquot of juice. It required 14.87 mL of 0.0981 M NaOH to reach the equivalence point. Analyze his results and answer the questions below. The reaction and the molar mass are given below. H3C6H5O7(aq) + 3 NaOH(aq) → Na3C6H5O7(aq) + 3 H2O(ℓ) Molar Mass of H3C6H5O7 = 192.12 g/mL How many grams of citric acid were present in the aliquot? How many mg of citric acid were present per mL of juice?
.0934g 11.0 mg/mL
What is the Molarity of the Ba(OH)₂ solution if a 12.5 mL aliquot of the Ba(OH)₂ solution required 17.50 mL of a 0.110 M H₂SO₄ solution to reach the equivalence point? Ba(OH)₂(aq) + H₂SO₄(aq) → BaSO₄(s) + 2H₂O(l)
.154
What is the mole to mole ratio between Cu(C₂H₃O₂)₂·H₂O and NaC₇H₄SO₃N·H₂O? Select one: 2 mol Cu(C₂H₃O₂)₂·H₂O : 3 mol NaC₇H₄SO₃N·H₂O 2 mol Cu(C₂H₃O₂)₂·H₂O : 1 mol NaC₇H₄SO₃N·H₂O 1 mol Cu(C₂H₃O₂)₂·H₂O : 2 mol NaC₇H₄SO₃N·H₂O 1 mol Cu(C₂H₃O₂)₂·H₂O : 1 mol NaC₇H₄SO₃N·H₂O
1 mol Cu(C₂H₃O₂)₂·H₂O : 2 mol NaC₇H₄SO₃N·H₂O
how do you determine how many mL of juice is needed to meet the DRI of vitamin C?
1.) first, find mg of C₆H₈O₆ by using the M and mL of DCIP, stoichiometry, and the molar mass of C₆H₈O₆ mg C₆H₈O₆ = M DCIP x L DCIP x (1 mol C₆H₈O₆ ÷ 1 mol DCIP) x (176.14 g C₆H₈O₆ ÷ 1 mol C₆H₈O₆) x (1000 mg / 1 g) 2.) then, divide the DRI (90 mg of C6H8O6) by mg of C6H8O6 per mL of juice to find the mL of juice. mL juice = DRI mg C₆H₈O₆ x (mL of juice ÷ mg C₆H₈O₆)
Cedrick & Astrid titrated a 25.00 mL aliquot of grapefruit juice with a 0.117 M NaOH solution to the end point. They calculated that there was 0.1462 g of citric acid in the juice sample. What is the amount mg of citric acid present per mL of juice? Select one: 3.65 mg/mL 6.85 mg/mL 9.60 mg/mL 5.85 mg/mL
5.85 mg/mL
Cooper, Zach and Nate did this experiment dissolving 3.80 g of Cu(C₂H₃O₂)₂·H₂O and adding 8.60 g NaC₇H₄SO₃N·H₂O (assume present in excess). What is the theoretical yield of the product, Cu(C₇H₄SO₃N)₂(H₂O)₄·2H₂O(s)? Molar Mass of Copper(II) acetate monohydrate = 199.65 g/mol Molar Mass of Sodium saccharinate monohydrate = 223.18 g/mol Molar Mass of product = 535.59 g/mol Select one: 10.32 g 10.45 g 12.40 g 10.19 g
10.19 g moles of Cu(ac)₂∙H₂O = g Cu(ac)₂∙H₂O x (1 mol Cu(ac)₂∙H₂O / mm Cu(ac)₂∙H₂O) = 3.80 x (1/199.65) = .01903 mol Cu(ac)₂∙H₂O mol of product = mole Cu(ac)₂∙H₂O x (1 mol Cu(sacch)₂(H₂O)₄∙2H₂O / 1 mol Cu(ac)₂∙H₂O) = .01903 x (1/1) = .01903 mol Cu(sacch)₂(H₂O)₄∙2H₂O mass of product = mol Cu(sacch)₂(H₂O)₄∙2H₂O x (mm Cu(sacch)₂(H₂O)₄∙2H₂O / 1 mol Cu(sacch)₂(H₂O)₄∙2H₂O) = .01903 x (535.59/1) = 10.19 g Cu(sacch)₂(H₂O)₄∙2H₂O theoretical yeild = 10.19 g of Cu(sacch)₂(H₂O)₄∙2H₂O (solution: to determine the theoretical yield of the product using the mass of the Cu(C₇H₄SO₃N)₂(H₂O)₄·2H₂O(s), its molar mass, and the reaction stoichiometry. To do this, you will need the molar mass of this reactant. This reactant has a water of hydration which is shown by the "·H2O" at the end of the chemical formula. So we must add the weight of water to the overall molar mass. Then calculate the moles of the product. Finally calculate the mass of the product by multiplying the moles of product by its molar mass.)
A calibration plot of absorbance vs. concentration (ppm) was obtained with standard known Red 40 dye solutions. The slope of the best-fit straight line of the plot is 0.061 ppm- 1 The absorbance of the dilute unknown sports drink was 0.73 What is the concentration of this dilute unknown sports drink? Select one: 12.0 ppm 4.45 ppm 83.6 ppm 22.5 ppm
12.0 ppm
Ozzie wanted to do another experiment using 8.75% (by mass) H₂O₂. What is the Molarity of this H2O2 solution ? ____ H₂O₂
2.57 M H₂O₂ In the last step of the calculations we did in the lab, we converted Molarity of H2O2 into % by mass by the equation below. For this problem, we need to work it in the reverse. % (by mass) = Molarity H2O2 x Molar Mass × (1 / (Density) × (1 L/1000 mL) × 100% *we assumed the density of water and the hydrogen peroxide solution were both 1.0 g/mL since the solution is mostly water). However, this time start with the % and convert it to Molarity by rearranging the equation above. Molarity = % (by mass) / 100 x Density × (1/Molar mass) x (1000 mL/1L) Now substitution the definition of % (by mass) into the equation above. % (by mass) = (mass of H2O2/ mass of solution) Molarity = [(g of H2O2/g of soln) / 100)] x (1.0 g soln/mL soln) x ( 1 mol H2O2/ 34.01 g H2O2) × (1000 mL/1L) So the g of H2O2 cancels, and the mass of solution cancels, so the units left are mol of H2O2 per mL soln which is the definition of Molarity.
C6H8O6(aq) + C12H7O2NCl2(aq) → C6H6O6(aq) + C12H9O2NCl2(aq) C12H7O2NCl2 is abbreviated as DCIP Molar Mass of C6H8O6(s) is 176.14 g/mol Daily Recommended Intake (DRI) for adults: 90 mg of Vitamin C Askor & Bich found that it required 14.81 mL of 1.07 x 10-3 M DCIP solution to reach the endpoint when titrating a 6.00 mL aliquot of pear juice. How many mg of ascorbic acid, C6H8O6, are present in the juice aliquot? ___ How many mL of this juice would Bich have to drink in order to meet the Daily Recommended Intake (DRI)? ___
2.79 194
In this problem, click on the correct answer above, then drag and drop it in the correct box. Cedrick & Astrid titrated a 15.00 mL aliquot of grapefruit juice with a 0.134 M NaOH solution to the end point. The initial buret reading was 1.04 mL and the final buret reading was 24.83 mL. H3C6H5O7(aq) + 3 NaOH(aq) → Na3C6H5O7(aq) + 3 H2O(l) What is the volume of NaOH titrated? ___ of NaOH What is the mass of citric acid in the juice sample? ___ of H3C6H5O7
23.79 mL NaOH 0.204 g H₃C₆H₅O₇ To calculate the Volume of NaOH titrated into the flask is: Final Buret volume - Initial Buret Volume = volume titrated To find mass of Citric acid in the juice, you need to: 1. Calculate the moles of NaOH M NaOH x Vol. NaOH titrated = moles NaOH reacted 2. Calculate the moles of Citric Acid (H3C6H5O7) mole NaOH reacted x [1 mol H3C6H5O7 / 3 mol NaOH] = mol of H3C6H5O7 3. Calculate the mass of H3C6H5O7 mol of H3C6H5O7 x Molar Mass of H3C6H5O7 = mass of H3C6H5O7
Ethyl, Ann & Ami reacted 3.33 g of NiCl₂·6H₂O with 8.45 mL of 4.00 M C₂H₈N₂. They recovered 2.47 g of product, (Ni(C₂H₈N₂)₃)Cl₂. The theoretical yield of the product is __ g. The percent yield of the product is __ %.
3.49 g 70.7 % To find the theoretical yield: If NiCl2· 6H2O is Limiting Reagent: Calculate moles of NiCl2· 6 H2O using mass of NiCl2· 6 H2O and dividing by its molar mass Use that number and the mole:mole ratio to find moles of product Calculate mass of product (theoretical yield) using previous number & Molar mass of the product. If ethylenediamine (C2H8N2) solution is Limiting Reagent: Calculate moles of C2H8N2 using Molarity x Liters of C2H8N2: Use that number and the mole:mole ratio to find moles of product Calculate mass of product (theoretical yield) using previous number & Molar mass of the product. Then compare the mass of product by each reagent, whichever is the SMALLER VALUE, is the theoretical yield To find percent yield: Divide the actual or experimental yield by the theoretical yield (smaller number), and multiply by 100%
Ozzie wanted to do another experiment with a stronger H₂O₂ solution to check the accuracy of the experiment by calculating the theoretical volume of O₂(g) it would produce. Then he could compare his experimental volume of O₂(g) to the theoretical volume of O₂(g). He used 4.50 mL of 5.88 M H2O2 and the partial pressure of O₂ was 0.9537 atm and the temperature was 296.35 K. What volume of O₂(g) could he theoretically produce (in mL)?
337 This question is just working the calculations in reverse from the lab. There are three parts to the calculations. Using Molarity and mL of H₂O₂ find moles of H₂O₂ By stoichiometry find moles of O₂ (g) By Ideal Gas Law find Volume of O₂ Part 1. Find moles of H₂O₂ Molarity H₂O₂ x mL H₂O₂ x 1L/1000 mL = mol of H₂O₂ Part 2. Find moles of O₂(g) moles H₂O₂ (from above) x 1 mol O₂/2 mol H₂O₂ = mol O₂ (from the balanced reaction) Now that you have moles of O₂ gas you can use the Ideal Gas Law. The H₂O₂ data is aqueous, so you can't use gas law calculations for solutions! Part 3. Find Volume of O₂(g) Use R = 0.08206 L atm/K mol and the PV=nRT ; Rearrange PV=nRT to solve for V since you have R, PO2 ,T, mol of gas (which is n) V = (nRT/P) Finally convert to mL of O₂
An 10.0 mL aliquot of a 125 ppm stock Red 40 dye solution was diluted to a final volume of 25.0 mL with deionized water. What is the concentration of the dilute Red 40 dye solution (in ppm)? Select one: 50.0 ppm 31.3 ppm 2.00 ppm 62.6 ppm
50.0 ppm
Zach and Nate calculated the theoretical yield as 6.57 g of Cu(C₇H₄SO₃N)₂(H₂O)₄·2H₂O(s). They recovered 4.53 g of Cu(C₇H₄SO₃N)₂(H₂O)₄·2H₂O(s). What is the percent yield? Select one: 98.0 % 45.1% 75.6 % 31.1% 68.9%
68.9% percent yield = (actual yield/theoretical yield) x 100 percent yield = (4.53/6.57) x 100 = 68.9
A 146 mg sample of Red 40 dye was dissolved and diluted to a final volume of 2.00 L with deionized water. What is the concentration of the Red 40 dye solution in ppm? parts per milliion (ppm) refers to 1 mg of solute in 1.0 L of solution; (mg/L) Select one: 73.0 ppm 29.2 ppm 0.0137 ppm 1.83 ppm
73.0 ppm
what is Beer's Law equation?
Absorbance = slope x constant
In today's experiment of determining the amount of Vitamin C in fruit juice,, which chemical species is the titrant?
C12H7O2NCl2 (2,6-dichloroindophenol or DCIP)
Problem After 2.05 g of NiCl2·6 H2O reacted with 7.50 mL of the 4.0 M C₂H₈N₂ (ethylenediamine), 1.87 g of the product, [Ni(en)3]Cl2 was obtained. The filtrate was pale pink and when tested with pH paper, the paper indicated a basic solution. how would you calculate the theoretical yield of [Ni(en)₃]Cl₂?
Calculate the theoretical yield of the product assuming the NiCl₂·6H₂O is the Limiting Reagent. Assuming NiCl₂·6H₂O is Limiting Reagent: mol NiCl₂·6H₂O = 2.05 g NiCl₂·6H₂O × (1 mol NiCl₂·6H₂O ÷ 237.69 g NiCl₂·6H₂O) mol NiCl₂·6H₂O = 0.008625 mol NiCl₂·6H₂O ↓ mass [Ni(en)₃]Cl₂ = 0.008625 mol NiCl₂·6H₂O × (1 mol [Ni(en)₃]Cl₂ ÷ 1 mol NiCl₂·6H₂O) × (309.90 g [Ni(en)₃]Cl₂ ÷ 1 mol [Ni(en)₃]Cl₂) ***mass [Ni(en)₃]Cl₂ = 2.67 g [Ni(en)₃]Cl₂ To CHECK that NiCl₂·6H₂O is the Limiting Reagent, assume en is the Limiting Reagent and calculate the theoretical yield. If the NiCl₂·6H₂O is indeed the Limiting Reagent (en is excess as indicated by the filtrate), then the theoretical yield from calculation (1) will be less than the theoretical yield from the calculation (2). Assume H₂NCH₂CH₂NH₂ (en) is Limiting Reagent: mol en = M en × L en = 4.00 M en × 7.50 mL en × (1 L ÷ 1000 mL) = mol en = 0.0300 mol en ↓ mass [Ni(en)₃]Cl₂ = 0.0300 mol en × (1 mol [Ni(en)₃]Cl₂ ÷ 3 mol en) × (309.90 g [Ni(en)₃]Cl₂ ÷ 1 mol [Ni(en)₃]Cl₂) ***mass [Ni(en)₃]Cl₂ = 3.10g[Ni(en)₃]Cl₂ ********The smaller mass of product is from calculation (1). Therefore the theoretical yield is 2.67 g**********
In this reaction, you observed several different Cu species which had different physical attributes. Select the appropriate chemical formula(s) that corresponds to a black solid. Select one or more: Cu(OH)₂ Cu CuSO₄ CuO
CuO
In this reaction, you observed several different Cu species which had different physical attributes. Select the appropriate chemical formula(s) that corresponds to a sky blue solution. Select one or more: Cu CuO Cu(OH)₂ CuSO₄
CuSO₄
Identify the reaction type for the generic reaction given. ABC → AB + C Select one: Oxidation-Reduction Double Displacement Single Displacement Combination Decomposition
Decomposition soltuion: • Decomposition: One species decomposes into two or more chemical species, generally as a result of heating. ABC→ AB + C CaCO₃(s) → CaO(s) + CO₂(g)
Will use of a lower Molarity value of DCIP affect the calculated mg of Vitamin C? Select one: Increases the value of mg of Vitamin C No effect Decreases the value of mg of Vitamin C
Decreases the value of mg of Vitamin C
An experimental method to estimate the density of a crystal is to compare it to a variety of solution of known densities. Solution K (density = 1.5 g/mL), Solution L (density = 1.2 g/mL) and Solution W (density = 1.0 g/mL) are poured into a test tube. The liquids are immiscible with each other and solid M is insoluble in each. When a few crystals of M are added to the test tube, it floats on top of solution W. What is the best approximate density of solid M? Select one: Density of M is greater than 1.2 g/mL but less than 1.5 g/mL. Density of M is greater than 1.0 g/mL but less than 1.2 g/mL. Density of M is greater than 1.5 g/mL. Density of M is less than 1.0 g/mL.
Density of M is less than 1.0 g/mL. The layer with the lesser density will float or rise to the top of the denser layer. (solution: solubility-the ability of a solid or gas to dissolve in a liquid ex. sugar & coffee solvent-the solution that does the dissolving ex. the coffee solute-the substance being dissolved ex. the sugar miscible-describes two liquids that mix completely with each other ex. cream & coffee immiscible-describes two liquids that do not mix completely and form layers ex. olive oil & vinegar)
Identify the reaction type for the generic reaction given. AC + BD → AD + BC Select one: Combination Decomposition Oxidation-Reduction Single Displacement Double Displacement
Double Displacement soltuion: • Double Displacement: AC + BD → AD + BC Cd(NO₃)₂(aq) + (NH₄)₂S(aq) → CdS(s) + 2 NH₄NO₃(aq) Zn(OH)₂(s) + 2HCl(aq) → ZnCl₂(aq) + 2 H₂O(ℓ)
How do you find the equivalent point volume from a conductimetric titration graph?
For conductimetric titrations, the equivalence point is the intersection of the two line segments. You can tap the point where the two segments intersect, and the equivalence point volume will be the volume in the lower box. For example in the book, it was at 10.0 mL.
Which chemical(s) used in this experiment is/are toxic? Check all appropriate chemicals that apply. Select one or more: H₂SO₄ H₂O₂ Cu Zn NaOH
H₂SO₄ NaOH
Piers & Ida performed this experiment. Both started with the exact same volume of hydrogen peroxide. However, some gas escaped out of the test tube because the rubber stopper wasn't tight in Ida's set-up. Piers did not have any errors with his experiment. Which of the following statement(s) are TRUE concerning Ida's results? I. Ida's experiment will give a lower volume of O₂(g) than Piers. II. Ida's experiment will give a lower partial pressure of O₂(g) than Piers. III. Ida's calculations will give a smaller molarity of H₂O₂ than Piers. Select one: Only I is true. I & II are true. II and III are true. I and III are true.
I and III are true.
how do you determine which reactant is the limiting reagent?
In some reactions, physical observations of the reaction indicate which reactant is in excess. In this experiment, each of the reactants has different physical properties. The Ni+2 complexes are green to blue-violet in color; whereas the ethylenediamine is a weak base. If the filtrate is blue to blue-violet, then the NiCl2·6 H2O is in excess. However, if the filtrate is pink and is basic, then ethylenediamine is in excess. EXAMPLE: After 2.05 g of NiCl2·6 H2O reacted with 7.50 mL of the 4.0 M ethylenediamine, 1.87 g of the product, [Ni(en)3]Cl2 was obtained. The filtrate was pale pink and when tested with pH paper, the paper indicated a basic solution. What is the limiting reagent for the reaction, theoretical yield, and percent yield of [Ni(en)3]Cl2? Solution: Since the filtrate is basic and pink, this indicates that ethylenediamine is in excess. Therefore, the NiCl2·6 H2O is the Limiting Reagent.
1.61 mL
In the picture, what is the correct volume reading? Read meniscus in 10 mL graduated cylinder Select one: 2.40 mL 2.31 mL 1.61 mL 1.69 mL
For the procedural error, indicate if the error will affect the actual yield of [Ni(C2H8N2)3]Cl2 product and if it does, will it raise or lower the actual yield: Dissolving the NiCl2·6 H2O in 12 mL of water initially Select one: Lowers the actual yield Raises the actual yield No effect
Lowers the actual yield
For the procedural error, indicate if the error will affect the actual yield of copper(II) saccharinate product and if it does, will it raise or lower the actual yield: Only leaving the beaker in the ice bath for 5 minutes instead of 25 minutes Select one: Lowers the actual yield Raises the actual yield No effect
Lowers the actual yield
For the procedural error, indicate if the error will affect the actual yield of copper(II) saccharinate product and if it does, will it raise or lower the actual yield: Washing the crystals with hot water Select one: Raises the actual yield Lowers the actual yield No effect
Lowers the actual yield
Will overtitrating the end point in Part II affect the value of Molarity of DCIP? Select one: Raises the value of Molarity of DCIP Lowers the value of Molarity of DCIP No effect
Lowers the value of Molarity of DCIP
Excess chemical - Appropriate waste container Reaction mixture - Appropriate waste container Used filter paper - Trash can Cracked or chipped beaker - Broken glass box
Match each item to its proper disposal method. Answers be used more than once. Excess chemical Reaction mixture Used filter paper Cracked or chipped beaker
What color is expected for the product, [Ni(C₂H₈N₂)₃]Cl₂(s), if C₂H₈N₂ is the limiting reagent? Select one: Pink-violet Blue-green Lime green Dark Purple Dark Blue White
Pink-violet
For the error, indicate if the error will affect the mg of citric acid present per mL of juice, and if it does, will it raise or lower the value of the mg citiric acid/ mL of juice? Not adding the phenolphthalein to the flask and stopping when the buret is empty
Raises the value of mg citric acid/mL juice
Here is useful information for the questions below. Select your answers from the pull-down menus. C12H7O2NCl2 is abbreviated as DCIP Molar Mass of C6H8O6(s) is 176.14 g/mol C6H8O6(aq) + C12H7O2NCl2(aq) → C6H6O6(aq) + C12H9O2NCl2(aq) Ace, Corby & Astrid dissolved 0.0248 g of C6H8O6 in a 50.0 mL volumetric flask and filled it with deionized water to the mark. *What is the Molarity of this standard solution of C6H8O6 (ascorbic acid/Vitamin C)?* A 12.0 mL aliquot of the standard C6H8O6 above required 21.36 mL of DCIP to reach the endpoint. *What is the Molarity of the DCIP solution (standardization of titrant)?* A 10.00 mL aliquot of pear juice required 18.37 mL of the same DCIP solution (from above) to reach the end point. *How many mg of C6H8O6 (ascorbic acid/Vitamin C) are present in the aliquot?*
Molarity of C6H8O6 = (mass C6H8O6 / Molar Mass C6H8O6) / total volume in L - answer is: 2.82 x 10^-3 M Molarity of DCIP = {M C6H8O6 (from above) x L C6H8O6 (aliquot) x (1 mol DCIP/1 mol C6H8O6) } / L of DCIP titrated - answer is: 1.58 x 10^-3 M mg of C6H8O6 in juice = M DCIP (from above) x L of DCIP titrated x (1 mol C6H8O6/1 mol DCIP) x Molar Mass C6H8O6 x (1000 mg/1 g) - answer is: 5.12 mg
In today's experiment, which chemical species is the titrant? Select one: (C₆H₅OH)₂C₂O₂C₆H₄ (phenolphthalein) H₃C₆H₅O₇ (citric acid) H₂O NaOH
NaOH
Which chemical(s) will be in the buret? Select one or more: (C₆H₅OH)₂C₂O₂C₆H₄ (phenolphthalein) Fruit juice H₂O NaOH
NaOH
For the procedural error, indicate if the error will affect the actual yield of [Ni(C₂H₈N₂)₃]Cl₂ product and if it does, will it raise or lower the actual yield: Rinsing the product with 10 mL of acetone instead of 5 mL in step 11 Select one: Lowers the actual yield No effect Raises the actual yield
No effect
In the reaction below, classify the chemical species as the type of electrolyte. Ba(OH)₂(aq) + H₂SO₄(aq) → BaSO₄(s) + 2H₂O(ℓ) •H₂O Select one: Strong electrolyte Nonelectrolyte Weak electrolyte
Nonelectrolyte
Predict the conductivity of CH3CH2OH, a molecular compound. Select one: Strong electrolyte Nonelectrolyte Weak electrolyte
Nonelectrolyte
A common error in titration is over-titrating. Which of the following statement(s) is/are TRUE for the citric acid titration? I. Over-titrating means too much titrant was added to the reaction. II. Since the volume of titrant is too large, dividing by a larger number in the calculation of citric acid making the mass of citric acid be too low. III. Over-titrating means too much aliquot was added to the reaction.
Only I is TRUE
Which of the following statements are TRUE if the titration of fruit juice (Part III) was over‑titrated? The molarity of DCIP would be lowered since the volume of titrant is in the denominator of the molarity of DCIP calculation. Too much DCIP was added in the titration, causing the Molarity of the ascorbic acid to be lowered. The amount of Vitamin C will be higher since the volume of DCIP is in the numerator in the Vitamin C calculation. Select one: Only I and II are TRUE. Only II and III are TRUE. Only III is TRUE. Only I and III are TRUE.
Only III is TRUE.
For the error, indicate if the error will affect the mg of citric acid present per mL of juice, and if it does, will it raise or lower the value of the mg citiric acid/ mL of juice? Not subtracting the initial buret volume (1.20 mL) from final buret volume (18.43 mL)
Raises the value of mg citric acid/mL juice
solubility is a _____ property
Solubility is a CHEMICAL property
Problem Determine if the concentration of H2O2 solution is 30% or 3% by mass. When 8.03 mL of the H2O2 solution reacted with the yeast at 23.7°C, 85.89 mL of gas was produced. The barometric pressure was 752.4 torr. The vapor pressure of H2O at 24°C is 22.4 torr.
Solution 1. Find the partial pressure of O2 using Dalton's Law of Partial Pressures P atm= P H₂O+P O₂ PO₂= P atm−P H₂O PO₂= 752.4 torr − 22.4 torr=730.0 torr 2. Find moles of O₂ using the Ideal Gas Law PV=nRT→n=PV/RT T=(23.7°C+273.15)K=296.85K R=0.08206 L⋅atm/K⋅mol mol O₂={PO2×VO2/(RT)} mol O₂=[(730.0 torr)×(1 atm/760 torr)×(0.08589 L)] ÷ [(0.08206 L⋅atm/K⋅mol)×(295.85 K)] mol O₂=0.003387 mol 3. Find the moles of H2O2 that produced this O2 using stoichiometry mol H₂O₂=mol O₂×(2 mol H₂O₂ ÷ 1 mol O2) mol H₂O₂=.003387 mol×(2 mol H₂O₂÷1 mol O₂) mol H₂O₂=0.006774 mol ↓ M H₂O₂=(mol H₂O₂)/(L H₂O₂) M H₂O₂=(0.006774 mol)÷(0.00803 L) M H₂O₂=0.8436 M 4. Convert M to % by mass using molar mass of H₂O₂. % H₂O₂(by mass)=(mass of H₂O₂÷mass of soln) ×100% % H₂O₂(by mass)=M H₂O₂ × MM H2O2 × (1÷Density)% % H₂O₂(by mass)=(0.8436 mol H₂O₂ ÷ 1 L)×(34.01 g H₂O₂ ÷ 1 mol H₂O₂)×(1 mL soln ÷ 1 g soln)×(1 L soln ÷ 1000 mL soln)×100% % H₂O₂(by mass)=2.87% The experimental value of 2.87% H2O2 solution is closer to 3% than 30%. So the solution is 3.00% H2O2.
In the reaction below, classify the chemical species as the type of electrolyte. Ba(OH)₂(aq) + H₂SO₄(aq) → BaSO₄(s) + 2H₂O(ℓ) •Ba(OH) Select one: Nonelectrolyte Strong electrolyte Weak electrolyte
Strong electrolyte
Predict the conductivity of KOH, a strong base. Select one: Strong electrolyte Weak electrolyte Nonelectrolyte
Strong electrolyte
What is the conductivity classification for an unknown G that ionizes completely in water? Select one: Strong electrolyte Nonelectrolyte Weak electrolyte
Strong electrolyte
soluble
Sugar disappears when added to hot tea. What describes the relationship between sugar and tea? Select one: soluble insoluble miscible immiscible
What is the conductivity classification for an unknown A that dissolves completely in water and partially dissociates into ions? Select one: Weak electrolyte Nonelectrolyte Strong electrolyte
Weak electrolyte
What is the conductivity classification for an unknown E that dissolves completely in water and only has a few of ions present in solution? Select one: Nonelectrolyte Weak electrolyte Strong electrolyte
Weak electrolyte
describes two liquids that do not mix completely and form layers ex. olive oil & vinegar)
What does the term immiscible mean?
the ability of a solid or gas to dissolve in a liquid ex. sugar & coffee
What does the term solubility mean?
the substance being dissolved ex. the sugar
What is a solute?
the solution that does the dissolving ex. the coffee
What is a solvent?
0.91 g/ml solution: mass(g) ÷ volume(mL) 2.98 g ÷ 3.29 mL = 0.91 g/mL
What is the density of a liquid that occupies a volume of 3.29 mL and weighs 2.98 g? Select one: 0.31 g/mL 0.91 g/mL 1.10 g/mL 1.30 g/mL
Consider an experiment where 4.55 mL of an unknown H₂O₂(aq) solution reacted with the yeast at 19.1°C to produce 98.10 mL of gas. The barometric pressure was 786.1 torr. The vapor pressure of H₂O is 16.5 torr at that temperature. Be sure to look at the units of the numbers when selecting your answers. What is the partial pressure of O₂ (in atm) in the collected gas? How many moles of O₂ were produced by the reaction? How many moles of H₂O₂ reacted to produce this amount of O₂? What is the Molarity of the H₂O₂ solution?
What is the partial pressure of O2 (in atm) in the collected gas? 1.0126 atm O2 How many moles of O2 were produced by the reaction? 0.004142 mol O2 How many moles of H2O2 reacted to produce this amount of O2 ? 0.008284 mol H2O2 What is the Molarity of the H2O2 solution? 1.82 M H2O2
Cooper and Zach did this experiment dissolving 3.10 g of Cu(C₂H₃O₂)₂·H₂O and adding 6.93 g NaC₇H₄SO₃N·H₂O. They recovered 6.28 g of product. In their calculations, they found that 8.32 g of product should be produced. Their calculation of (6.28 g / 8.32 g) x 100% is equal to 75.5%. What is the percent yield? What is the actual yield? What is the theoretical yield?
What is the percent yield? 75.5 What is the actual yield? 6.28 What is the theoretical yield? 8.32 (solution: you're simply identifying which calculations are which - there is no math required for this question)
When everyone in the lab has put away all chemicals and glassware.
When is it okay to remove your chemical splash proof goggles? Select one: When the goggles fog up When washing glassware When not working with acids or bases When everyone in the lab has put away all chemicals and glassware.
Consider the data below. Liquid A is immiscible with Liquids B & C. Liquid B is immiscible with Liquid C. The liquids were poured into the beaker in order of increasing density. beaker with 3 colored layers Select the appropriate label for each layer. Liquid A: 28.1 g, 20.0 mL Liquid B: 21.8 g, 20.0 mL Liquid C: 9.2 g, 10.0 mL Which chemical species is in the bottom layer? Which chemical species is in the middle layer? Which chemical species is in the top layer?
Which chemical species is in the bottom layer? Liquid A Which chemical species is in the middle layer? Liquid B Which chemical species is in the top layer? Liquid C
T-shirt Scrub pants Jeans Lab Apron
Which of the following items of clothing are appropriate to wear in CHEM 1212 Lab to protect your skin ? Select all that apply. Select one or more: T-shirt Scrub pants Jeans Lab Apron Knee length shorts Leggings Tank top Midriff top
Which definition corresponds to aliquot? Select one: a. A small portion of a chemical species. b. The amount of product that could be produced using a given amount of reactant. c. The reactant that is completely consumed during the reaction. d. A chemical species that contains a metal ion that is coordinated to ligands. e. The reactant that is present as extra material when the reaction is complete.
a. A small portion of a chemical species.
Which statement best explains what happens to the percent yield if the crystals (product) were washed with hot water? Select one: a. The actual yield will decrease since part of the product will dissolve and not be recovered as a solid. Since actual yield is in the numerator, the percent yield will also decrease. b. The actual yield will increase since the original amount of product will now absorb water. Since the actual yield is in the numerator, the percent yield will also increase. c. The actual yield will decline since part of the product will dissolve and not be recovered as a solid. But since the actual yield is in the denominator, the percent yield will increase. d. The actual yield will increase since the original amount of product will now absorb water. And since the actual yield is in the denominator, the percent yield will decrease.
a. The actual yield will decrease since part of the product will dissolve and not be recovered as a solid. Since actual yield is in the numerator, the percent yield will also decrease.
Which definition corresponds to excess reagent? Select one: a. The reactant that is present as extra material when the reaction is complete. b. The reactant that is completely consumed during the reaction. c. A chemical species that contains a metal ion that is coordinated to ligands. d. A small portion of a chemical species. e. The amount of product that could be produced using a given amount of reactant.
a. The reactant that is present as extra material when the reaction is complete.
What should you do if some NaOH comes into contact with your skin? Select one: a. Wash the skin for 15 minutes with water and notify the TA. b. Put a bandage (from the first aid kit) on the skin and notify the TA. c. Wash your eyes for 15 minutes with water from the eyewash and notify the TA. d. Dry the skin with a chemical absorbent pad from the chemical spill kit.
a. Wash the skin for 15 minutes with water and notify the TA.
What should you do if some of the H₂O₂-H₂SO₄ solution comes into contact with your skin? Select one: a. Wash the skin for 15 minutes with water and notify the TA. b. Wash your eyes for 15 minutes with water from the eyewash and notify the TA. c. Put a bandage (from the first aid kit) on the skin and notify the TA. d. Dry the skin with a chemical absorbent pad from the chemical spill kit.
a. Wash the skin for 15 minutes with water and notify the TA.
Which statements explain the shape of the titration curve observed in lab? I. Initially, the conductivity is zero since the reaction has not begun. II. During the reaction, the conductivity decreases due to the increase of Ba+2 and OH- ions. III. At the equivalent point, the conductivity is nearly zero since all the Ba+2 and OH- ions have reacted to produce nonelectrolyte products. IV. After the endpoint, the conductivity increases due to the excess titrant. Select one or more: a. I & II are TRUE. b. III and IV are TRUE. c. II & III are TRUE. d. I & III are TRUE.
b. III and IV are TRUE.
how can you classify a compound?
by either their conductivity/ionization - the more ions a solution has the better its conductivity will be
What will result when liquid W (density = 0.75 g/mL), liquid S (density = 0.65 g/mL, immiscible with liquid W), and liquid T (density = 0.85 g/mL, miscible with liquid W and immiscible with liquid S) are poured into test tube? Select one: a. Two layers will form in the test tube. The top layer is a mixture of liquid S & W and the bottom layer is liquid T. b. Two layers will form in the test tube. The top layer is a mixture of liquid T & W and the bottom layer is a liquid S. c. Two layers will form in the test tube. The top layer is liquid S and the bottom layer is a mixture of liquid W & T. d. Three layers will form in the test tube. The top layer is liquid S, the middle layer is liquid T and the bottom layer is liquid W. e. Three layers will form in the test tube. The top layer is liquid T, the middle layer is liquid W and the bottom layer is liquid S.
c. Two layers will form in the test tube. The top layer is liquid S and the bottom layer is a mixture of liquid W & T.
From the description of the reaction below, balance the equation and also select the classification of the reaction. When silver Mg(s) is ignited in the presence of O₂(g), it forms a white, powdery solid, MgO(s). __ Mg(s) + __ O₂(g) → __ MgO(s) coefficient in front of MgO(s) coefficient in front of O₂(g) coefficient in front of Mg(s) Classification of reaction
coefficient in front of MgO(s) - 2 coefficient in front of O₂(g) - 1 coefficient in front of Mg(s) - 2 Classification of reaction - Combination
What solution color will be observed initially in the Erlenmeyer flask?
colorless
how do you determine the dilute concentration of an analyte in an unknown?
concentration dilute = (concentration stock x volume stock) ÷ volume dilute
Which definition corresponds to an analyte? Select one: a. A chemical species that has a pH value < 7.0 b. A chemical species that changes colors at different pH values. c. The chemical species being added in a titration. d. The chemical species of interest.
d. The chemical species of interest.
density is a ________ property.
density is a PHYSICAL property
to go from g → moles
divide by molar mass
Which definition corresponds to theoretical yield? Select one: a. A small portion of a chemical species. b. The reactant that is completely consumed during the reaction. c. The reactant that is present as extra material when the reaction is complete. d. A chemical species that contains a metal ion that is coordinated to ligands. e. The amount of product that could be produced using a given amount of reactant.
e. The amount of product that could be produced using a given amount of reactant. The theoretical yield is the amount of product that could be produced from the reactant if the reaction completely converts reactants into products.
How do you calculate the percent composition (by mass) of H2O2 from volume of O2(g)?
find the partial pressure of O2 using dalton's law....find moles O₂ using ideal gas law.......find moles/Molarity H₂O₂ that produced said O₂ using the stoichiometry.....finally, convert Molarity to percent using H₂O₂ molar mass.
how do you find the molarity of an unknown solution from titration data?
first, find the moles of Ba(OH)₂ by using the Molarity and mL of H₂SO₄ and stoichiometry....then, find Molarity moles of Ba(OH)₂ = M H₂SO₄ x mL H₂SO₄ x (1 L ÷ 1000 mL) x (1 mole Ba(OH)₂ ÷ 1 mole H₂SO₄) M of Ba(OH)₂ = (mol Ba(OH)₂) x (1 ÷ mL Ba(OH)₂) x (1000 mL ÷ 1 L)
In this experiment, you will measure the volume of the oxygen. Before doing calculations, the volume must be converted to what unit of measure? Select one: milliliter Celsius Kelvin liter
liter
It required 16.24 mL of DCIP to titrate 5.00 mL of lime juice to its end point. Molarity (M) of DCIP is 1.11 * 10⁻³. Calculate how many mg of ascorbic acid is present in the juice. How many mL of juice is needed to satisfy the DRI (90 mg of C₆H₈O₆) of Vitamin C?
mg C₆H₈O₆ = M DCIP x L DCIP x (1 mol C₆H₈O₆ ÷ 1 mol DCIP) x (176.14 g C₆H₈O₆ ÷ 1 mol C₆H₈O₆) x (1000 mg / 1 g) mg C₆H₈O₆ = (1.11 * 10⁻³) x 0.01624 x (1 mol C₆H₈O₆ ÷ 1 mol DCIP) x (176.14 g C₆H₈O₆ ÷ 1 mol C₆H₈O₆) x (1000 mg ÷ 1 g) mg C₆H₈O₆ = 3.175 mg mL juice = DRI mg C₆H₈O₆ x (mL juice ÷ mg C₆H₈O₆) mL juice = 90 mg C₆H₈O₆ x (5 mL juice ÷ 3.175 mg C₆H₈O₆) ***mL juice = 141.7 mL → 142 mL
In this experiment, you will be using and measuring various chemical species by a variety of units. Which units of measure is used for the volume of O₂(g)? Select one: torr grams milliliters Kelvin
milliliters
miscibility is a _____ property
miscibility is a CHEMICAL property
In this experiment, you will be using the gas constant, R. Which unit is NOT found in R? Select one: atmosphere molarity Kelvin liter
molarity
What is the Molarity of a Ba(OH)₂ solution that took 8.20 mL of 0.100 M H₂SO₄ solution to reach the equivalence point when titrating 10.0 mL of the Ba(OH)₂ solution? (See balanced reaction below.) Ba(OH)₂(aq) + H₂SO₄(aq) → BaSO₄(s) + 2 H₂O(ℓ)
moles Ba(OH)₂ = M H₂SO₄ x mL H₂SO₄ x (1 L ÷ 1000 mL) x (1 mole Ba(OH)₂ ÷ 1 mole H₂SO₄) moles Ba(OH)₂ = 0.100 M H₂SO₄ x 8.20 mL H₂SO₄ x (1 L ÷ 1000 mL) x (1 mole Ba(OH)₂ ÷ 1 mole H₂SO₄) moles Ba(OH)₂ = .00082 mole M Ba(OH)₂ = (mol Ba(OH)₂) x (1 ÷ mL Ba(OH)₂) x (1000 mL ÷ 1 L) M Ba(OH)₂ = .00082 mol Ba(OH)₂ x (1 ÷ 10 mL Ba(OH)₂) x (1000 mL ÷ 1 L) M Ba(OH)₂ = .082 M
to go from moles → g
multiply by molar mass
Problem After 2.05 g of NiCl2·6 H2O reacted with 7.50 mL of the 4.0 M C₂H₈N₂ (ethylenediamine), 1.87 g of the product, [Ni(en)3]Cl2 was obtained. The filtrate was pale pink and when tested with pH paper, the paper indicated a basic solution. How would you calculate the percent yield of [Ni(en)₃]Cl₂ if the theoretical yield was calculated to be 2.67 g [Ni(en)₃]Cl₂?
percent yield = (actual yield ÷ theoretical yield) x 100% percent yeild = (1.87 g [Ni(en)3]Cl2 ÷ 2.67 g Ni(en)₃]Cl₂) x 100% = 75.7%
Density is an example of a ________ property.
physical property
the densities of the liquids (________ property) and to determine the solubility of a solid in several liquids (__________ property) and the miscibility of the liquids with each other (_________ property).
the densities of the liquids (PHYSICAL property) and to determine the solubility of a solid in several liquids (CHEMICAL property) and the miscibility of the liquids with each other (CHEMICAL property).
what are the calculations of the theoretical yield based on?
the limiting reagent
describes two liquids that mix completely with each other ex. cream & coffee
what does the term miscible mean?
The general classifications are described with both a generic reaction and an example of a specific reaction(s).
• Combinations: Two elements or species come together to form one compound. A + B → AB 2Mg(s) + O₂(g) → 2MgO(s) • Decomposition: One species decomposes into two or more chemical species, generally as a result of heating. ABC→ AB + C CaCO₃(s) → CaO(s) + CO₂(g) • Single Displacement: One element (B) displaces an atom (A) in a compound (AC) to form a new compound (BC), while the displaced atom (A) is converted to its elemental form. AC + B → A + BC Cd(NO₃)₂(aq) + Zn(s) → Zn(NO₃)₂(aq) + Cd(s) • Double Displacement: AC + BD → AD + BC Cd(NO₃)₂(aq) + (NH₄)₂S(aq) → CdS(s) + 2 NH₄NO₃(aq) Zn(OH)₂(s) + 2HCl(aq) → ZnCl₂(aq) + 2 H₂O(ℓ) • Oxidation-Reduction: Two half-reactions added together. A → A⁺ⁿ + n e⁻ (loss of electrons-oxidation) B + me⁻ → B⁻ᵐ (gain of electrons-reduction) ------------ A + B → A⁺ⁿ + B⁻ᵐ Mg → Mg⁺² + 2e⁻ (oxidation) Co⁺² + 2e⁻ → Co (reduction) ------------- Mg + Co⁺² → Mg⁺² + Co ***In oxidation half reaction, one species (A) will lose electrons. Then these electrons are used in the reduction half reaction, where the other species (B) gains electrons. The number of electrons transferred must be the same in each half reaction.