Title: Spectrophotometric analysis of a two-component mixture Aim: i. To prepare working standards of dichromate and permanganate ii. To measure the absorbance of the prepared working standards of dichromate and permanganate using a spectrophototometer iii. To determine the concentrations of permanganate in a mixture of unknown. Abstract: Working standards of dichromate and permanganate were prepared and absorbance for each found. This was done in order to plot
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Determination of the Composition of Cobalt Oxalate Hydrate Experiment 12 Robbie Kinsey Partner: Debnil Chowdhury Chem. 1312-D TA’s: Russell Dondero & Sylvester Mosley February 9‚ 2000 Purpose The purpose of this lab was to determine the percent cobalt and oxalate by mass‚ and with that information‚ the empirical formula for cobalt oxalate hydrate‚ using the general formula Coa(C2O4)b.cH2O. Procedure The powdered cobalt oxalate
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Equation HX + NaOH NaX + H2O 1 mole 1 mole 1 mole 1 mole Aim To analyze the Vitamin C from a rival company and compare with that of my company and find out the best value for the money spent by the consumer. Controlled Variable * Mass of the tablet and hereby vitamin C in rival company tablets * Consistency of components in the Vitamin C tablet * Concentration of NaOH Apparatus 1. Burette (50ml) 2. Conical flask (50ml) 3. Vitamin C tablet (1 tablet
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CHEMISTRY (Theory) Time allowed : 3 hours General Instructions: (i) All questions are compulsory. Maximum Marks : 70 (ii) Marks for each question are indicated against it. (iii) Question numbers 1 to 8 are very short-answer questions and carry 1 mark each. (iv) Question numbers 9 to 18 are short-answer questions and carry 2 marks each. (v) Question numbers 19 to 27 are also short-answer questions and carry 3 marks each. (vi) Question numbers 28 to 30 are long-answer questions and carry 5 marks
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13.80 mL x = 0.01380 L Moles of potassium acid phthalate = = = 0.000735 Moles of NaOH solution = Moles of (HKC8H4O4) x = 0.000735 mol x = 0.000735 mol Molarity (NaOH) = = = 0.053 M TRIAL II Volume of NaOH solution = Final reading of buret - Initial reading of buret =27.20 mL– 13.80 mL = 13.40 mL Converting the Volume (mL) to Volume (L) Volume (liters) = Volume (mL) x = 13.40 mL x = 0.01340 L Moles of potassium acid phthalate
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Last Name ____________________First Name _______________________ Ignore: 5‚ 7‚ 19‚ 30 1. | How many grams of CaCl2 (molar mass = 111.0 g/mol) are needed to prepare 4.44 L of 0.500 M CaCl2 solution? | | A) 369 g B) 271 g C) 258 g D) 296 g E) 246 g | 2. | An aqueous solution of ammonium sulfate is allowed to react with an aqueous solution of lead(II) nitrate.The complete ionic equation contains which of the following species (when balanced in standard form)? | | A)
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water = (Mass of hydrate) – (mass of dehydrate) Mass of water = (answer to #1) – (answer to #3) Mass of water = (5.000g) – (4.500g) Mass of water = 0.500g 5. Convert the mass of water to moles of water. To do this‚ we need the molar mass (from 2.04 and 3.09-molar mass is the mass‚ in grams‚ of one mole of your
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9-12 Unit 7 II: Unit Title: Mole Concept III. Unit Length: 7 days (on a 90 min. per day block schedule) IV. Major Learning Outcomes: Students should be able to: Mole Concept • Calculate formula mass. • Convert representative particles to moles and moles to representative particles. (Representative particles are atoms‚ molecules‚ formula units‚ and ions.) • Convert mass of atoms‚ molecules‚ and compounds to moles and moles of atoms‚ molecules‚ and compounds
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appropriate response. 1. What is a mole ratio? 2. What piece of information is needed to convert grams of a compound to moles of the same compound? 3. Phosphine (PH3) can be prepared by the hydrolysis of calcium phosphide‚ Ca3P2: Ca3P2 + 6 H2O 3 Ca (OH)2 + 2 PH3 a) One mole of Ca3P2 produces 2 mol of PH3. b) One gram of Ca3P2 produces 2 g of PH3. c) 3 moles of Ca(OH)2 are produced for each 2 mol of PH3 produced. d) The mole ratio between phosphine and calcium
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corrodes and rust is a byproduct. (16 pts.) 5. Determine the mass of iron consumed in the reaction. Show your work. (16 pts.) 7.75-5.5g=2.25g 6. Calculate the number of moles of iron consumed. Show your work. (16 pts.) 2.25G X 1MOL FE/55.85g= 0.040286 = 0.040286=0.0403 moles Fe Formula: Mass Fe used X 1 mol Fe = Moles of iron used Molar mass of Fe 7. Determine the mass of product formed. Show your work. (16 pts.) 3.52g-.086g=2.66g Formula: (Mass of
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