needed to form the I3- complex. In this type of analysis‚ excess iodide is added to the oxidizing agent‚ and the triiodine is titrated with stand thiosulfate. This indirect analysis finds the number of moles of ascorbic acid based on the known number of moles of IO3- and subtracting half the amount of moles of the thiosulfate solution. II: Equations: Iodate with Iodide: IO3- + 8I- + 6H+ 3I3- + 3H2O Thiosulfate with Triiodide 2S2O32- + I3- S4O62- + 3I- Ascorbic Acid with Triiodide
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Mg2+ (aq) + 2Cl- (aq). The major findings of this experiment were that there was a chemical reaction between hydrochloric acid and magnesium‚ and the reaction produced the hydrogen gas. The results also indicated how many moles of hydrogen gas were equal to the amount of moles of magnesium consumed. Procedure Step 1:
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concentration = moles / volume 0.10=moles/0.010 Moles of Cu (NO3)2 (aq) = 0.001 moles Mass of empty bottle = 6.00grams Mass of empty bottle +copper metal =6.05grams Mass of copper metal recovered after the experiment = 0.050 grams Finding moles of copper: Moles = mass/ Mr = 0.050 / 63.55 =0.00079 moles
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Processing Raw Data : Part 1 Equation of the reaction; Mg(s) + 2HCl(ℓ) → MgCl2(aq) + H2(g) no of moles : no of moles of Mg = mass/molar mass = 0.31g___ 24.31g/mol = 0.01275 moles no of moles of HCl = molarity x volume = 1.0M x (25/1000)mL = 0.025 moles heat of reaction will be 0.025 moles as HCL act as the limiting reactant. Enthalpy change of the reaction‚H2 Volume of HCL solution =
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rules by a divine being. Example: Iran 26. Laissez-faire - "let do" example: America 27. Pragmatism - being practical and someone who tries to each their goals. Example: people who follow their dreams 28. Tea party - US political party. Example: Adrian Smith13. Radical - favors social reform. Example: KKK 14. Reactionary - wants change‚ but also wants to retreat to the past. Example: Mohamed Siyaad Barre 15. Conservative - doesn’t want change. Example: being against gay marriage 16. Liberal -
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204.22 g/mol No. of moles of KHP = Mass of KHP used / Molar mass = 0.42 g / 204.22 g/mol = 0.0021 moles Concentration of NaOH = No. of moles / Volume = [0.0021 mol / {(22.50 + 25) / 1000} L] * 100 = 4.4 M Trial 2 Mass of KHP transferred = 0.4139 g Volume of Distilled water = 25 mL Volume of NaOH used = 22.80 mL Molar mass of KHP = 204.22 g/mol No. of moles of KHP = Mass of KHP used / Molar mass = 0.4139 g / 204.22 g/mol = 0.0020267 moles Concentration of NaOH = No. of moles / Volume = [0.0020267
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Stoichiometry of a Precipitation Reaction March 20‚2013 Amber McCollum Introduction Stoichiometry is a branch of chemistry that deals with the quantitative relationships that exist among the reactants and products in chemical reactions To predict the amount of product produced in a precipitation reaction using stoichiometry‚ accurately measure the reactants and products of the reaction‚ determine the actual yield vs. the theoretical yield and to calculate the percent yield. The equation
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and good-natured Mole loses patience with spring cleaning. He flees his underground home‚ heading up to take in the air. He ends up at the river‚ which he has never seen before. Here he meets Ratty (a water rat)‚ who at this time of year spends all his days in‚ on and close by the river. Rat takes Mole for a ride in his rowing boat. They get along well and spend many more days boating‚ with Rat teaching Mole the ways of the river. One summer day shortly thereafter‚ Rat and Mole find themselves near
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Baroque Orchestra and McGill Cappella Antica I attended the concert of McGill Baroque Orchestra and McGill Cappella Antica on Wednesday‚ February 19‚ 2014‚ at 7:30 p.m. What special about this concert was the guest conductor and solo violinist Adrian Butterfield. The performed pieces were Welcome to all the pleasures‚ Leclair’s Violin Concerto in A major‚ Locatelli’s Introduzione teatrale in G major‚ C. P. E. Bach’s Sinfonia in C major and My heart is inditing. The venue was Redpath Hall of McGill
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solvent kg of solvent = 10 mL x 1 g/mL x 1kg/1000g = 0.01 kg molality = moles of solute / kg solvent moles of solute = molality x kg solvent Isopropyl alcohol trial 1 moles of solute = 5.91 m x 0.01 kg = 0.0591 moles Isopropyl alcohol trial 2 moles of solute = 5.38 m x 0.01 kg = 0.0538 moles Ethyl alcohol trial 1 moles of solute = 4.30 m x 0.01 kg = 0.0430 moles Ethyl alcohol trial 2 moles of solute = 3.76 m x 0.01 kg = 0.0376
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