60+9.53)/3 = 9.66mL Sample Calculations: (9.85+9.60+9.53)/3 = 9.66mL The average volume of NaOH used. Calculations: 1. Moles NaOH = M x V = (0.1M) (0.00966L) = 0.000966 moles 2. Moles HCl = moles NaOH 0.000966 moles -> 9.66x10-4 3. NaOH + HCl = NaCl + H2O Moles NaOH = M x V = (0.1M)(0.00966L) = 0.000966 moles Moles HCl = moles NaOH [HCl] = moles/volumes = (0.000966)/ (0.0096L) [HCl] = 0.1M Follow-up Questions: 1. It will have no effect because the phenolphthalein
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Experiment 9 Aim A. To investigate the existence of hydrogen bonds between ethanol molecules. B. To measure the strength of hydrogen bond formed between ethanol molecules C. To investigate the formation of hydrogen bonds between molecules of ethyl ethanoate and trichloromethane. D. To measure the strength of hydrogen bond formed between molecules of ethyl ethanoate and trichloromethane. Procedure A. 1. 10 cm3 of ethanol was added into an insulated 50 cm3 beaker by
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II - Mole Calculations/ Limiting and Excess Reagent – Lecture Notes 1. Given the balanced equation N2(g) + 3H2(g) 2NH3(g) How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with hydrogen? 2. Given the equation: SiO2 + HF SiF4 + H2O a. Calculate the number of moles HF that would completely react with 2.5 moles of SiO2. b. Calculate the number of moles SiF4 formed by completely reacting 2.5 moles SiO 2 with HF. c. Calculate the mass of water formed by reacting 2.5 moles SiO
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are written with water separate by a dot (.). This (.) means for hydrated metal salt there are how many number of moles of water bound to each mole of that metal salt. Research Question: The aim of this experiment is to calculate the number of moles of water bounded to each mole of cobalt chloride ( CoCl2. H2O). This cobalt chloride hydrated may be monohydrate with 1 mole of water attracted to cobalt chloride. It may be dihydrate‚ trihydrate‚ tetrahydrate or pentahydrate; your task is
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50‚ 1.35 x 10‒5 and 8.70 moles respectively in a 12-L flask. Calculate the equilibrium constant 7 Kc. (1.08 x 10 ) 2. The number of moles of H2‚ O2‚ and H2O present at equilibrium for the reaction: 3. The equilibrium constant Kc for the reaction: 2HCl(g) ⇄ H2(g) + Cl2(g) is 0.0213 at 400 C. If 20.0 o moles of HCl(g) are heated at 400 C‚ what amounts of HCl(g)‚ H2(g) and Cl2(g) would be present in the equilibrium mixture? (H2 = Cl2 = 2.26 moles; HCl = 15.48 moles) o 4. The equilibrium
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Number | MS204-S/03 | Manufacturer | METTLE TOLEDO | Type | New Classic MF | Boiling chips detail:- VWR scientific‚ Inc. Porous boiling chips VWR cat. # 26397 - 409 Given sample: - 2-propanol (CH3CHOHCH3) Molecular weight: - 60.08 gm/mole Sample volume taken: - 2 mL | Flask 1 | Flask 2 | Flask 3 | Mass of flask and condensed vapour | 77.8735 gm | 77.8768 gm | 77.8750 gm | Mass of empty flask apparatus | 77.6050 gm | 77.6059 gm | 77.6055 gm | Mass of condensed vapour | 0.2685
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C2H3O2NO2 2/3 Co(acac)3 + 2 C2H3O2NO2 ( 2/3 Co(acac-NO2)3 + 2 C2H3O2 • Moles of reagent used - Cu(NO3)2·3H2O‚ MW: 241.59 g/mol [pic] - Acetic anhydride‚ MW: 102.1 g/mol‚ D: 1.080 g/mL [pic] - Co(acac)3‚ MW: 356.24 g/mol [pic] • Limiting Reagent From the stoichiometry of the reactions‚ acetic anhydride is in excess‚ and 2/3 (0.67) mole Co(acac)3 should reaction with 1 mole copper nitrate trihydrate. The actual mole ratio when 0.49 g Co(acac)3
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Formula: Number of Moles = Concentration x Volume Therefore‚ No of Kl Moles = Concentration x Volume = 0.30 x 0.01 ± 0.0005 = 0.003mol ± 0.000015mol No. of Pb(NO3)2 Moles = Concentration x Volume = 0.50 x 0.01 ± 0.0005 = 0.005mol ± 0.000025mol Also‚ The Theoretical Mole Ratio would be: 21=2 The Experimental Mole Ratio would be: 0.0030
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chlorine gas‚ Cl2‚ is 70.9 g/mol‚ what mass of chlorine gas contains 0.652 moles of chlorine gas? Question 5 options: 109 g 92.4 g 46.2 g 23.1 g 9.20 g Save Question 6 (2.7 points) What is the molar mass of dinitrogen pentoxide? Question 6 options: 108 g/mol 30 g/mol 150 g/mol 94 g/mol 44 g/mol Save Question 7 (2.7 points) How many nitrogen atoms are present in 3.00 × 10–3 moles of nitrogen gas‚ N2? Question 7 options: 1.81 × 1019 3.61 ×
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glycolitic enzymes during carbohydrate metabolism. Hydrogen is released and glucose is metabolized to pyruvic acid. In the presence of oxygen‚ the pyruvic acid is converted into acetyl coenzyme A. (Acetyl CoA). 1 mole of glucose produce 2 moles of ATP or 1 mole of glycogen produces 3 moles of ATP. * In the Kreb Cycle system‚ once the acetyl CoA is formed‚ it enters the krebs cycle‚ a complex series of chemical reactions that permit the complete oxidation of acetyl coenzyme A. in this process‚
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