What is the theoretical yield of isopentyl acetate if 0.50mL of 10.0M acetic acid reacts with 0.85mL of isopentanol? CH3CO₂H (aq) + HOCH₂CH₂CH(CH3)2(aq) → CH3CO₂CH₂CH₂CH(CH3)2(aq) + H₂O(1) Acetic acid isopentanol isopentyl acetate
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- 10.06 cm³ of the stock 0.100 mol dm-3 solution of NaOH was required to fully neutralise 49.5 mg of a diprotic acid (H2A). Write a balanced equation for this reaction and hence determine the molar mass (Mr) of the acid.H2SO4 (aq) + 2 NaOH (aq) Na2SO4 (aq) + 2 H2O (l) calculate the molarity of the H2SO4 solution if 14.92 mL of NaOH was necessary to reach the endpoint of a titration. The molarity of the NaOH solution was 0.63 M and 25.18 mL of H2SO4 was added to the Erlenmeyer flask.(a) Find the pH of a 1.00 L solution prepared with 12.43 g of base (CH2OH)3CNH2 (MW =121.135 g/mol) plus 4.67 g of its conjugate acid (CH2OH)3CNH3Cl (MW = 157.596 g/mol).(b) How many mL of 0.500 M NaOH should be added to 10.0 g of (CH2OH)3CNH3Cl to give a pH of 7.60in a final volume of 250 mLConsider the pKa of (CH2OH)3CNH3+ = 8.072
- Part A A solution of 0.158 M NaOH is used to titrate 23.5 mL of a solution of H,SO4: H2SO4 (aq) + 2N2OH(aq) → 2H2O(1) + NazSO. (aq) If 39.4 mL of the NaOH solution is required to reach the endpoint, what is the molarity of the H2SO4 solution? Express your answer with the appropriate units. HA ? molarity = Value Units Submit Request Answer Previous al Draft Speec...pdf MacBook AirWhat volume of carbon dioxide gas at 0.980 atm and 293 K could be made from reacting 10.0 grams of sodium carbonate with 100. mL of 6.0 M HCl? This is a limiting reactant problem.Given: Na2CO3 (s) + 2 HCl (aq) -> 2 NaCl (aq) + H2O (g) + CO2 (g)A How many mL of 0.40 M NaOH would be fequired to titrate 22.50 mL of 0.250 M H2C2O4 to the equivalence point? The balanced chemical equation has been provided. (3 points) H2C204 (aq) + 2 NaOH (aq) Na2C204 (aq) + 2 H2O (1) 22.50 mL 0.250 M 0.40 M
- What is the [CH3CH2CH2NH3+] of a solution which is 0.1710 M in CH3CH2CH2NH2 and 0.1550 M in CsOH? CH3CH2CH2NH2(aq) + H2O(l) → CH3CH2CH2NH3+(aq) + OH−(aq) Kb = 5.100e-4cedric and 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 24.83ml. H3C6H5O7(aq) + 3 NaOH(aq) yeilds Na3C6H5O7(aq) + 3 H2O)(l) The volume of NaOH titrated is 23.79ml 24.83ml - 01.04ml = 23.79ml of NaOH ***What is the mass of citric acid in the juice sample? 0.204g of H3C6H5O7 ( can you please show how to calculate this answer)Adding acid to the buffer, NH3-NH4*, will produce this (net ionic) reaction: H*(aq) +OH(aq)=H₂O(1) H*(aq)+NH4*(aq) NH3(aq)+H₂(g) O H*(aq) + NH4+ (aq)=NH5²+ (aq) O H*(aq)+NH3(aq)=NH4+ (aq)
- An unkown solid acid is either citric acid or tartaric acid. To determine which acid you have, you titrate a sample of the solid with NaOH. The appropriate reactions are as follows: Citric acid: H3C6H5O7 (aq) + 3 NaOH (aq) --> 3 H2O (l) + Na3C6H5O7 (aq) Tartaric acid: H2C4H4O6 (aq) + 2 NaOH (aq) --> 2 H2O (l) + Na2C4H4O6 (aq) A 0.956 g sample requires 29.1 ml of 0.513 M NaOH for titration to the equivalence point. What is the unknown acid?How would you make 500 mL of 0.100 M NaOH(aq) solution? How would you standardize this NaOH solution? Given: Solid NaOH 100 M HCl(aq) Solid KHP Digital BuretteWhat volume of 0.25M NaOH solution is needed to react with 50mL of 0.125M H2SO4? Reaction:2NaOH(aq)+H2SO4(aq)>Na2SO4(aq)+H2O(1)