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- How many grams of Ca3(P04)2 will dissolve in a) 2.0 L of pure water? b) 500 mL of 1.0M Ca(NO3)2? Ksp for Ca3(PO4)2 = 2.0 x 10-29ERIMENTAL GENERAL CHEMISTRY /ill Q Time left 0:19:44 What is the molar concentration of NaCIO in 6.00% bleach? (molar mass of NaCIO = 74.5 g/mol, density of bleach solution= 1.08 g/mL). C Fi t of O a. 1.16 M O b. 1.01 M O c. 1.30 M O d. 0.725 M O e. 0.870 M Next page7:36 .l LTE 7 Unanswered •1 attempt left • Due on Oct 6, 7:00 PM 5.0 mL of 3.6 M K2CO3 solution is diluted to make a 0.45 M solution, what is the volume of the diluted solution? А 5.0 mL 10. mL C 20. mL D 40. mL 80. mL Submit B
- -2 The PPm Concentration of 4.5 *la M Solution off Lif is I1940fやPm D 196. PPm 3 1950 PPm7:36 .l LTE Unanswered • 1 attempt left • Due on Oct 6, 7:00 PM What volume of a 0.300 M LiOH is required to neutralize 48.0 mL of a 0.200 M H2SO4 solution? А 16.0 mL 32.0 mL C 64.0 mL D 96.0 mL 128. mL Submit B4. What was the molarity of a volumetric solution of disodium edetate if 28.50 mL of the solution were consumed by 185 mg of calcium carbonate? (MW CaCO3 = 100.09 g/mole) A. 0.04 M B. 0.05 M C. 0.06 M D. 0.07 M E. None of the choices 5. Calculate the weight necessary to prepare 1L of 0.05 N Potassium Permanganate (MW: 158.034 g/mol). A. 1.58 g B. 3.16 g C. 3.95 g D. 7.90 g E. None of the choices
- In order to form Ba(IO3)2, 500 mL of 0.5000 M Ba(NO3)2 was mixed with 500 mL of 0.0500 M NaIO3. Ksp= 1.57x10^-9How many millimoles of Ba(NO3)2 are needed to completely react with NaIO3?What is the limiting reagent?What is the excess reagent?In order to form Ba(IO3)2, 500 mL of 0.5000 M Ba(NO3)2 was mixed with 500 mL of 0.0500 M NaIO3. Ksp= 1.57x10^-9 How many millimoles of Ba(NO3)2 are needed to completely react with NaIO3? 500 mmol 250 mmol 25 mmol 12.5 mmol What is the concentration of the excess reagent? 0.1125 M 0.3219 M 0.1195 M 0.2375 M What is the molar solubility of Ba(IO3)2 in this solution? 04x10^-5 M 07x10^-5 M 15x10^-5 M 30x10^-5 M3. Which of the following statement/s is/are correct? 1. 6.00 N H2CO3 is equivalent to 3.00M H2CO II. 5.29 % (w/v) H2SO4 is equivalent to 0.539 N HSO4 II. 1.15 N Mg(OH)2 is equivalent to 16.3 g Mg(OH)2 in 500 ml solution. (MH2504 98.086 g/mol; Miacos 62.026 g/mol; MOH2 58.326 g/mol) O l and IlI O l only O l and II Oland I
- A gas mixture contains 3 ppmv of CO₂. At 25C and latm, determine the μα concentration of CO₂ inA 0.2985 g sample of an antibiotic powder was dissolved in HCl and the solution diluted to 100.0 mL. A 20.00 ml aliquot was transferred to a flask and followed by 25.00 ml of 0.01886 M KBRO3. An excess of KBr was added to form Br2, and the flask was stoppered. After 10 min, during which time the Br2 brominated the sulfanilamide, an excess of KI was added. The liberated iodine titrated with 12.88 ml of 0.1235 M sodium thiosulfate. Calculate the percent sulfanilamide (NH2C,H4SO2NH2) in the powder. Bro, + 5Br + 6H* 3Br, + 3H,0 NH2 NH2 Br. Br + 2B12 + 2H++ 2Br SO,NH2 SO,NH2 sulfanilamide Br, + 51 2Br + 12 excess 2 + 25,0,- 25,0,2 + 21- - MM: NH2C6H4SO2NH2 = 172.21 KBRO3 = 167.00 KBr 119.00The thiourea in a 1.455 g sample of organic material was extracted into a dilute sulfuric acid solution and titrated with 37.31 mL of 0.009372 M Hg2+ via reaction: 4(NH2)2CS + Hg2+ →[(NH2)2CS]4 Hg2+ P.S. Answer only the last two letters of the following questions. (Only C and D) a. Is this an example of total analysis technique or concentration technique? Explain. b. Calculate the percent (NH2)2CS ( 76.12 g/mol) in the sample. c. What is classification of the analysis based on the amount of sample and amount of analytes present? Explain. d. If the true value is 10.00%, calculate the absolute and relative error.