Question 6 The % w/w in a 0.6712-g sample was determined by Volhard titration. After adding 50.00 mL of 0.05619 M AgNO3 and allowing the precipitate to form, the remaining silver was back titrated with 0.05322 M KSCN, requiring 35.14 mL to reach endpoint. Agl (s) + NO3 AgNO3 + Ag+SCN-AgSCN How many millimoles of AgNO3 reacted with ? O 17.76 O 0.9400 119.3 O0.6860
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- The % w/w I– in a 0.6712-g sample was determined by Volhard titration. After adding 50.00 mL of 0.05619 M AgNO3 and allowing the precipitate to form, the remaining silver was back titrated with 0.05322 M KSCN, requiring 35.14 mL to reach end point. AgNO3 + I– → AgI (s) + NO3– Ag+ + SCN– → AgSCN How many millimoles of AgNO3 reacted with I–? How many milligrams of I– (MM=126.9 g/mol) were present in the sample? What is the % w/w I– in the sample?0.0585 g Na2C2O4 10 mL to adjust KMnO4 solution prepared in 0.1 M'pure water, 2 M H2SO4 added, heating process and 8.4 mL titrant as a result of titration it's spin out. Calculate the actual concentration of potassium permanganate accordinglyThe % w/w I– in a 0.6712-g sample was determined by Volhard titration. After adding 50.00 mL of 0.05619 M AgNO3 and allowing the precipitate to form, the remaining silver was back titrated with 0.05322 M KSCN, requiring 35.14 mL to reach endpoint. AgNO3 + I– → AgI (s) + NO3– Ag+ + SCN– → AgSCN How many millimoles of AgNO3 reacted with I–?
- 3 L contaminated air 50 mL 0.0116 M in an air pollution analysisCarbon dioxide (CO2) BaCO3 is passed through Ba (OH) 2 solution.is precipitated as. Excess of base, next to phenol phthalate (f.f.) indicatorIt is titrated with 23.6 mL of 0.0108 M HCl. CO2 in this air sampleCalculate its concentration in ppm. (Density of CO2Take it as 1.98 g / L. C = 12, O = 16 g / mol).A 0.7120 g of iron ore was brought into solution and passed through a Jones reductor. Titration of Fe(II) produced required 39.21mL of 0.02086M KMnO4. Express the results of analysis in terms of (a) percent Fe (MM 55.85) and (b) percent Fe2O3 (Molar mass= 159.69). 5Fe+2 + MnO4- +8H+ → 5Fe+3 + Mn+2 + 4H2OThe %w/w Cl- (36.45 g/mol) in a 0.5785 g sample was determined by Volhard titration. After adding 65.00 mL of 0.0250 M AgNO3 and allowing the precipitate to form, the remaining silver was back titrated with 0.0365 M KSCN, requiring 15.00 mL to reach the end point. Report the %w/w Cl- in the sample
- Calculate the concentration of Fe2O3 in a mineral sample after solubilization of 0.4891g and availability of Fe as Fe2+. The released Fe2+ ions were titrated with a 0.02153 mol/L K2Cr2O7 solution requiring 36.92 mL to reach the end point of the titration. Show the calculations and express the concentration in % m/m. K2Cr2O7; + 6Fe2+ + 14H+ → 2Cr+3 + 2K+ + 6Fe3+ + 7H20Suitable treatment of a 2.566 g soil sample converts all arsenic to AsO43-. 50.00 mL of 0.082 M Ag+ was added to form Ag3AsO4 precipitate. The excess Ag+ was back-titrated with 21.8 mL of 0.024 M SCN- to reach the equivalence point Ag+ + SCN- → AgSCN(s) Solve for the % As (MW = 74.92 g/mol) in the sample.0.7050-g of pure KHP (FW = 204.2) was dissolved and titrated with 35.00mL of NaOH solution. The excess NaOH was backtitrated with 5.00mL of HCl solution. In a separate titration it was found that 30.00mL of NaOH will react with 28.00mL of HCl. Find the molarity of NaOH and HCl.
- A 0.1214-g sample of impure Na2CO3 was analyzed by the Volhard method. After adding 50.00 mL of 0.07011 M AgNO3, the sample was back titrated with 0.06021 M KSCN, requiring 27.15 mL to reach the end point. State the net ionic equation.1. 0.646 g of sample containing BaCl2.2H2O (244.26 g/mol) was dissolved and enoughpotassium chromate was added. After filtering, the precipitate was dissolved in acid andenough KI was added and titrated with thiosulfate. Since 48.7 mL of 0.137 M thiosulfateis used for this, what is the percentage of BaCl2.2H2O in the sample?K2Cr2O7 + 7H2SO4 + 6KI 4K2SO4 + Cr2(SO4)3 + 7H2O + 3I2A 5.00 mL aqueous sample containing hydrogen peroxide was diluted to 25 mL and analyzed by titration with permanganate: 2MnO4- + 5H2O2 + 6H+ → 5O2 + 2Mn2+ + 8H2O The sample required 42.8 mL of 0.0175 M permanganate to reach the end point. What is the concentration of hydrogen peroxide in the original sample?