Find the amount of NH4CI that must be added to give 0.5 dm3 solution with pH = 4.25? [Kb (NH3) = 1.8 X 10-5] Mw(NH4CI)=53.49 g/mol 152 g 2.3 x 10-3 15.81 g 750.8 g 4.7 g
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![Find the amount of NH4CI that must be added to give 0.5 dm3 solution with pH = 4.25?
[Kb (NH3) = 1.8 X 10-5] Mw(NH4CI)=53.49 g/mol
152 g
2.3 x 10-3
15.81 g
750.8 g
4.1 g](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4387b4f9-7d67-4692-bda0-cc0deb015fcc%2Fb6aa77b1-892f-4d27-b14a-3801fad8d701%2Fv56lj6_processed.png&w=3840&q=75)
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- Consider the titration illustrated in Figure 8.] Anhydrous sodium carbonate (Na₂CO₃) is a primary standard. When 0.364 grams of the substance is placed in a conical flask, then 20.00 cm³ of sulphuric acid (H₂SO₄) solution is required to reach the end point of the titration. What is the concentration in mol·dm⁻³ of the sulphuric acid solution? [Give the answer to 3 decimal places. Do not type in the unit. Use a decimal point.] *To a 0.5 g sample of impure NaCl are added 0.784 g of pure AgNO3 (MW = 169.88) crystal. This is in excess of the amount required to ppt. All the Cl as AgCl. After dissolving and filtering out the pptd. AgCl, the filtrate requires 25.50 ml of 0.028M KCNS. What is the percentage of NaCl (MW=58.44) in the sample? A. 55.40 % B. 50.10% C. 45.60% D. 100%The solubility of borax, which is made up of sodium tetraborate (Na2B4O5(OH)4 8H2O), was analyzed. The dissolution of borax is: Na2B4O5(OH)4 • 8H2O(s) ⇌ 2 Na+(aq) + B4O5(OH)42–(aq) + 8 H2O(l) A 50 mL saturated solution was prepared. After filtration of solution, 5 mL aliquot was transferred to a flask and titrated using 0.432 M HCl. The endpoint was found to be 4.73 mL of the titrant. Tetraborate anion (B4O5(OH)42-) is a weak base which reacts with HCl like the following reaction: B4O5(OH)42–(aq) + 2 H+(aq) + 3 H2O(l) ⇌ 4 H3BO3(aq) What is Ksp expression for the dissolution? What is the tetraborate ions concentration in the filtrate? What is the molar solubility and Ksp of borax if the titration was done at room temperature (298 K)?
- An impure sample of calcium carbonate with a mass of 7.95 g was reacted with 50.00 cm3 of 1.00 mol dm hydrochloric acid (an excess). The resulling solution was transferred to a volumetric flask and titrated with 11.10cm3 of 0.300 mol dm-3 sodium hydroxide solution. Determine the percentage purity by mass of the calcium carbonate sample.CaCO3 + HCl -> CaCl2 + H2O + CO2 HCl + NaOH -> NaCl +H2O a. Determine how many moles of hydrochloric acid were used.b. Determine how many moles of excess HCI was titratedc. Determine how much in moles calcium carbonate present in the sample.d. Calculate the mass of calcium carbonate presente. Determine the percentwge calcium carbonate is in the sample.2 what is the purpose of BSC Level 1 & 2, proper layout, performance, and its application in biosafety assessmentWhat is the Ksp of Zn3(PO4)2 (MW = 320.8 g/mol) if its solubility in water at 25 °C is 4.9 x 10-5 g/L? format for the scientifc notation 1.0 x 10^-1
- A first-stage recovery of magnesium from seawater is precipitationof Mg1OH22 with CaO:Mg2+(aq) + CaO(s) + H2O(l)---->Mg(OH)2(s) + Ca2+(aq) What mass of CaO, in grams, is needed to precipitate 1000lb of Mg(OH)2?Gravimetric analysis of Fe3O4 (MW = 232 g/mole) may be undertaken with the following reactions: Fe3O4 → Fe2O3 → Fe (OH)3. Weight of sample containing 8.00% Fe3O4 that must be taken to obtain a precipitate of Fe(OH)3 (MW = 107 g/mole) that weighs 150 mg is . a. 0.108 g b. 0.325 g c. 1.355 g d. 4.065 g Amount of Fe2O3 (MW = 160 g/mole) from which 150 mg of Fe(OH)3 (MW = 107 g/mole) may be obtained is . a. 0.112 g b. 0.224 g c. 0.448 g d. none of the other choicesThe data below was obtained with a Cl- ISE to prepare a plot of mV vs. log[Cl-]. Assume the unknown saltwater solution was diluted as in the procedure before being measured. What is the ppm Cl- of the undiluted unknown solution? Cl- Concentration (ppm) Potential (mV) 10 160 100 94 1000 40 Diluted Unknown Saltwater 132
- To a 0.5 g sample of impure NaCl are added 0.784 g of pure AgNO3 (MW = 169.88) crystal. This is in excess of the amount required to ppt. All the Cl as AgCl. After dissolving and filtering out the pptd. AgCl, the filtrate requires 25.50 ml of 0.28M KCNS. What is the percentage of NaCl (MW=58.44) in the sample?Determine the volume of glacial acetic acid to prepare 500 ml of a 0.020 M acetic acid solution (glacial acetic acid; strength 99.8%, density 1.05 g/ml, MW 60.05 g/mole).1. What is the MW of the analyte?2. What is the meq of the analyte?3. What is the weight (in grams) of the analyte?4. What will be the percentage purity of the sample if 43mL of the titrant was consumed?5. Did the sample conform to the USP requirement? Yes or No?
