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- Density of solution:Trial 1: 1.2 g/mLTrial 2: 1.2 g/mLTrial 3: 1.2 g/mL Average density = 1.2 g/mL What is the relative average deviaion, %?Sttoul SuluiIUN A client requires BI). wund eleasing with 3 eleasing Shrength hydrrgen Peroseide Aorder to prepare enough solutim to do turo wo und trieatments you will Prepare 500 mL of sulutiuusing a diluent. ofescribe themethod for miscing this sulutturj Nomal salineA 786 mLmL NaClNaCl solution is diluted to a volume of 1.19 LL and a concentration of 3.00 MM . What was the initial concentration C1C1C_1?
- 5. How many milligrams of H3BO3 should be used in compounding the following prescription? Rx Phenacaine HCI Chlorobutanol (E = 0.1657) (E = 0.1836) (E = 0.5259) 1.0%. %3D H;BO3 %3D Purified water ad 60.0 mL Make isotonic sol. Sig. One drop in the eye 6. How much H3BO; should be used in compounding the following prescription? 'D' 1% solution 0.62°C Rx Tetracaine HCI 0.33% Zinc sulfate 0.17% 0.62°C H3BO3 Purified water ad 30.0 mL Make isotonic sol. Sig. One drop into each eyecium chloride per liter. Express the centration in terms of milliequivalents of calcium chloride (CaCl2-2H20-m.w 147) per liter. con- 27. How many milliequivalents/of potassium would be supplied daily by the usual dose (0.3 mL three times a day) of saturated potassium iodide solution? Saturated po- tassium iodide solution contains 100 g of potassium iodide per 100 mL.oluce 28. An intravenous solution calls for the addi- tion of 25 mEq of sodium bicarbonate. How many milli 4%Q1) A Sada- Lime Sample is g0% NaoH and 10% Cao. f 3 gm is dissolved in 250 ml, what is Hthe tutal normality of the Selutton as abase ? How many mililiters of 0.5N H2S4 woutd be required to titrate ioo me of the Solution? Ans. ( 0.309N, 60.59 ml)
- Zonasha Berrett concentrdion of 5.50mollL.what. A stock solution has a volune ef this stock Solution is required to melke (50.omL. of Solution with a concentration ef lrowmollL? (V2)=650.0ml CM2) =1.00 mb/L=l.00M Cvij =? CMIJ= s.50 ndlL Vi =118.2 mC %3D • • Volume of stock Solution required is ll8.2 mLSALT SOLUTION: Trial 1: Volume: 25 mLMass of 25mL solution: 29.824 gDensity: 1.2 g/mL Trial 2:Volume: 25 mLMass of 25 mL solution: 29.855 gDensity: 1.2 g/mL Trial 3:Volume: 25 mLMass of 25 mL solution: 29.816 gDensity: 1.2 g/mL METAL BAR: Method I - Trial 1:Mass of metal bar: 57.063 gVolume of metal bar: 6.7 cm^3Density of bar: 8.5 g/cm^3 Trial 2:Mass of metal bar: 57.063 gVolume of metal bar: 6.0 cm^3Density of bar: 9.5 g/cm^3 Trial 3:Mass of metal bar: 57.063 gVolume of metal bar: 6.0 cm^3Density of bar: 9.5 g/cm^3 Method 2 - Trial 1:Dimensions: diameter = 1.20 cm, height = 4.91 cmMass of metal bar: 57.063 gVolume of metal bar: 5.55 cm^3Density of bar: 10.3 g/cm^3 Trial 2:Dimensions: diameter = 1.20 cm, height = 4.95 cmMass of metal bar: 57.063 gVolume of metal bar: 5.60 cm^3Density of bar: 10.2 g/cm^3 Trial 3:Dimensions: diameter = 1.20 cm, height = 4.95 cmMass of metal bar: 57.063 gVolume of metal bar: 5.60 cm^3Density of bar: 10.2 g/cm^3 QUESTION - From your data, calculate the…How much of the suspending agent is needed to prepare a 60-mL preparation of Aluminum hydroxide gel based on this formulation? Ammonium alum 800 g Sodium carbonate 1000g Peppermint Oil 0.01% (0.01 g/100mL) Sodium benzoate 0.1% (0.1 g/100mL) Purified water, qs ad. 2000 mL a. 0.06 g b. 30 g c. 24 g d. 0.006 g
- 1-Pentanol to 1-bromopentane Chemicals: - 60ml Conc. Sulfuric Acid - 100ml Saturated Sodium bicarbonate - 65ml 1-Pentanol - 78g sodium bromide - Distilled water - 58.42g 1-Bromopentane 1-Pentanol Sodium Bromide Sulfuric Acid 1-Bromopentane Formula C5H12O NaBr H2SO4 C5H11Br MW (g/mol) 88.15 102.894 98.078 151.04 Density (g/mL) 0.811 3.21 1.84 1.218 Boiling point (*C) 138 1,396 337 130 NaBr(aq) + H2SO4(aq) -> NaHSO4(aq) + HBr(aq) CH3(CH2)4OH(aq) + H+ Br- (aq) CH3(CH2)4OH2 (aq) + Br-(aq) CH3(CH2)4OH2 (aq) + Br-(aq) CH3(CH2)4Br(aq) + H2O(aq) How do I calculate the percent yield and identify the limiting reagent?Expeniment and name Chemical formula Cobalt (11) nitzate Co(NOs)2 Buldatance Aane and dhemical farmula mm, q/mo! 182.95g lmol Fruihal volume , L Iniial concentrahion measure d, mal/L Number of mole's and massof mass Of solute 500ML 0.876 mollL 0 -438 mol Colw0z to Concentration after eventahom affer evapo 1.461 mal L Can you please explatn to calculate how the 180.19) cduo wsas arrived at ? Thanks ass6.3017g sample was weighted Analytically and made up volumetrically to 500mL. A 25mL alliqout of this solution required a total valume of 41.32mL of 0.0987MHCl to reach bromocresol green endpoint Another 25ml alligout of this salution was treated with 25mL of 0.1014M NaOH react with excess of BaCl₂ solution und left over NaOH required 21.83mL of 0.0987MHCl to reach the phenolpthelin end point. Determine the percentage of Na₂CO3 and. NaHCO3
