n(Na2CO3) = C(Na2CO3) × V(Na2CO3) = 0.035 mol L-1 × 0.25 L = 0.00875 mol.m(Na2CO3) = n(Na2CO3) × M(Na2CO3) = 0.00875 mol × (2×23 + 12 + 3×16) = 0.9275...
1 Answers 1 viewsTo prepare 100 ml of 0.02 normal Na3PO4 solution, it is necessary to dissolve 0.109 g of dry Na3PO4 salt in 100 ml of water.
1 Answers 1 viewsMolecular weight of isotonic saline = 58.50.92×58.5 = 53.8253.92/100×100= 53.92%
1 Answers 1 views1) CH3-CH2-CH2-CH3 + Cl2 ---gt; CH3-CH2-CH2-CH2-Cl + HCl CH3-CH2-CH2-CH2-Cl + KOH (dry) --gt; CH3-CH2-CH2=CH2 CH3-CH2-CH2=CH2 + Br2/H2O2 --gt; CH3-CH2-CH2-CH2-Br CH3-CH2-CH2-CH2-Br + KOH (aq) --gt; CH3-CH2-CH2-CH2-OH + KBr CH3-CH2-CH2-CH2-OH + NaBr/H2SO4...
1 Answers 1 views1st you have to prepare 250 ML approx 0.1 N NaOH solution by dissolving 1 gm NaOH in 250 ml distilled water. Then titrate NaOH with exact 0.1 N oxalic...
1 Answers 1 views1) 1 L of 0.5 M HCl solution contain n(HCl) = c(HCl)*V(1 M HCl solution) = 1 L * 0.5 M = 0.5 mol of HCl. The volume of 6...
1 Answers 1 viewsSolution:70 mg NH4Cl per 1 mL??? mg NH4Cl per 100 mLHence,(70 mg NH4Cl / 1 mL) × (100 mL) = 7000 mg = 7 gAnswer: 7 grams of NH4Cl is...
1 Answers 1 viewsSolution:sodium hydroxide = NaOHThe molar mass of NaOH is 40 g/molHence,(3 g NaOH) × (1 mol NaOH / 40 g NaOH) = 0.075 mol NaOHn(NaOH) = 0.075 molCalculate the molarity...
1 Answers 1 viewsLet's assume the solution density is 1 g/ml. Therefore, 100 ml = 100 g.% = (msolute / msolution) x 100%5% = (msolute / 100) x 100%msolute = 5 gThe...
1 Answers 1 viewsGiven the density of the solution = 1.1 gram/ mlSo the mass of 150 ml of aqua solution = 150 x density of the solution = 150 ml X 1.1 gram...
1 Answers 1 views