Application Information

This drug has been submitted to the FDA under the reference 019284/001.

Names and composition

"OCL" is the commercial name of a drug composed of POLYETHYLENE GLYCOL 3350 and POTASSIUM CHLORIDE and SODIUM BICARBONATE and SODIUM CHLORIDE and SODIUM SULFATE.

Forms

ApplId/ProductId Drug name Active ingredient Form Strenght
019284/001 OCL POLYETHYLENE GLYCOL 3350; POTASSIUM CHLORIDE; SODIUM BICARBONATE; SODIUM CHLORIDE; SODIUM SULFATE SOLUTION/ORAL 6GM per 100ML and 75MG per 100ML and 168MG per 100ML and 146MG per 100ML and 1.29GM per 100ML

Similar Active Ingredient

ApplId/ProductId Drug name Active ingredient Form Strenght
019284/001 OCL POLYETHYLENE GLYCOL 3350; POTASSIUM CHLORIDE; SODIUM BICARBONATE; SODIUM CHLORIDE; SODIUM SULFATE SOLUTION/ORAL 6GM per 100ML and 75MG per 100ML and 168MG per 100ML and 146MG per 100ML and 1.29GM per 100ML
090769/001 CLENZ-LYTE POLYETHYLENE GLYCOL 3350; POTASSIUM CHLORIDE; SODIUM BICARBONATE; SODIUM CHLORIDE; SODIUM SULFATE FOR SOLUTION/ORAL 236GM per BOT and 2.97GM per BOT and 6.74GM per BOT and 5.86GM per BOT and 22.74GM per BOT

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Answered questions

2NaOH + CA (OH)2 + 2Cl2 → Ca(OCL)2 + 2NaCl + 2H2O.?
a) what masses in grams of chlorine and sodium hydroxide react with 1067 g of Ca(OH)2 b) what mass in g of Ca(OCL)2 is produced according to this equation? c) what masses of the products of Nacl and H2O are produced? Asked by Anya Gelvin 1 year ago.

a) 2NaOH + Ca(OH)₂ + 2Cl₂ → Ca(OCl)₂ + 2NaCl + 2H₂O Molar mass of Ca(OH)₂ = (40.1 + 16.0×2 + 1.0×2) g/mol = 74.1 g/mol Number of moles of Ca(OH)₂ reacted = (1067 g) / (74.1 g/mol) = 14.4 mol According to the above equation, mole ratio NaOH : Ca(OH)₂ : Cl₂ = 2 : 1 : 2 Number of moles of NaOH reacted = Number of moles of Cl₂ reacted = (14.4 mol) × 2 = 28.8 mol Molar mass of NaOH = (23.0 + 16.0 + 1.0) g/mol = 40.0 g/mol Mass of NaOH reacted = (28.8 mol) × (40 g/mol) = 1150 g Molar mass of Cl₂ = 35.5 × 2 g/mol = 71.0 g/mol Mass of Cl₂ reacted = (28.8 mol) × (71.0 g/mol) = 2040 g ==== b) According to the above equation, mole ratio Ca(OH)₂ : Ca(OCl)₂ = 1 : 1 Number of moles of Ca(OCl)₂ produced = (14.4 mol) × 1 = 14.4 mol Molar mass of Ca(OCl)₂ = (40.1 + 16.0×2 + 35.5×2) g/mol = 143.1 g/mol Mass of Ca(OCl)₂ produced = (14.4 mol) × (143.1 g/mol) = 2060 g ==== c) According to the above equation, mole ratio Ca(OH)₂ : NaCl : H₂O = 1 : 2 : 2 Number of moles of NaCl produced = Number of moles of H₂O produced = (14.4 mol) × 2 = 28.8 mol Molar mass of NaCl = (23.0 + 35.5) g/mol = 58.5 g/mol Mass of NaCl produced = (28.8 mol) × (58.5 g/mol) = 1680 g Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol Mass of NaCl produced = (28.8 mol) × (18.0 g/mol) = 518 g Answered by Dirk Freyre 1 year ago.


How many moles of OCl- are necessary to react with 0.272 moles of I-?
THANK YOU!!! I thought that is what it was but wanted confirmation... Asked by Man Futral 1 year ago.

OCl-(aq) + 2I-(aq) +2H3O+ -> I2(aq) + Cl-(aq) +3H2O(l) According to the balanced equation above 1 mole of OCl- reacts completely with 2 moles of I- (see the coefficients of the same above). Therefore by simple arithmetic 0.272 moles of l- will react with 0.272/2 = 0.136 moles of OCl-. Answered by Francoise Sciallo 1 year ago.


Calculate the [OCl-] of a 2.37×10-3 M solution of the weak acid HClO (make an exact calculation assuming that?
Calculate the [OCl-] of a 2.37×10-3 M solution of the weak acid HClO (make an exact calculation assuming that initial concentration is not equal to the equilibrium concentration). HClO = OCl- + H+ Constant Ka=3.50×10-8 very confused please help!:) thankyou Asked by Anja Beilfuss 1 year ago.

HClO(aq) + H2O(l) <=> H3O^+(aq) + OCl^-(aq) Ka = [H3O^+][OCl^-]/[HClO] Let [H3O^+] = [OCl^-] = x, and [HClO] = 0.00237 - x x^2/(0.00237 - x) = 3.50 x 10^-8 x^2 = (3.50 x 10^-8)(0.00237 - x) = 8.895 x 10^-11 - 3.50 x 10^-8(x) x^2 + 3.50 x 10^-8(x) - 8.895 x 10^-11 = 0 >> Now solve using the quadratic equation Write the numbers as a = 1, b = 0.0000000350, and c = -0.00000000008895 x = 9.41 x 10^-6 M = [H3O^+] = [OCl^-] [HClO] = 0.00237 - x = 0.00237 - 0.00000941 = 0.00236 M That's how its done. Hope this is helpful. JIL HIR Answered by Georgianna Dobler 1 year ago.


What the is reaction between hypochlorite ion (OCl-) from household bleach and thiosulfate ion (S2O3)2- ?
What is the reaction if I mix NaOCl with NaS2O3? Asked by Moises Pettigrove 1 year ago.

OCl- + 2 (S2O3)2- +H2o ----> Cl- + (S4O6)2- +2 OH- Thiosulfate ion is used to remove chlorine (hypochlorite) from water or from solutions used to bleach paper pulp...in Paper manufacturing and dechlorination of water....... Answered by Elmo Files 1 year ago.


Can someone plz help me, I'm stressing over this: Oxidation of Cr 3+ to CrO4 2- by OCl- in alkaline solution?
(OCl- is converted to Cl-) Asked by Coreen Rehberger 1 year ago.

Cr3+ + 4 H2O = CrO42- + 8 H+ + 3e- OCl- + 2 H+ + 2e- = Cl- + H2O 2 Cr3+ + 3 OCl- + 5 H2O= 2 CrO42- + 3 Cl-+ 10 H+ add 10 OH- 2 Cr3+ + 3 OCl- + 10 OH- = 2 CrO42- + 3 Cl- + 5 H2O Answered by Jeane Schweikert 1 year ago.


Given that ka for HOCl is 3.5*10^-8, calculate the ph of a 0.102 M solution of Ca(OCl)2?
Given that ka for HOCl is 3.5*10^-8, calculate the ph of a 0.102 M solution of Ca(OCl)2? Thanks! Asked by Yoko Auer 1 year ago.

a 0.102 M solution of Ca(OCl)2 releases twice that : 0.204 Molar (OCl)-1 ions which are the conjugate base of the acid HOCl conjugate pairs, have the relationship Ka times Kb = K water so Kb = (1 X 10^-14) / (3.5*10^-8) Kb = 2.86 X 10^-6 for the equilibrium: (OCl)-1 in water) <=> HOCl & OH- Kb = [HOCl] [OH-] / [OCl-] 2.86 X 10^-6 = [X] [X] / [0.204] X2 = 5.83 X 10^-8 X = [OH-] = 2.41 X 10^-4 pOH = 3.62 since pH + pOH = 14 the pH = 10.38 Answered by Rosaria Woolcott 1 year ago.


Given that Ka for HOCl is 3.5e-8, how do you calculate the pH of a 0.146M solution of Ca(OCl)2?
Reaction of the hypochlorite ion with water yields hypochlorous acid. Asked by Jesusa Yanko 1 year ago.

0.146 molar Ca(OCl)2 releases twice = 0.292 molar OCl- --------------------- OCl- & water --> HOCl & OH- K hydrolysis = K water / K a = 1e-14 / 3.5e-8 = 2.86e-7 Khydrolysis = [HOCl] [OH-] / [OCl-] 2.86e-7 = [x] [x] / [O0.292] x2 = 8.35e-10 x = [H+] = 2.89e-5 youir answer: pH = 4.54 Answered by Lela Habowski 1 year ago.


What is the concentration of OCl- in a 0.60 M solution of HOCl? Ka= 3.1 x 10^-8?
what is the concentration of OCl- in a 0.60 M solution of HOCl? Ka= 3.1 x 10^-8 Asked by Hang Onken 1 year ago.

Write the Ka expression Ka = [OCl-][H+]/[HOCl] Let [OCl-] and [H+] = x Insert your known values into the Ka expression: 3.1x10^-8 = (x)(x)/.60 Solve for x: (3.1x10^-8 * .6)^1/2 = x 1.36x10^-4 = x ---> [OCl-] = 1.4x10^-4 moles/L Answered by Quinn Hase 1 year ago.


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