CH3-C=O-NH2 +HCl----> CH3-C=O-OH+ NH4Cl
Answer:12mL of 50% solution and 24mL of 20% solution must be mixed to produce 36mL of a 30% acid solutionLet x = 50% acid solutiony = 20% acid solution Equations:x + y = 36mL ----equation (1)0.5x + 0.2y = 0.3 * 360.5x + 0.2y = 10.8multiplying by 105x + 2y = 108 ----equation (2)eliminating equations (1) and (2)-2(x + y = 36)-2-2x -2y = -725x +2y = 108=========3x = 36x=12substitute x=12 to equation (1)12 + y = 36y = 36 - 12y = 24thus 12mL of 50% solution and 24mL of 20% solution must be mixed to produce 36mL of a 30% acid solution
Answer:170mL of 5% acid solution and 30mL of 25% solution must be mixed to produce 200mL of a 8% acid solutionLet x = 5% acid solutiony = 25% acid solution Equations:x + y = 200mL ----equation (1)0.05x + 0.25y = 0.08 * 2000.05x + 0.25y = 16multiplying by 1005x + 25y = 1600 ----equation (2)eliminating equations (1) and (2)-5(x + y = 200)-5-5x -5y = -10005x +25y = 1600=========20y = 600y=30substitute x=30 to equation (1)x + 30 = 200x = 200 - 30x = 170thus 170mL of 5% solution and 30mL of 25% solution must be mixed to produce 200mL of a 8% acid solution
Answer:525mL of 75% solution and 420mL of 30% solution must be mixed to produce 945mL of a 50% acid solutionLet x = 75% acid solutiony = 30% acid solution Equations:x + y = 945mL ----equation (1)0.75x + 0.3y = 0.5 * 9450.75x + 0.3y = 472.5multiplying by 10075x + 30y = 4725 ----equation (2)eliminating equations (1) and (2)-30(x + y = 945)-30-30x -30y = -2835075x +30y = 4725============45x = 23625x=525substitute x=525 to equation (1)525 + y = 945y = 945 - 525y = 420thus 525mL of 75% solution and 420mL of 30% solution must be mixed to produce 945mL of a 50% acid solution
Its an equation you can use to find the pH of a solution. it is.... --- pH = pKa + log (Base/Acid) --- these may help too Ka = 10^-pKa Kw = Ka*Kb
x = amount of acid to add. y = final volume. 5 gal + x = y original amount of acid + acid added = final amount of acid (5 X .2) + x = O.5y Subtract the second equation from the first one. 5 - 1 + x - x = y - 0.5y 4 = 0.5y 8 = y Therefore the final volume is 8 gallons. 5 gal + x = 8 x = 3 gal. the amount of pure acid to add. Check the answer 5 X .2 = 1 gal of acid in original solution. 3 gallons added = 4 gallons total acid in solution. 4 gallons total acid in final solution of 8 gallons total solution = 50% acid.
The chemical equation for the reaction between acetamide (CH3CONH2) and sulfuric acid (H2SO4) is: CH3CONH2 + H2SO4 → CH3COOH + NH4HSO4 This reaction results in the formation of acetic acid (CH3COOH) and ammonium hydrogen sulfate (NH4HSO4).
The general equation for the hydrolysis of a lipid molecule is: Lipid + water → fatty acid(s) + glycerol
Acetamide is a weak base. It can undergo protonation to form the conjugate acid, acetic acid, in acidic solutions.
The hydrolysis of salicin involves the breakdown of the glycosidic bond, resulting in the formation of salicylic acid and glucose. The general equation for this reaction is: Salicin + H2O → Salicylic acid + Glucose
If using acid-catalyzed hydrolysis of starch you can tell the hydrolysis is complete with the solution no longer gives a bluish/purple color with iodine solution. The color should be colorless.
Acidic hydrolysis uses an acid to break down chemical compounds, while alkaline hydrolysis uses a base. Acidic hydrolysis typically results in the formation of an acid and alcohol, while alkaline hydrolysis results in a salt and alcohol. The choice between acidic and alkaline hydrolysis depends on the specific compound being treated and the desired reaction products.
Phosphoric acid (H3PO4) undergoes acid hydrolysis to form H2PO4- and H3O+. The balanced chemical equation for this reaction is: H3PO4 + H2O -> H2PO4- + H3O+.
An acetamide is an amide of acetic acid, chemical formula CH3CONH2.
The iodine test is performed to indicate the completion of the hydrolysis of starch by acid. When starch is completely hydrolyzed, the blue-black color of the iodine-starch complex will disappear, turning the solution colorless.
write chemical equation for the reaction between viscose solution and sulphuric acid
Sulfuric acid is a strong acid that helps to break down proteins effectively during hydrolysis. It provides a stable pH environment for the hydrolysis reaction, ensuring efficient protein breakdown. Additionally, sulfuric acid is readily available and cost-effective for use in laboratory settings.
Acid hydrolysis using sulphuric acid and water (equilibrium reaction). The ester splits into a carboxylic acid and alcohol, protons donated from the acid. The solution can then be distilled and the remaining acid can be checked using UV indicator. Acid hydrolysis using sulphuric acid and water (equilibrium reaction). The ester splits into a carboxylic acid and alcohol, protons donated from the acid. The solution can then be distilled and the remaining acid can be checked using UV indicator.