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Molarity = moles of solute/Liters of solution ( 1500 mL = 1.5 Liters ) Molarity = 0.800 moles NaOH/1.5 Liters = 0.533 M sodium hydroxide ...
To test presence of Sodium ions, do a flame test: color observed: golden yellow. To test presence of carbonate or hydrogen carbonate ions: Take some of the sodium hydrogen carbonate in a dry test tube. Heat the test tube and bubble the gas through limewater. Limewater turns milky. Carbonate or hydrogen carbonate ions present. To distinguish between carbonate and hydrogen carbonate: Add indicator solution. If colour of solution turns green, pH is 7-8, hydrogen carbonate ions are present. If colour of solution turns blue, pH is 12-13, carbonate ions are present. -Iberuz
n=CV/1000 Where n = number of moles, C= Molarity concentration of the solution, and V= volume of the solution in milliliter's We are given, n=0.150, V=8.16 liters(8160 milliliters)and C is not given. Solving for C from the equation above we get: C=(n*1000)/V C= (0.150*1000)/8160 C= 0.018 The molarity of the solution is 0.018
precipitated Fe(OH)3 = ferric hydroxide or Fe(OH)4-(aq) = ferrate anion in solution, when in excess of hydroxide
So you have 0.25 molar Na3PO4. But for every one molecule of Sodium phosphate, you have 3 molecules of sodium. Therefore you have 0.25*3 or 0.75 molar sodium.
Molarity = moles of solute/Liters of solution ( 1500 mL = 1.5 Liters ) Molarity = 0.800 moles NaOH/1.5 Liters = 0.533 M sodium hydroxide ...
30 gm of sodium hydroxide desolved in 1 litre distilled water.
Solid ferric hydroxide, Fe(OH)3, precipitates out of solution. The resulting sodium sulphate remains dissolved.
Calcium Hydroxide & Ammonia Solution & Sodium Hydroxide. Are the Common Alkalis you find in a Lab
To test presence of Sodium ions, do a flame test: color observed: golden yellow. To test presence of carbonate or hydrogen carbonate ions: Take some of the sodium hydrogen carbonate in a dry test tube. Heat the test tube and bubble the gas through limewater. Limewater turns milky. Carbonate or hydrogen carbonate ions present. To distinguish between carbonate and hydrogen carbonate: Add indicator solution. If colour of solution turns green, pH is 7-8, hydrogen carbonate ions are present. If colour of solution turns blue, pH is 12-13, carbonate ions are present. -Iberuz
precipitated Fe(OH)3 = ferric hydroxide or Fe(OH)4-(aq) = ferrate anion in solution, when in excess of hydroxide
n=CV/1000 Where n = number of moles, C= Molarity concentration of the solution, and V= volume of the solution in milliliter's We are given, n=0.150, V=8.16 liters(8160 milliliters)and C is not given. Solving for C from the equation above we get: C=(n*1000)/V C= (0.150*1000)/8160 C= 0.018 The molarity of the solution is 0.018
Add sodium hydroxide solution; aluminium hydroxide will dissolve but Mg(OH)2 remain.
So you have 0.25 molar Na3PO4. But for every one molecule of Sodium phosphate, you have 3 molecules of sodium. Therefore you have 0.25*3 or 0.75 molar sodium.
Phosphoric Acid dissociates to give off 3H+ ions, meaning that one mole of Phosphoric acid reacts with three mols of sodium hydroxide. Using the equation n = c x v n = 0.1 x 0.05 =0.005 mols of OH ions in the solution therefore there are 0.005/3 = 0.00167 So the volume of phosphoric acid - v = n/c v = 0.00167/0.2 v = 0.00835 l = 8.4ml of Phosphoric Acid reacts completely with Sodium hydroxide
The chemical formula (not symbol) of sodium hydroxide is NaOH.
C2H4O2 + NaOH = H2O + C2H3O2Na Acetic acid (60 gm) + sodium hydroxide ( 40 gm) = 100 gm water (18 gm) + sodium acetate (82 gm) = 100 gm Ratio reactants to products = 1:1 Molarity = moles / L, 3M = 3 moles / 1 L Acetic acid = 60 gm / total reactant 100gm = 1.8 moles Multiply by 3 = 1.8 moles or 180 grams Sodium Hydroxide = 40 gm / total reactant 100 mg = 1.2 moles or 120 grams. 180 grams acetic acid + 120 grams sodium hydroxide = 300 grams. 300 grams divided by 1 liter = 3M So in order to make 3 M sodium acetate combine solution, add 180 grams acetic acid and 120 grams sodium hydroxide with 1 liter of water.