Wednesday, May 6, 2020

The Effect Of The Amount Of Naoh On The Gas Volume At 2...

1). Title The Effect of the Amount of NaOH on the Gas Volume at 2 Minutes 2). IV, DV, 3 SVs IV: The amount of base (NaOH) DV: The the volume of gas in mL after 2 minutes. SV: The amount of Hydrogen Peroxide, the amount of reaction time (2 minutes), and the amount of enzyme (catalase). 3). Hypothesis If different amounts of a base (2 mL, 4 mL, 6 mL) were added, then the reaction would produce the most gas at 6 mL because bases have more OH and the extra oxygen will make the reaction take less time because of the enzyme in the raw liver. 4). Materials Safety Goggles, 50 mL granulated cylinder, 10 mL granulated cylinder, glass vial with stopper and pipet, thermometer, plastic shoe box (tub), catalase from chicken liver, pH paper 0.1M NaOH, 3% Hydrogen Peroxide, tweezers, filter paper disks. 5). Procedure Fill bin with water Fill granulated cylinder with water by submerging and make sure no bubbles at the top. Place filter paper disks dipped in catalase on one side of glass vial. Fill the glass vial with 10 mL of 3% hydrogen peroxide and NaOH (2mL, 4mL, or 6mL, depending) on the other side of the glass vial. Quickly cap the tube with the rubber stopper and submerge and make sure dropper is under the granulated cylinder and that the bubbles are going up the cylinder. Record volume of gas (mL) after 2 minutes. 6). Data Table Enzyme Experiment Trials IVs 1 2 3 Averages Control 40 mL 80 mL 70 mL 63.3 mL Base 2 mL 70 mL 55 mL 63 mL 62.67 mL Base 4 mL 50 mLShow MoreRelatedChemical Reaction and Solution2470 Words   |  10 Pagessolubility of Cu(OH)2(s) is 1.72 x10–6 g/100. mL of solution at 25 ° C. (i) Write the balanced chemical equation for the dissociation of Cu(OH)2(s) in aqueous solution. Cu(OH)2 Cu 2+ + 2 OH – (ii) Calculate the solubility (in mol/L) of Cu(OH)2 at 25  °C. (1.72 x10–6 g/0.100 L)(1 mol/97.5 g) = 1.76 x10–7 mol/L (iii) Calculate the value of the solubility-product constant, Ksp, for Cu(OH)2 at 25  °C. 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