Swirl the solution to ensure that the NaOH is properly dissolved in the deionised water. B) Prepare an Oxalic Acid Solution (approximately 0,05M) 1. Place a clean, dry glass beaker on the electronic scale. 2. Determine the mass of the glass beaker.
If the concentration of the H+ ion is equal to the concentration of the OH- ion, the solution is said to be ________. a. Acidic c. alkaline (basic) b. Neutral d. saturated 11. What is the concentration of a solution that contains 10.0 moles of HCl in 5.50 L of water?
The compound’s empirical formula was determined to be FeK3(C2O4)3•3H2O. With the molecular formula and the balanced equation for the synthesis of potassium trioxalatoferrate (III) trihydrate, stoichiometry revealed potassium oxalate monohydrate was the limiting reactant. The theoretical yield of 6 grams of potassium oxalate monohydrate was
There is one mole of OH- in the solution since NaOH goes to Na+ and OH-. Trial 1: 25.65mL NaOH x 0.100mol/1000mL = 2.57 x 10-3 mol NaOH = 2.57 x 10-3 mol HA = 2.57 x 10-3 mol H+. The equivalent mass is 0.356g Acid / 2.57 x 10-3 mol H+ = 139g/mol H+ Trial 2: 49.57mL NaOH x 0.100 mol / 1000 mL = 4.96 x 10-3 mol NaOH = 4.96 x 10-3 mol HA = 4.96 x 10-3 mol H+. The equivalent mass is 0.644g Acid / 4.96 x 10-3 mol H+ = 130.g/mol H+ Average = (139g Acid / 1 mol H+) + (130.g Acid / 1 mol H+) / 2 = 135g/mol H+. The average equivalent mass for the acid is 135g/mol H+.
This helps to indicate whether or not the reaction follows Markovnikov’s Rule, which states that the electrophile (E+) will add to the carbon involved in a double bond that produces the most stable carbocation. If the rule is followed, the reaction will proceed according to the mechanism in Figure 1. In the silver nitrate test, the alkyl bromide is added to AgNO3. The rate of precipitation with 2° should be faster than the solution with the 1° alkyl halide. In the sodium iodide test, the alkyl halide is added to sodium iodide in acetone.
Balanced Chemical Equation: Cu(s) + 4HNO3(aq) —> Cu(NO3)2 (aq) + 2NO2 (g) + 2H2O (l) Reaction 2: when sodium hydroxide (NaOH) is added to copper (II) nitrate (Cu(NO3)2), a double displacement reaction will occur. Copper and sodium will displace each other to create copper (II) hydroxide and sodium nitrate. Balanced Chemical Equation: Cu(NO3)2 (aq) + 2NaOH (aq) —> CuOH2 (s) + 2NaNO3 (aq) Reaction 3: When copper (II) hydroxide is heated, a decomposition reaction will occur. The reaction will decompose forming two compounds, Copper (II) oxide, and water. Balanced Chemical Equation: Cu(OH)2 (s) + Heat —> CuO (s) + H2O (g) Reaction 4: when a sulphuric acid is added to the solution that contains copper (II) oxide, a double displacement reaction will occur.
In the second step, the addition of sodium borohydride reduced the imine into another derivative, which was yellowish lime color. The solution turned clear when acids and anhydrides was added, which indicated the precipitate were dissolved. However, after refluxing for a while, yellow precipitates begin to form near the top of the flask. It was assumed that the remaining starting material was concentrated from a decrease volume to reappeared in solution. Nevertheless, this may have been a sign of contamination that will negatively affect the entire reaction.
Nonetheless, the light yellow solid was purified by using the recrystallization technique. The formation of o-nitroacetanilide is inevitable and in order to eliminate it, 95% ethanol is used as the solvent of choice. The ortho isomer is soluble in the cold alcohol solution whereas p-nitroacetanilide in insoluble. As a result, the ortho isomer remains in the liquid solution and the final product, the p-nitroacetanilide is isolated with a final vacuum
In order to determine the value of X, the hydrate is heated on a burner to undergo decomposition reaction to be decomposed into CuSO4 and water vapor. Water vapor is evaporated during the reaction, leaving CuSO4 crystals, which is supposed to be white, in remain. By weighing the mass of CuSO4 and the mass difference of substance before and after the reaction, the mole of CuSO4 and H2O can be calculated. The value of X can thus be determined by calculating the mole ratio of CuSO4 and H2O. In the lab, through calculation, the value of X is determined to equal to 5.361211229, which is close to 5.
Shayna Salloway AP Chemistry A Snyder 11 September 2014 Title: Finding Mole Ratios of Reactants in a Chemical Reaction Purpose: Experiment using the method of continuous variations to figure out mole ratios of reactants. Procedure: 1. Take the NaClO and the sodium thiosulfate solutions and measure the temperature of each solution. Record in the data table. 2.