Calculate the molarity of the original vinegar solution and its concentration in gdm-3, given that it reacts with NaOH in a 1:1 ratio. 7. 2.5 g of a sample of ethanedioic acid, H2C2O4.nH2O, was dissolved in water and the solution made up to 250 cm3. This solution was placed in a burette and 15.8 cm3 were required to neutralise 25 cm3 of 0.1 moldm-3 NaOH. Given that ethanedioic acid reacts with NaOH
Aim: To determine the presence of the cations Na+, K+, NH4+, Ag+, Cu2+ and Bi3+ in the solution under analysis. Procedure: Please refer to page 26-27 of CHM152/154/2252 Laboratory Manual. Results: Procedure No and Ion Test Reagent or Technique Evidence of Chemical Change Chemical(s) Responsible for Observation Equation(s) for Observed Reaction 1. Na+ Perform flame test by dipping a test loop in test solution and putting in Bunsen flame after cleaning the loop with HCl(aq) A brilliant yellow persistent flame occurs Presence of sodium in the test solution - 2. K+ Perform flame test by dipping a test loop in test solution and putting in Bunsen flame after cleaning the loop with HCl(aq).
Litmus test: Red turned blue and blue had no reaction Dilute HCl: Fizzes and substance turns yellowish while bubbling the back to white. Dilute NaOH: Dissolves and substance became a bit thick Substance: CuCO3 Color: Green Odor: Yes Effect of Heat: Turned into black powder Solub. or Reaction w/cold H2O: No reaction. Solub. or Reaction w/hot H2O: No reaction Litmus test: No change on blue or red.
Once a light blue solution remained, 5mL of distilled water was added. This solution was then placed onto the magnetic stir plate and magnetic stir bar was placed in liquid solution. Then to the solution .75g of zinc was added to solution. Then once the solution was colorless, the solution was removed from magnetic stir plate and the liquid was decanted away from the copper. Then 5mL of HCl was added to copper to completely remove all traces of zinc.
|Cloudy milk liquid|Rapid |Red-center |Fizzled |No change | | | | |Powder remains. Odor noted.| |boiling/slight odor|white with |Chemical | | | | | | | | |faded blue |Reaction | | | | | | | | | |Powder to |
Let’s call this number X. Then the structural formula of BaCl2 hydrate can be written as BaCl2•XH2O. The reaction of dehydration is |BaCl2•XH2O ( BaCl2 + XH2O |(5) | According to the stoichiometry of the reaction (5) |[pic] |(6) | Where N1 is the number of moles of BaCl2 formed in the reaction (5) and N2 is the number of moles of water lost in the reaction (5). To find out N1 you need to divide the mass of BaCl2 after the reaction by its molar mass. To find out N2 you have to determine the mass of water produced in the reaction.
Rf = Ds/Df Ds = Distance traveled by the compound Df = Distance traveled by the solvent Materials: coffee filter, tape, isopropyl alcohol (rubbing alcohol), water, 3 different color magic markers (not permanent); 3 identical tall, narrow drinking glasses; metric ruler, 3 pencils Pre-lab Questions: List one extensive and one intensive property of marker ink. Intensive - color Extensive - volume Define mixture. a substance made by mixing other substances together. Is tap water a heterogeneous or homogenous mixture? Explain your answer.
For part 2, do the same thing as part 1 except use an unknown hydrate and calculate the percent mass of water in an unknown hydrate. Observations Part 1 At first the hydrate was a crystal red color. When it was placed on the hot plate and its state changed it began to change color. As it heated it started to become liquidly and bluish in color. Once it cooled down it became a purplish black color.
To observe the change in properties of various synthesized copper compounds in order to identify the reactions and products being formed. Secondly, to become familiar with methods of separating compounds through decanting, heating with Bunsen burner and suction filtration system. In doing so, be able to recover copper metal and interpret the resulting percentage yield by applying the law of conservation of matter. Upon combining 20.0mL of NaOH to 10.0mL of Cu(NO3)2 and stirring the solution, there was a darkening of the initial blue color. Adding distilled water to the solution lowers the initial intensity of the blue color, bringing back to a lighter blue.
3 x (C H5 N) = C3H15N3 Hydrated compounds Solving process: 1st- the difference between the initial mass and that of the dry sample is the mass of water that was driven off. Mass of hydrate minus mass of dry sample equals the mass of water 10.407 – 9.520 = 0.887 g 2nd- The mass of dry BaI2 and the mass of water are converted to MOLES. 9.520 g BaI2 x 1 mol BaI2 ∕ 391 g BaI2 = 0.0243 mol BaI2 anhydrate 0.887 g H2O x 1 mol H2O / 18.0 g H2O = o.o493 mol H2O 3rd: Dividing both results by the amt of 0.0243 mol, we get a ratio of 1 to 2.03, or 1 to 2, since the formula must have full numerical integers of water molecules, in other words no fractions of a water molecule. Thus, for every 1 mole of BaI2, there are two moles of water. The formula for the hydrate is written as BaI2 • 2H2O And it is named barium iodide dihydrate.