Name: ___________________________________ Date: ______________ Practice Test #3 ____ 1. When a precipitation reaction occurs, the ions that do not form the precipitate A) evaporate B) are cations only C) form a second insoluble compound in the solution D) are left dissolved in the solution E) none of these 2. An aqueous solution of potassium chloride is mixed with an aqueous solution of sodium nitrate. The complete ionic equation contains which of the following species (when balanced in standard form)? A) B) C) D) E) ____ 3.
3) Write equations to indicate what you consider to have happened in each case in which there was precipitate formed. Use ions to represent the species in the reacting solutions, but for those products that were precipitates write a formula for the compound. Place (aq) after those species in solution and (s) after the precipitates. Be sure to write the equations so that both atoms and charge are conserved. For example: Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq) ( AgCl(s) + Na+(aq) + NO3-(aq) 4) Rewrite the equations, leaving out the ions not involved in the reaction (spectators).
2-propanol (bp=82 degrees C) 3. tetrahydofuran (bp=65 degrees C) 4. 1-butanol (bp=118 degrees C) 5. butanone (bp=80 degrees C) Give a better separation for the mixture to be distilled tetrahydofuran (bp=65 degrees C) because it is farthest from 100 degrees C Which alkyl halide would react fastest in a nucleophilic substitution using silver nitrate in ethanol (weak nucleophile, protic solvent)? 3-bromo-3-methylpentane (most
When the hydrate is heated, it easily loses water molecules attached and becomes an anhydrous salt. The corresponding chemical reaction for hydrated magnesium sulfate can be written as |MgSO4•7H2O ( MgSO4 + 7H2O |(1) | where MgSO4 is the anhydrous salt. Usually,
With the use of this technique we placed chlorine, bromine, and iodine into solutions containing chloride, bromide, and iodide. In the reaction the free halogen (X2) oxidizes the other halide ion (Y-) and gets reduced by gaining electron(s). In table 3, chlorine was the strongest oxidizing agent and iodine was the weakest oxidizing agent. Since chlorine was the strongest oxidizing agent it will react more and the weak agent will react less. This explanation can be demonstrated in table 3 also because the results of the reactions demonstrates that chloride reacted more by the color of the product compared to the color of chloride in the mineral oil.
However majority of the solid Iodine was left at the bottom of the test tube; which leads to the conclusion that it’s not reactive on water. 2. Solid iodine in 1 mL of potassium iodide is slightly more soluble. The liquid in the small test tube turned a dark brown color. 3.
Assignment 03.06 Covalent Bonding and Lewis Structures Part I: Lab Ciara f Insert completed data tables for each part of the lab. Be sure that the data tables are organized and include units when necessary. Melting Point (4 points) |Substance |Melting Point | |Substance A |Higher than 300 C | |Substance B |210 C | |Substance C |Higher than 300 C | |Substance D |130 C | Conductivity (4 points) Part II |Substance A |Conductive |Substance C |Conductive | |Solid |No |Solid |No | |Liquid |Yes |Liquid |Yes | |Aqueous |Yes |Aqueous |Yes | |Substance B |Conductive |Substance D |Conductive | |Solid |No |Solid
Also include any observations that you made over the course of part II. (4 points) Unknown (A,B,C,D) Mass of Metal Volume of water Initial temp. in calorimeter Initial temp. in beaker Final temp. of mixture Metal C 25.605g 24.6mL 25.2°C 100.5°C 28.7°C Calculations: Show your work and write a short explanation with each calculation.
Even though the results under hexane and toluene are similar, the distances of original mixture, first and second fraction are different from hexane and toluene solvent because they have different polarity. As first fraction containing fluorene, which is much non-polar than fluoreone, fluorene in first fraction is much easier to carry by the moving non-polar solvent. Thurs, as the more non-polar the solvent is the longer distance that fluorene will move. Since hexane has larger non-polar carbon-hydrogen single bond groups than toluene, it is much non-polar than toluene. As the result of this, it can explain why the distance of fluorine in hexane is longer (1.3 cm) than the one in toluene (0.5 cm) and due to less non polar toluene has.
The precipitate(s) for the above reaction is/are: A) Silver nitrate B) Aluminum bromide C) Silver bromide D) Aluminum nitrate E) There is no precipitate 18 – 37: For each of the two REDOX reactions below, perform the following: A) Balance the following equation and place the final answer in the boxes provided; show all work on the exam pages. B) Identify the oxidizing and reducing reagents. C) Assign the oxidation numbers to the elements in bold on both sides of the equation. D) Identify the oxidation and reduction half-reactions by placing LEO and GER, respectively, above the arrows in each half-reaction. E) NOTE that illegible writing will not be graded.