What are the identities of the substances found after electrolysis and heating? The identities of the substances found were Iodine, Iodide ion (I-), Triodide ion(I3-), Zinc metal and Zinc ion(Zn2+) Post Lab questions: 1. What did this experiment convincingly show about the composition of the white solid? I put that the experiment showed us the white substance was comprised of Zinc ions and iodide ions, and that the two ions and be separated back into their respective elements 2. During the electrolysis, a gray solid formed on the negative wire of the battery and the dark red solution formed at the positive wire of the batter.
Computer Additivity of Heats of Reaction: Hess’s Law 18 (1) Solid sodium hydroxide dissolves in water to form an aqueous solution of ions. (2) Solid sodium hydroxide reacts with aqueous hydrochloric acid to form water and an aqueous solution of sodium chloride. NaOH(s) + H+(aq) ) + Cl–(aq) → H2O(l) + Na+(aq) + Cl–(aq) ∆H2 = ? OBJECTIVES • • • • In this experiment, you will Combine equations for two reactions to obtain the equation for a third reaction. Use a calorimeter to measure the temperature change in each of three reactions.
Lesson written by Carolina Sylvestri Experiment: Reaction Between Ions in Aqueous Solutions The Monster Mash Background: Ionic solids dissolve in water to form aqueous solutions which conduct electricity. These solutions contain both positive and negative ions in such numbers that their net electric charge is zero. In this experiment, you will mix various ionic solutions, two at a time, to determine which combinations form precipitates. Knowing which ions are present makes it possible to deduce which of the possible ion combinations are responsible for the precipitates. From your data table, it will then be possible to generate a solubility table.
As an example, Model Science (2011) provided “sodium burns orange, potassium -purple/blue, barium - green, and lithium – red”. After this experiment is completed, the following questions will be answered. * Why do different compounds have different colors in their visible emissions? * Would you expect the emissions to vary if metal fluorides were used rather than metal
This is done by a procedure called refluxing. Refluxing is the process of heating a product to the boiling point and re-condensing the vapor continuously. The procedure halogenation is the addition of a halogen to a π bond forming a halo alkane. In this synthetic reaction bromine was used in the process called bromination. The bromine is acting first like an electrophile, and then after bromine has broken the π bond, a carbocation has formed, and a bromide ion has been created, the bromide ion then acts as the nucleophile and forms a bond with the carbocation.
Wan Huang Oct 26, 2012 Flame Tests Lab Introduction By placing atoms of a metal into a flame, electrons can be induced to absorb energy and jump to an excited energy state, a quantum jump. They then return to their ground state by emitting a photon of light (the law of conservation of energy indicates that the photon emitted will contain the same amount of energy as that absorbed in the quantum jump). The amount of energy in the photon determines its color; red for the lowest energy visible light, increasing energy through the rainbow of orange yellow green blue indigo, and finally violet for the highest energy visible light. Photons outside the visible spectrum may also be emitted, but we cannot see them. Hypothesis If we can identify the color of flame after burning the atoms of metal, then we can determine the amount of energy in the photon.
Objectives: The purpose of this lab is to observe the reaction of crystal violet and sodium hydroxide by looking at the relationship between concentration and time elapsed of the crystal violet. CV+ + OH- CVOH To quantitatively observe this reaction of crystal violet, the rate law is used. The rate law tells us that the rate is equal to a rate constant (k) multiplied by the concentration of crystal violet to the power of its reaction order ([CV+]p) and the concentration of hydroxide to the power of its reaction order ([OH-]q). Rate = k[CV+]p[OH-]q To fully understand the rate law, concentrations of the substances must be looked at first. The concentration is measured in molarity.
Distinguish between an element and a compound with 2 examples for each An element is a substance that cannot be broken down into any simpler substances. Some examples of Elements are Hydrogen and Helium. A Compound is when two or more elements combine. Examples of this are Water (H2O) and Salt (NaCl) Explain the relationship between the group number and the number of electrons in the valence shell The relationship between the group number and the numebr of
1. Atomic radius increases 2. Electron shielding increases 3. The nuclear attraction between the electrons and the nucleus therefore increases, making it harder for the outer shell to gain an electron into the p-subshell. (d) describe the term disproportionation as a reaction in which an element is simultaneously oxidised and reduced, illustrated by: (i) the reaction of chlorine with water as used in water purification, Cl2(aq) + H2O(l) -----> HClO(aq) + HCl(aq) (ii) the reaction of chlorine with cold, dilute aqueous sodium hydroxide, as used to form bleach, Cl2(aq) + 2NaOH(aq) ----> NaCl(aq) + NaClO(aq) +
It is produced by reacting nitric acid and sulphuric acid simultaneously. The lone pair from oxygen breaks the OH bond when it comes in contact with sulphuric acid. Furthermore, NO2OH2 break and dissociate to form H2O water along with the electrophiles NO2. The electrons from the benzene ring attack NO2 causing the double bond to break. Multiple resonance is created from the movement of electrons.