Chemistry-Precipitation Essay

My obscure chemical is atomic number 11 iodide. This was determined by testing the chemical with cations such as capital process, atomic number 27 (II) treat, and atomic number 29 nitrate. It was similarly tested with anions atomic number 11 iodide, sodium carbonate, sodium oxalate, and sodium phosphate. It was tested to see if the product will yield a precipitate or unwavering (cloudiness, power, or crystals) and any food coloring changes that argon very uniform to the products/changes when solely the known cations and anions were tested for changes.First of all, sodium iodide is an anion, it was determined that the unknown chemical was an anion because it did not react with any of the anions. Since the unknown did not react with the anions, it did not produce a unharmed or precipitate. This is chemically true because the new products will some(prenominal)(prenominal) contain NO3, or nitrate, and NO3 is sedimentary with all cations.When the unknown was tested with si lver nitrate a solid was produced, but also the color was a light sensationalistic that is quite similar to the comment of the product of sodium iodide added to silver nitrate (a dull but light yellow). This color was not exactly similar to the colors organise from other combinations as they were slightly yellow for the regulation 2AgNO3(aq)+NaCO3(aq) >AgCO3(s)+2NaNO3(aq) , white for 2AgNO3(aq)+NaC2O4(aq) >AgC2O4(s)+2NaNO3(aq) , and yellowish for 3AgNO3(aq)+Na3PO4(aq) >Ag3PO4(s)+3NaNO3(aq). A solid was formed in both situations due to a cloudy substance that formed with the silver nitrate + sodium iodide and the unknown + silver nitrate. It makes sense that a solid was formed becauseAgNO3(aq)+NaI(aq) >AgI(s)+NaNO3(aq)On the backward of the periodic table it states that if the anion I- is part of the sharpen and then cations desire Ag+ and Pb2+ will form a solid with it. Since it is AgI, a solid is clearly formed.When the unknown was tested with cobalt (II) nitrate a solid was not produced and there was no color change it was the kindred light pink as truely, similar to the description of the product of sodium iodide added to cobalt (II) nitrate which was same light pink as the original color of the mixture. A solid was not formed in both situations becauseCo(NO3)2(aq)+2NaI(aq) >CoI2(aq)+2NaNO3(aq)According to the back of the periodic table it states that if the anion iodine is part of the compound then cations like Ag+, Pb2+, Hg22+, and Cu+ will form a solid with it, but iodine with all other cations form an aqueous solution. Therefore, since cobalt is not listed as one of the cations that iodine forms a solid with, no solid would be formed which is exactly what happened when sodium iodide was added to cobalt (II) nitrate and no other combination with cobalt (II) nitrate. Since sodium iodide in combination with cobalt (II) nitrate was the only one involving cobalt (II) nitrate that did not have a reaction it proves that the unknown is one o f the two. But, since it has been proved that unknown is an anion, sodium iodide is our only option.When the unknown was tested with copper (II) nitrate a solid was produced, but also the color was a deplorable o head for the hills-yellow that was not too cloudy. This description is basically the same as the description of the product of sodium iodide added to copper (II) nitrate which was dark yellow/range tone, copper color, slightly cloudy. A solid was formed in both situations because it was slightly cloudy in both situations.Cu(NO3)2 (aq)+2NaI(aq) > CuI2(s)+2NaNO3(aq)According to the back of the periodic table it states that if the anion iodine is part of the compound then cations like Ag+, Pb2+, Hg22+, and Cu+ will form a solid with it. Therefore, since copper is listed as one of the cations that iodine forms a solid with, a solid would be formed which is exactly what happened when sodium iodide was added to copper (II) nitrate. Since sodium iodide in combination with copp er (II) nitrate was the only combination to form a dark yellow/orange color of all 4 anions in combination with the cation copper (II) nitrate sodium iodide seems the only option for the unknown. It is also not probable that the unknown is copper (II) nitrate because if there is a precipitate and a color change, it would not be the same color because different formulas yield different reactions like the colors formed. They are all unique.