White precipitate shows the presence of chloride (Cl-). Chloride anion equation: HCl(aq) + AgNO3 (aq) → HNO3 (aq) + AgCl(s). The nitrate anion test involves cooling a mixture containing 1 mL of test solution and 3mL 18M H2SO4. 2mL is poured down the inner test tube side and the presence of a brown ring shows nitrate (NO3-) to be present. The carbonate anion test mixes 1 mL of test solution and drops of 6M HCl.
h) A way to make hard water softer is to put an sodium nitrate and create a precipitate to mellow out the reaction. Another way of making it softer is by removing the calcium ions one way of doing that is by boiling the solution to take out some of the ions. Conclusion: Overall, we determined that sodium carbonate, Na2CO3, is the anion that can be used to precipitate the most metal cations. Also, we learned that the anion sodium chloride, NaCl, could be used to remove silver ions from solutions. The stuff that I found interesting was that how many colours you can get when you mix the cations and anions
Add 1 mL of deionized water to the small test tube containing the precipitate and mix it and centrifuge it for 60 seconds. Then, add the supernatant into the boiling test tube and repeat this step one more time with another 1 mL of deionized water. Acquire a pair of metal test tube holders and heat the boiling test tube to evaporate the water for 15 minutes. Let is cool after and weigh it. Then, calculate a percent yield of zinc iodide and write a balanced chemical equation and determine the limiting
The reaction that occurred with this step was displacement and metathesis in the form of gas formation. The balanced equation of this step looks as follows: CuSO4aq+Zns→Cus+ZnSO4(aq) Once this step was finished, the remaining copper was retrieved. First, to recover the copper HCl was added to remove all the zinc. When this happened, a yellow tint was observed in the liquid, as well as bubbling as the zinc was broken down. Once the copper dried out, it was weighed and came to a total of 240 mg.
Halides Lab: Background information: Halide ions are reactive and useful. Salts are positively charged ions (metals) combined with any negative ions (nonmetal), and when placed in a solution (water) it separates into the cations and anions that made it up. The Purpose of this lab is to find out how the Halides react with the indicators, and to determine the identity of the two unknown solutions (A and B). Color of solutions prior to experiment: NaF | NaCI | KBr | KI | Unknown A | Unknown B | clear | clear | clear | clear | clear | clear | Color of indicator prior to experiment: 5% Bleach (NaOCI) | 0.2 M Na2S2O3 | 0.1 M AgNO3 | 0.5 M Ca(NO3)2 | clear | clear | clear | clear | Halide solutions | NaF | NaCI | KBr | KI | unknown A | unknown B | Test 1: Ca(NO3)2 | Cloudy White (Nothing) | Clear | Nothing | light yellow (Nothing) | Nothing | Nothing | Test 2, Part A: AgNO3 | clear (Nothing) | Milky White | Gold (Cloudy yellow) | milky green (Cloudy yellow) | turned white, film developed on top layer | Milky | Test 2, Part B: add Na2S2O3 to test tube from part A | Dark Orange/brown | Clear | Dark Gold(precipitate yellow then clear) | milky green (no change) | white precipitation, settled on bottom | Milky | Test 3: NaOCI (Bleach) | Clear (Nothing) | Nothing | Nothing | Orange (Clear) | Nothing | Orange | Unknown A is identified as NaCI (Sodium Chloride), because in test#1 the solution turned a cloudy white color when Ca(NO3)2 (Calcium nitrate) was added. In the first part of test#2, when AgNO3 (Silver nitrate) is added, the solution turned white, with a thin layer of film developing on the surface.
Take sample of Zinc Iodide and dissolve in solution. Then, take the battery, with exposed wire tips attached, and place the tips into the solution. In about 1-2 minutes, there should be a red-brown color coming out of one of the wires, and a dark substance depositing on the other wire, these should iodine and zinc respectively. 2. What are the identities of the substances found after electrolysis and heating?
William Flores-Paz Monitoring Acid-Base Titrations with a pH meter October 30, 2013 Introduction The purpose of this experiment was to record the volume of HCl and acetic acid with a NaOH with a known molarity. This data would allow us to create a graph so we could compare the two titration curves. We would then use these curves to calculate the unknown molarities. HCl + NaOH >H2O + NaCl This equation shows the relationship between the acid and the base then the reaction goes forward and they are titrated creating salt and water. This particular reaction is a strong acid and a strong base which means that when the reaction reaches the equivalence point, the moles of the acid and the base are equal and the solution is neutral so the pH should be around 7.0 depending on the final volume of each solution.
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.
| Centrifuge: used to isolate the solid from the solution | . Focus Questions: When acidified water is not used in the zinc and iodine synthesis, zinc hydroxide is formed. How can it be identified based on the amphoteric nature. When zinc and iodine react without the presence of acetic acid in the water, zinc hydroxide forms. This is evident because the substance tested positive for both the acid test and the base test for zinc hydroxide.
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.