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    Discussion The purpose of this lab is to display the influence of retronasal olfaction when it comes to flavor. Humans have two different types of olfaction. Olfaction is the detection of chemicals in the air. These chemicals which are meant to activate odor receptors‚ are called odorants. Humans have approximately 20 million olfactory receptors‚ which is only a fraction compared to bloodhounds who have 4 billion. The first type of olfaction is called orthonasal olfaction which is inhaling. Here

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    test tube D and E. The metal spreader was then sterilized again for any future uses. These plates for C‚ D‚ E‚ and the streaking plate were left upside down‚ taped and labeled with the section and lab ID number and then incubated for 2 days at 30 degrees Celsius and refrigerated until the following lab

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    Single Displacement Reaction The objective of this experiment was to observe evidence of a reaction and to determine the chemical equation‚ if there is a reaction. In order for a reaction to occur‚ a metal strip must have evidence of black or grey deposits on the surface of the metal once dipped into the solution. The reactivity of an element is related to its tendency to lose or gain electrons. Three metal strips that were used are copper‚ lead and zinc as well as the solutions of silver nitrate

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    mistake was made. Results Reaction time to a second word or non-word is influenced by the previous word or non-word. The null hypothesis stated that all pairs would have the same mean reaction time‚ regardless of whether the first string was a word or non-word. The alternative hypothesis stated that the mean reaction time for word-non-word pairs would be slower than non-word-word pairs‚ which would be equal to the mean reaction time for non-word-non-word pairs.

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    b) Write the anode and cathode half-reactions‚ and the net cell reaction for the above diagram. [ T/I / 2 ] Oxidation: Zn (s) Zn2+ (aq) + 2e- Reduction: Cu2+ (aq) +2e- Cu(s) Zn (s) + Cu2+ (aq) +2e- Zn2+ (aq) + 2e- + Cu(s) Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu(s) c) Write the cell notation for the above galvanic cell drawn. [ T/I / 1 ] Cu(s) | Cu2+(aq) || Zn 2+(aq) | Zn (s) d) Calculate the percent error of the cell potential for the above galvanic cell. Show all

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    In order to first perform the reaction‚ 1.5 mL of 4-methylcyclohexanol was added to a pre-weighed 5mL conical vial. The conical vial was then weighed to find the actual mass of 4-methylcycloheaxanol. Then‚ 0.4mL of 85% phosphoric acid was added to the conical vial using a plastic pipet. Six drops of concentrated sulfuric acid was then added to the vial using a glass pipet. A spin vane was then added before adding the Hickman head‚ water condenser and a drying tube packed with calcium chloride to

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    this reaction at 25(C.  A.  1020 kJ/mol B.  -1.22 ( 103 kJ/mol C.  2.00 ( 103 kJ/mol D.  -1.42 ( 103 kJ/mol E.  -198 kJ/mol 2. For the reaction H2(g) + S(s) ( H2S(g)‚ (H( = -20.2 kJ/mol and (S( = +43.1 J/K·mol. Which of these statements is true?  A. The reaction is only spontaneous at low temperatures. B. The reaction is spontaneous at all temperatures. C. (G( becomes less favorable as temperature increases. D. The reaction is spontaneous

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    From the results recorded it was identified that activity A and C had exothermic reactions‚ whilst activity B had an endothermic reaction. Each of the activities had initially begun with 23.4/8/9oC‚ however‚ within one minute of adding the 2nd reactants‚ the results began showing‚ as seen from the graph. Activity A and C had risen in temperature‚ with A gaining 4oC and in total of 2 minutes having the temperature of 27.9oC along with C increasing by 2.2oC and within 2 minutes of the experiment having

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    Melissa Carrithers AP Chemistry 12/17/14 Rate Law Determination Pre-Lab Purpose: The purpose of this lab is to observe the reaction between sodium hydroxide and crystal violet so that the order of the reaction could be determined along with the rate constant‚ k‚ and the half-life for this reaction. Hypothesis: If the reaction between crystal violet and sodium hydroxide reacts appropriately‚ then the order will be first order. Procedure: Mix 10 mL of sodium hydroxide and crystal violet solution

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    Step 1: Pb (NO3) 2 (aq) + CaCl2 (aq)  PbCl2 (aq) + Ca (NO3) 2 (aq) (double displacement reaction) According to the solubility guidelines lead (II) chloride (PbCl2) is a possible precipitate. This is because even though most chlorides are soluble‚ lead chloride is considered insoluble (p.2‚ Lesson 17). Step 2: PbCl2 ↔ Pb2+ + 2Cl- Q = [Pb2+] [Cl-] 2 Step 3: Ksp = 1.2 x 10-5 (from table 17.1‚ p.5) Step 4: V2 = 20.0 mL (volume of Pb (NO3)2) + 45.0 mL (volume of CaCl2) = 65.0

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