Quan Chapter 15: Redox Titrations 1. (A) Which of the following is true of a redox titration? (i) Since the redox reaction is spontaneous‚ the equilibrium constant does not have to be large for an effective titration. (ii) The titration effectiveness is increased when the two half-reaction potentials are far apart. (iii) Without an indicator the equivalence point cannot be detected. (iv) For an effective redox titration‚ the reducing agent must always be in the buret. (v) The equivalence
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forming when 0.5 M CaCl2 was added. This made us conclude the limiting reactant was in fact CaCl2. Introduction Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. Doing stoichiometry can calculate masses‚ moles‚ and percent’s with a chemical equation. The use of stoichiometry is how we were able to find the limiting reagent in this lab. We know that the limiting reagent is the chemical
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Unit 4: Solutions and Solubility Lesson 13 Key Questions: 42. a) Cloudy pond water – Heterogenous‚ light cannot pass though‚ can see different parts. b) Apple Juice – Solution‚ organic apple particles(solute)‚ and water(solvent). c) Rainwater – Solution‚ CO2‚ hydrogen(solute)‚ and water(solvent). d) 14-karat gold in jewelry – Heterogenous‚ light cannot pass through. Just because light can’t pass through‚ doesn’t mean it’s not a solution! Gold is a solid solution – it is different metals
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Ashley Silva Lab 7: Three Stooges in Chemical Reactions Objective: The purpose of this lab is to experimentally determine the equilibrium constant‚ Kc ‚ for the following chemical reaction: Fe3+ (aq) + SCN-(aq) ↔FeSCN2+(aq) Background Information: A system is at equilibrium when the rate of the forward reaction is the same as the rate of the reverse reaction. There is no change in concentration for the reactants or products at chemical equilibrium. When the system is disturbed there
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which ratio of NaClO to solution “B” was optimal. The ratio with the highest temperature change was 40mL of NaClO to 10 mL of solution “B”. Once simplified‚ this means the ratio is 4:1. Now that the mole ratio was determined it can be used in stoichiometry conversions. Introduction: In order to determine the mole ratio of the two reactants‚ NaClO and Solution B‚ the method of continuous variations was used. Method of continuous variations involves changing the ratios of the two reactants to find
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Chapter 6 practice MULTIPLE CHOICE 1. The formula for acetic acid‚ CH3CO2H‚ is an example of a(n) a.|condensed formula.| b.|mathematical formula.| c.|structural formula.| d.|molecular formula.| ANS: A OBJ: Goal #3.1: Interpret‚ predict‚ and write formulas for ionic and molecular compounds 3. Which of the following statements are correct? 1.|Metals generally lose electrons to become cations.| 2.|Nonmetals generally gain electrons to become anions.| 3.|Group 2A metals
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course that covers fundamental concepts of chemistry‚ including measurements‚ states of matter‚ atomic theory and structure‚ chemical bonding‚ properties of matter‚ solutions‚ acids and bases‚ gases‚ intermolecular forces‚ chemical reactions‚ stoichiometry‚ and thermochemistry. Attendance: Attendance at all lecture and laboratory sessions is mandatory. Students will be dropped from course for excessive absences without valid excuses. Also‚ you are expected to be on time everyday. Grading:
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including magnesium‚ methanol‚ and fire. This was to make what we are doing on paper in class a real life circumstance. Analysis questions 1) A) Part A 2Mg+O2 à(delta/heat) 2MgO Part B 2CuCO2 à(delta) 2Cu+CO2+O2 Part C Zn+HCl à H2 Part D CuCl2+2NaOHà 2NaCl+ Cu(OH)2
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0M NH3 .Cu(OH)2(s) (( Cu2+(aq) + 2OH-(aq) Ksp Cu(OH)2 4(NH3(( NH4 + OH-) Kb NH3 x 4 Cu2+ + 4 NH3(( [Cu(NH3)4]2+ 1/Kf[Cu(NH3)4]2+ Net Keq = (Ksp Cu(OH)2 x (Kb NH3 x 4))/ Kf[Cu(NH3)4]2+ = 4.619 x 10-53 f. 1.0M Na3PO4 3(Cu(OH)2(s) (( Cu2+(aq) + 2OH-(aq)) Ksp Cu(OH)2 2(PO43- + H2O((HPO42- + H3O+) 1/Ka3 H3PO4 2(HPO43- + H2O((H2PO42- + H3O+) 1/Ka2 H3PO4 2(H2PO43- + H2O((H3PO42- + H3O+) 1/Ka1 H3PO4 3Cu2+ + 2PO43-(( Cu3(PO4)2 1/Ksp Cu3(PO4)2
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5mL of CuSO4 in another test tube. Using a pipette‚ transfer the solution of the first test tube into another clean one. Observe the first one and the fresh one. D. Double Replacement: Pour 2mL of zinc acetate in a clean test tube and add 2mL of Na3PO4. Observe what happens. Pour 5mL of Na2S in a clean test tube and add 1mL of HCl. Observe what happens.
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