something + 2Fe+2 Equation 2: 8H+ + 5Fe+2 + MnO4- --> 5Fe+3 + Mn+2 + 4H2O Equation 3: 6H+ + 2MnO4- + 5H2C2O4 --> 2Mn+2 + 10CO2 + 8H2O Conclusion: Therefore the concluded reaction would be: NH3OH+ + 2Fe+3 --> N2O + 2Fe+2 This was obtained by using stoichiometry half reactions the product of that reaction was determined to be N2O. Some systematic errors could be if the wrong molarity was determined for the permanganate because then that would though off the calculations for the Fe+2 and the rest of the
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Name: Brian James|Date:3/10/13| Exp 9: Stoichiometry of a Precipitation Reaction|Lab Section: 73426| Data Tables: Step 3: Show the calculation of the needed amount of Na2CO3 CaCl2.H2O(aq)= m/M =1/147 =0.0068 mol CaCO3(s)=0.0068*1/1 =0.0068 mol CaCO3(s)= CaCO3 (s)= CaCO3 mol *CaCO3 g =0.0068 mol*100.01 g =.68 g Step 4: Mass of weighing dish _0.6_g Mass of weighing dish and Na2CO3 .72_g Net mass of the Na2CO3 .12_g Step 6: Mass of filter paper
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ANSWER KEY FOR TEST 2 DECEMBER 2011 ECON 0110: MACROECONOMICS DR. KENKEL 25 questions worth 10 points each. Total Points = 250 1. Assume the following data apply to a certain country using 1996 as the base year: Year Consumer Nominal Price GDP Index (Trillions) 2005 144.9 12.2 2006 155.6 14.8 2007 169.5 17.5 The growth rate of real GDP between 2005 and 2006 is a. Less than 5.0% b. 5.0% to less than 6.0% c. 6.0% to less than 7.0% d. 7.0% to less than
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moles-stoichiometry-practice-problems Now you’re ready to use what you know about conversion factors to solve some stoichiometric problems in chemistry. Almost all stoichiometric problems can be solved in just four simple steps: 1.Balance the equation. 2.Convert units of a given substance to moles. 3.Using the mole ratio‚ calculate the moles of substance yielded by the reaction. 4.Convert moles of wanted substance to desired units. These "simple" steps probably look complicated at first
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combining the elements‚ instead of having it naturally do so‚ made a significantly fully more bright color. In Theophilus’s “recipe” for vermillion‚ it is stated that there should be far more sulfur than a stiochiometric reaction should require. Stoichiometry
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Understanding Academic Integrity Answer Key Speakers: Narrator‚ Student This answer key reviews correct and incorrect feedback for the Understanding Academic Integrity Tutorial scenario questions. Scenario 1 Student: I’m taking this course a second time‚ so I’m just going to reuse my paper from the last time I took it. I wrote the paper‚ so it doesn’t make sense for me to write the same paper all over again. What should I do? (Option 1) I should probably ask my instructor if I can reuse my paper
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BIOE 301/362 Name: Exam 2 March 13‚ 2008 The exam consists of 10 questions. Show all work to receive credit. Clearly organize your work and draw a box around your final answers. NEATNESS COUNTS! Good Luck! Problem 1 (5): Problem 2 (18): Problem 3 (12): Problem 4 (15): Problem 5 (8): Problem 6 (12): Problem 7 (12): Problem 8 (18): Extra Credit (4): Total (100): PROBLEM 1: Development of technology
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Bio 101 Review Sheet Test #3 (Chapters 7‚8‚9) Chapter 7 1. 3 effects of mutations a. Good‚ bad‚ silent i. What silent is in terms of amino acids 2. Point mutation 3. Frameshift mutation 4. Main causes of mutation of DNA 5. Which mutations are heritable 6. Definition of allele b. How process of mutation in replication leads to new alleles 7. Transgenic organism c. What it is d. How its created e. Definition of recombinant
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Title: Stoichiometry of a Precipitation Reaction Purpose: The purpose is to predict the amount of precipitation using stoichiometry. Another purpose is to measure and calculate percentage yield. Procedure: 1. Weigh out 1.0g of CaCl2*2H2O and put it into a 100mL beaker. 2. Add 25mL of distilled water and stir. 3. Using stoichiometry to determine how much Na2CO3 is needed for a full reaction. 4. Weigh the calculated amount and put it in a small paper cup. Add 25mL distilled water and stir
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STOICHIOMETRY OF GASOLINE. AN INTRODUCTION The internal combustion engines burn fuel to create kinetic energy. The burning of fuel is basically the reaction of fuel with oxygen in the air to form water and Carbon dioxide as the major end product . The amount of oxygen present in the cylinder is the limiting factor for the amount of fuel that can be burnt that is to say it determines the level of burning in our combustion engine. If there’s too much fuel present‚ not all fuel will be burnt and un-burnt
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