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CHEM 103 Lab Exam 3 (100 out of 100) Questions and Verified Answers (Portage Learning)
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CHEM 103 Lab Exam 3 (100 out of 100) Questions and Verified Answers (Portage Learning)
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This study source was downloaded by 100000880877832 from CourseHero.com on 02 -15-2024 05:18:10 GMT -06:00 https://www.coursehero.com/file/199926914/CHEM -103-Lab-Exam -3docx/ Question 1 pts Question 2 pts Question 3 pts CHEM 103 Lab Exam 3 (100 out of 100) Questions and Verified Answers (Portage Learning) Scientific Calculator Answer the following two questions: 1. Why was the hydrate heated with the Fisher burner in part 2 of lab 3? 2. For what purpose was SEM carried out in lab 3? 1. The hydrate was heated with the Fisher burner in lab 3 to cause loss of water. 2. The SEM was carried out in lab 3 to detect the elemental composition. 2. The SEM was carried out in lab 3 to detect the elemental composition. (We did not use it for imaging) Scientific Calculator Answer the following question. A 1.1000 gr am hydrate sample chosen from Na 2CO3∙10H 2O, AlCl 3∙6H 2O, MgC l2∙6H 2O and BaCl2∙2H2O was heated and found to lose 0.6920 gr am of H2O. (1) Show the calculation of the % H2O in the unknown hydr ate sample. (2) Show the calculation of the % H2O in each of the hydrate compounds and identify the unknown hydrate from the list. Atomic weights: H = 1.008, O = 16.00. M Ws: Na2CO3∙10H2O = 286.15, AlCl3∙6H 2O = 241.43, MgC l2∙6H 2O = 203.301 and BaCl2∙2H2O = 244.462 % H2O in unknown = (0.6920 / 1.1000) x 100 = 62.91% % H2O in Na2CO3∙10H2O = 180.16 / 286.15 x 100 = 62.96% unknown is Na2CO3∙10H2OThe benefits of buying summaries with Stuvia:
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