Chemistry 103 lab 3 exam Study guides, Class notes & Summaries

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CHEM 103 Lab Exam 3 – Portage Learning. A 1.1000 gram hydrate sample chosen from Na2CO3∙10H2O, AlCl3∙6H2O, MgCl2∙6H2O and BaCl2∙2H2O was heated and found to lose 0.5841 gram of H2O. (1) Show the calculation of the % H2O in the unknown hydrate 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. MWs: Na2CO3∙10H2O = 286.15, AlCl3∙6H2O = 241.43, MgCl2∙6H2O = 203.301 ...

Chem 103 portage learning Lab Exam 3

CHEM 103 Module 2 Exam (Portage Learning) Show the calculation of the molecular weight for the following compounds, reporting your answer to 2 places after the decimal. 1. Al (SO ) 2. C H NOBr 2 4 3 7 5 1. Al (SO ) Al = 2 x 26.98 = 53.96 S = 3 x 32.07 = 96.21 O = 12 x 16.00 = 192 2 4 3 M2: Exam - Requires Respondus LockDown Browser + Webcam: General Chemistry I w/Lab-2021- Kozminski 3/18 Total Molecular Weight for Al (SO ) = 342.17 2. C H NOBr C = 7 x 12.01 = 84.07 H = 5 x 1.008 ...

CHEM 103 Module 2 Exam (General Chemistry) – Portage Learning. Al (SO ) Al = 2 x 26.98 = 53.96 S = 3 x 32.07 = 96.21 O = 12 x 16.00 = 192 2 4 3 M2: Exam - Requires Respondus LockDown Browser + Webcam: General Chemistry I w/Lab-2021- Kozminski 3/18 Total Molecular Weight for Al (SO ) = 342.17 2. C H NOBr C = 7 x 12.01 = 84.07 H = 5 x 1.008 = 5.04 N = 1 x 14.01 = 14.01 O = 1 x 16.00 = 16.00 Br = 1 x 79.90 = 79.90 Total Molecule Weight for C H NOBr= 199.02

CHEM 103 Module 2 Exam (General Chemistry) – Portage Learning. Show the calculation of the molecular weight for the following compounds, reporting your answer to 2 places after the decimal. 1. Al (SO ) 2. C H NOBr 2 4 3 7 5 1. Al (SO ) Al = 2 x 26.98 = 53.96 S = 3 x 32.07 = 96.21 O = 12 x 16.00 = 192 2 4 3 M2: Exam - Requires Respondus LockDown Browser + Webcam: General Chemistry I w/Lab-2021- Kozminski 3/18 Total Molecular Weight for Al (SO ) = 342.17 2. C H NOBr C = 7 x 12.01 ...

CHEM 103 Lab Exam 3 – Portage Learning. A 1.1000 gram hydrate sample chosen from Na2CO3∙10H2O, AlCl3∙6H2O, MgCl2∙6H2O and BaCl2∙2H2O was heated and found to lose 0.5841 gram of H2O. (1) Show the calculation of the % H2O in the unknown hydrate sample. (2) Show the calculation of the % H2O in each of the hydrate compounds and identify the unknown hydrate from the list.

The sample was coated with carbon in the SEM experiment. Answer the following question. A 1.1000 gram hydrate sample chosen from Na2CO3∙10H2O, AlCl3∙6H2O, MgCl2∙6H2O and BaCl2∙2H2O was heated and found to lose 0.5841 gram of H2O. (1) Show the calculation of the % H2O in the unknown hydrate 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. MWs: Na2CO3∙10H2O = 286.15,...

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CHEM 103 Lab Exam 3 – Portage Learning. Why was the crucible cooled in a desiccator? 2. With what was the sample coated in the SEM experiment? 1. The crucible cooled in a desiccator to prevent water from being picked up. 2. The sample was coated with carbon in the SEM experiment. Answer the following question. A 1.1000 gram hydrate sample chosen from Na2CO3∙10H2O, AlCl3∙6H2O, MgCl2∙6H2O and BaCl2∙2H2O was heated and found to lose 0.5841 gram of H2O. (1) Show the calculation of t...