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

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...

CHEM 103 LAB EXAM 3 QUESTIONS AND ANSWERS Question 110 / 10 ptsScientific Calculator Answer the following two questions: 1. Why was H2SO4added to the metal carbonate in part 1 of lab 3? 2. Why were gloves used during the ninhydrin spray procedure? 1. H2SO4was added to the metal carbonate in lab 3 to cause the loss of CO2 2. Gloves were used during the ninhydrin spray procedure to prevent the reagent from reacting with the dead skin amino acids and causing purple marks on the skin. ...

Chem 103 Lab Final Exam with Complete Solutions **What physical properties were used to differentiate between alkanes and alkenes?** Boiling points and density. **What color does potassium permanganate turn in Baeyer's test?** Purple. **What color does bromine appear in the bromine test?** Yellowish-orange. **Which type of hydrocarbon gives a positive result in Baeyer's and/or bromine test?** Alkenes. **What solvent was utilized for the extraction of curcumin?** I...

Chem 103 Lab Exam 3 Questions & Answers Oxygen, O2 - ANSWERis a colorless, odorless gas that supports combustion. A glowing splint will brighten or burst into flame when thrust into the conatiner. Hydrogen, H2 - ANSWERis a colorless, odorless gas that burns in air. It is less dense than air and can be captured in an inverted vessel. A confined mixture of this and air (in an inverted test tube) will ignite with a "pop" when exposed to a flame. By contrast, pure __ will burn quietly with a b...

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. 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 2 4 3 7 5 7 5 1. 2Al + 3S + 12O = 342.17 2. 7C + 5H + N + O + Br = 199.02 Question 2 10 / 10 pts You may find the following resources helpful: Scientific Calculator ( Periodic Table ( Equation Table ( downloa...

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 (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 ...