Chem 103 lab exam 5 - Study guides, Class notes & Summaries

In the Boyle’s law experiment, how was the volume of gas at different pressures determined? 2. (2 points) According to Boyle’s law, as the volume of a gas decreases, the pressure _________. 3. (6 points) A sample of gas occupies a volume of 10.5 L when the pressure is 14.3 psi. Use Boyle’s Law to calculate the volume (L) when the pressure is decreased to 12.8psi. Show the calculation. Answer: 1. A syringe was used, and the volume was read from the graduation markings. 2. increase...

CHEM 103 Lab Exam 5 – Portage Learning. In the Boyle’s law experiment, how was the volume of gas at different pressures determined? 2. (2 points) According to Boyle’s law, as the volume of a gas decreases, the pressure _________. 3. (6 points) A sample of gas occupies a volume of 10.5 L when the pressure is 14.3 psi. Use Boyle’s Law to calculate the volume (L) when the pressure is decreased to 12.8psi. Show the calculation. Answer: 1. A syringe was used, and the volume was read f...

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 5/Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem 103 Lab Exam 5Chem ...

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 (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 (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 ...