CHEM 103 Chemistry (CHEM103)

CHEM 103 Lab Exam 5 – Portage Learning . 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.

CHEM 103 Lab Exam 4 – Portage Learning. Does the following reaction give off heat or absorb heat? Explain briefly. C D H = +263Kcal/mol 3. (4 points) In one experiment in the video, room temperature water was mixed with water at about 38 degrees. To what temperature (about) did the mixture of the two water samples come to after setting for a few minutes? Answer: 1. The heat depends on the amount of substance. If the amount of substance is doubled, then the heat doubles. Other propertie...

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 ...

Show the calculation of the molecular weight for the following compounds, reporting your answer to 2 places after the decimal. 1. (NH4)2CO3 2. C8H6NO4Br

Chem 103 portage learning Lab Exam 3

Chem 103 portage learning exam 1

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 portage learning 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 1 to 6 Exam Answers 2021/2022 Portage learning. MODULE 1 EXAM Question 1 Click this link to access the Periodic Table. This may be helpful throughout the exam. 1. Convert 845.3 to exponential form and explain your answer. 2. Convert 3.21 x 10-5 to ordinary form and explain your answer. 1.Convert 845.3 = larger than 1 = positive exponent, move decimal 2 places = 8.453 x 102 2.Convert 3.21 x 10-5 = negative exponent = smaller than 1, move decimal 5 places = 0. Question 2 ...