Mole conversions - Study guides, Class notes & Summaries

CHEM 103 MODULE 1 EXAM 2023/2024 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 = 0.0000321 Question 2 Click this link to...

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 ...

CHEM120 INTRODUCTION TO GENERAL, ORGANIC & BIOLOGICAL CHEMISTRY 2024| LAB REPORT FOR WEEK 3 SCORED A+ OL Lab 3: Stoichiometry: Avogadro’s number and molecular calculations / Solution Preparation: From salt to solution Learning Objectives: • Explain the relationship between mass, molecular weight, and numbers of atoms or molecules and perform calculations deriving these quantities from one another • Perform mass-to-mass stoichiometric calculations via conversions to moles • Ident...

Exam CHEM 103 Final Exam (Version-1, Latest-2023)/ CHEM103 Final Exam / CHEM 103 General Chemistry Final Exam/ CHEM103 General Chemistry Final Exam: Portage Learning |100 % Correct Q & A| Final Exam - Requires Respondus LockDown Browser + Webcam Question 1 Complete the two problems below: 1. Convert 0.0000726 to exponential form and explain your answer. 2. Convert 5.82 x 10 3 to ordinary form and explain your answer. Answer: 1.Convert 0.0000726 = smaller than 1 = negative exponent,...

General Chemistry 1103 Exam 1 With 100% Correct Answers 2023 Mega- 1.0x10^6 (M) Kilo- 1.0x10^3 (k) Deci- 1.0x10^-1 (d) Centi- 1.0x10^-2 (c) Milli- 1.0x10^-3 (m) Micro- 1.0x10^-6 (μ) Nano- 1.0x10^-9 (n) Pico- 1.0x10^-12 (p) Conversion Factors 1kg = 2.205lbs, 1oz. = 28.35g 1mi = 1.609km, 1in = 2.54cm 1gal = 3.785L, 1L = 1.06qts 1g = 1ml = 1cm^3 1hr = 3600s Kelvin to Celsius Conversions K = C+273.15 C= K-273.15 C...

CHEM 103 FINAL 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 10 2 2. Convert 3.21 x 10 -5 = negative exponent = smaller than 1, move decimal 5 places = 0.0000321 Question 2 Click this link to access the Periodic ...

CHEM 103 FINAL EXAM.Click this link to access the Periodic Table. This may be helpful throughout the exam. Do the conversions shown below, showing all work: 1. 246oK = ? oC 2. 45oC = ? oF 3. 18oF = ? oK 1. 246oK - 273 = -27 oC oK → oC (make smaller) -273 2. 45oC x 1.8 + 32 = 113 oF oC → oF (make larger) x 1.8 + 32 3. 18oF - 32 ÷ 1.8 = -7.8 + 273 = 265.2 oK oF → oC → oK Question 3 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the ca...

Portage Learning CHEM 103 Final Exam Study Guide Latest Updated 2023 Question 1 Show the calculation of the percent of each element present in the followingcompounds. Report your answer to 2 places after the decimal. 1. (NH4)2CrO4 2. C8H8NOI 1. %N = 2 x 14.01/152.08 x 100 = 18.43% %H = 8 x 1.008/152.08 x 100 = 5.30% %Cr = 1 x 52.00/152.08 x 100 = 34.20% %O = 4 x 16.00/152.08 x 100 = 42.08% 2. %C = 8 x 12.01/261.05 x 100 = 36.80% %H = 8 x 1.008/261.05 x 100 = 3.09% %N = 1 x 14.01...

CHEM 103 FINAL EXAM.Click this link to access the Periodic Table. This may be helpful throughout the exam. Do the conversions shown below, showing all work: 1. 246oK = ? oC 2. 45oC = ? oF 3. 18oF = ? oK 1. 246oK - 273 = -27 oC oK → oC (make smaller) -273 2. 45oC x 1.8 + 32 = 113 oF oC → oF (make larger) x 1.8 + 32 3. 18oF - 32 ÷ 1.8 = -7.8 + 273 = 265.2 oK oF → oC → oK Question 3 Click this link to access the Periodic Table. This may be helpful throughout the exam. Show the ca...

● Stoichiometry is the use of quantitative relationships between substances that are involved in chemical reactions which determine the amounts of the reactants or the products (Bauer, Birk, & Mar ks, 2019). Stoichiometry - the process of determining the amounts of substances in a chemical reaction (Bauer, Birk, & Marks, 2019). ● It is used to relate the reactants and products by using calculations and balanced equations (Bauer, Birk, & Marks, 2019). ● By knowing the amount of a reactant o...