Mole conversions - Study guides, Class notes & Summaries

Balancing Chemical Equations WEEK 3 1. Sodium bromide reacts with calcium chloride to form sodium chloride and calcium bromide. Which of these chemical reactions represents the balanced reaction? a. NaBr + CaCl2 → NaCl + CaBr2 b. 2 NaBr + CaCl2 → 2 NaCl + CaBr2 c. NaBr + 2 CaCl2 → NaCl + 2 CaBr2 d. NaBr + CaCl → NaCl + CaBr Balanced chemical equations have the same number of each atom on each side of the arrow. This equation is correct: 2 NaBr + CaCl2 2 → NaCl + CaBr2 2. The...

UCF-CHM2045C % comp - Answer- = (looking for)/ compound Percent Abundance - Answer- set to x and 1-x Molecular formula from empirical - Answer- Mass of molecular/ mass of empirical convert three elements - Answer- find the molar mass and divide by lowest mol Formula for conversions - Answer- Mass=(1/amu)=Mole=(Avo #/Mole)=Atom (2(compound)/Compound)=Atoms Atomic Number - Answer- Protons Mass # - Answer- Protons and neutrons Isotopes are what? - Answer- different neutrons...

Question 1 Complete the two problems below: 1. Convert 0. to exponential form and explain your answer. 2. Convert 5.82 x 103 to ordinary form and explain your answer. Answer: 1.Convert 0. = smaller than 1 = negative exponent, move decimal 5 places = 7.26 x 10-5 2.Convert 5.82 x 103 = positive exponent = larger than 1, move decimal 3 places = 5820 Question 2 Do the conversions shown below, showing all work: 1. 358oK = ? oC 2. 53oC = ? oF 3. 158oF = ? oK Answer: 1. 358oK - 273 = 85 ...

14.2 - As the relative costs of the reactor increase, which of the following are generally expected to be true for positive order reactions? (a) The reactor volume will need to be larger (b) Higher conversions will generally be optimal (c) A CSTR will be desirable over a PFR (d) Lower conversions will generally be optimal (e) None of the above are true - correct answer (d) Lower conversions will generally be optimal. If reactor costs are relatively expensive as compared to the othe...

Introduction to Chemical Reactions II Balancing Chemical Equations III Types of Reactions IV Oxidation and Reduction V The Mole and Avogadro’s Number VI Mass to Mole Conversions VII Mole Calculations in Chemical Equations VIII Mass Calculations in Chemical Equations IX Percent Yield X Limiting Reactants

Detailed notes on Unit 5 - Formulae, Equations and Amounts of Substance covering all the specification points. This allowed me to achieve an A* in chemistry. Covers formulae, definitions, empirical formula, calculations of molar mass, ionic compounds and equations, chemical equations and balancing, mole calculations, conversions, gas calculations, mass to mass calculations, mass to volume calculations, Avogadro's calculations, percentage yield calculations, atom economy, stoichiometry, limiting ...

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...

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...

In the chromatography of the reaction mixture, water absorbed on cellulose functioned as the stationary phase. What was the principal factor determining the migration of individual components in the sample? A) Hydrogen bonding B) Solute concentration C) Stationary phase concentration D) Thickness of paper - Answer A) Hydrogen bonding The answer to this question is A because the relative amount of hydrogen bonding to the stationary phase will determine the relative rate of migration of ...

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...