Solutions
Chemistry the partial vapour pressure of each component of the solution is Another important class of solutions consists of solids dissolved in liquid ...
D:TEXTBOOKSRATIONALISED 20222-23NehaP85
V. 0.130. = 3.17 V – 0.21V = 2.96 V. Solution. Rationalised 2023-24. Page 10. 40. Chemistry.
Section 4.1: Types of Chemical Bonds
Page 12. Copyright © 2012 Nelson Education Ltd. Chapter 4: Chemical Bonding 7. Answers will vary. Sample answer: Coordinate covalent bonding is like you ...
Chem 12 SM Ch5 Review final new ok revised
Δ = −. Statement: The enthalpy change ΔH
Chem12 SM Ch05 Section5 2 final ok revised
Tutorial 3 Practice page 304. 1. (a) Solution: Step 1: Write the balanced chemical equation without the energy term. 2 C2H2(g) + 5 O2
Grade 12 Chemistry: A Foundation for Implementation
Jan 21 2011 ... textbook. The successful implementation of Grade 12 Chemistry depends ... solutions based on this result? Explain. 3. Are there any reactions ...
Chem12 SM Ch7 Section7.5 final ok revised
Solution: Step 1. Convert the given initial amount of cyclopropane gas to (c) The chemical system is not at equilibrium. We have been given K = 1.2 × 102 ...
Section 8.4: Calculations Involving Acidic Solutions
Section 8.4: Calculations Involving Acidic Solutions. Tutorial 1 Practice page x2 ≈ 6.2 × 10−12 x = [H+(aq)]. ≈ 2.49 × 10−6 mol/L. pH = −log(2.49 × 10 ...
Chem12 SM Ch8 Section8e1 final ok revised
solution; an Arrhenius base forms hydroxide ions in aqueous solution. (b) A Brønsted–Lowry acid is a proton donor; a Brønsted–Lowry base is a proton.
Grade 12 Chemistry: A Foundation for Implementation
Jan 21 2011 promotes the idea that all answers are enshrined in a textbook. The successful implementation of Grade 12 Chemistry depends on a ...
Chem12 SM Ch7 Section7.6 final revised
Chapter 7: Chemical Equilibrium 7.6-1 Write a balanced equation for the solution equilibrium. Ca3(PO4)2(s) ! "!# !! ... Ksp of CuI(s) = 1.3 × 10?12.
Chem12 SM Ch05 Section5 2 final ok revised
Chapter 5: Thermochemistry. 5.2-3. Solution: Step 1: Determine the change in temperature AT . final initial. 27.8 °C 19.8 °C.
Chem 12 SM Ch5 Review final new ok revised
(a) Answers may vary. Sample answer: Hydrogen gas has a high enthalpy of combustion releasing about 2.5 times the quantity of energy per gram than methane but
Solutions
Chemistry. Type of Solution. Solute. Solvent. Common Examples. Gaseous Solutions. Gas. Gas. Mixture of oxygen and nitrogen gases.
Chem12 SM Ch7 Section7.5 final ok revised
Chapter 7: Chemical Equilibrium Solution: N2O4(g) ! "!# !! 2 NO2(g) ... Using the balanced chemical equation construct an ICE table for calculating.
Chem12 SM Ch05 Section5.4 final ok revised
An enthalpy diagram of the reaction is: Page 2. Copyright © 2012 Nelson Education Ltd. Chapter 5: Thermochemistry. 5.4-2. 2. (a) Solution: Step 1: Label the
chemistry grade 12 studen textbook
CHEMISTRY GRADE 12. 2. MAIN CONTENTS. 1.1 Homogeneous and Heterogeneous Mixtures. 1.2 Types of Solutions. 1.3 The Solution Process.
EngLinks
Grade 12 Chemistry Review Workbook. 1 of 22. 1. Organic Chemistry. General IUPAC rules for naming organic compounds: 1) The lowest numbers possible must be
Chem12 SM Ch05 Section5 3 final ok revised
combustion of ethene gas to gaseous carbon dioxide and liquid water is an exothermic reaction. 3. Solution: Step 1: Use the balanced chemical equation:.
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Copyright © 2012 Nelson Education Ltd. Chapter 5: Thermochemistry 5.3-1 Section 5.3: Bond Energies Tutorial 1 Practice, page 312 1. There are 3 C-H bonds and 1 C-Cl bond in chloromethane, CH3Cl(g). Given: CH
3 moln
; CH413 kJ/molD
; CCl1 moln
; CCl339 kJ/molD
Required: HΔ
Analysis:
!H=n #D bonds brokenSolution: bonds broken
CHCC l
3 mol1 mol
3 mo l
HnD DD413 kJ
mol×1 mo l+
339 kJ
mol1578 kJHΔ=
Statement: The energy required to separate 1 mol of chloromethane into free atoms is 1578 kJ. 2. Solution: Step 1: Write the balanced chemical equation for the combustion of 1 mol of ethene gas. C2H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2O(l) Step 2: Determine the bonding of each substance by drawing structural formulas for each molecule in the reaction. Step 3: Determine the number of moles of reactants and products, the number of moles of bonds broken or formed, and the molar bond energy for each. Organize this information in a table. Substance Number of bonds per mole (nsubstance) Amount of bonds in reaction Bond energy per mole reactants C2H4 4 mol C-H bonds 1 mol C=C bonds 4 mol 1 mol 413 kJ/mol 614 kJ/mol O2 1 mol O=O bonds 3 mol 495 kJ/mol products CO2 2 mol C=O bonds 4 mol 799 kJ/mol H2O 2 mol O-H bonds 4 mol 467 kJ/mol
Copyright © 2012 Nelson Education Ltd. Chapter 5: Thermochemistry 5.3-2 Step 4: Calculate the enthalpy change, ΔH, of the reaction.
!H=n #D bonds broken $n #D bonds formed =4 mol#D C$H +1 mol#D C=C +3 mol#D O=O4 mol#D
C=O +4 mol#D O$H =4 mol#413 kJ
mol +1 mol#614 kJ
mol +3 mol#495 kJ
mol4 mol#
799 kJ
mol +4 mol#467 kJ
mol 0 0 =1652 kJ+614 kJ+1485 kJ$(3196 kJ+1868 kJ) !H=$1313 kJStatement: Since enthalpy change, ΔH, of the reaction is negative, the complete combustion of ethene gas to gaseous carbon dioxide and liquid water is an exothermic reaction. 3. Solution: Step 1: Use the balanced chemical equation: N2H2(g) + F2(g) → N2(g) + 2 HF(g) Determine the bonding of each substance by drawing structural formulas for each molecule in the reaction. Step 2: Determine the number of moles of reactants and products, the number of moles of bonds broken or formed, and the molar bond energy for each. Organize this information in a table. Substance Number of bonds per mole (nsubstance) Amount of bonds in reaction Bond energy per mole reactants N2H2 2 mol N-H bonds 1 mol N=N bonds 2 mol 1 mol 391 kJ/mol 418 kJ/mol F2 1 mol F-F bonds 1 mol 154 kJ/mol products N2 1 mol N≡N bonds 1 mol 941 kJ/mol HF 1 mol H-F bonds 2 mol 565 kJ/mol
Copyright © 2012 Nelson Education Ltd. Chapter 5: Thermochemistry 5.3-3 Step 3: Calculate the enthalpy change, ΔH, of the reaction. ()
bonds brokenbonds formedNHNN FF
NNHF2 mol1 mol1 mol
1 mol 2 mol
2 mo l
HnDnD DDD DD391 kJ
mol×1 mo l
418 kJ
mol×1 mo l
154 kJ
mol1 mo l
941 kJ
mol×2 mo l
565 kJ
mol782 kJ418 kJ154 kJ (941 kJ1130 kJ)
717 kJH
Statement: When enthalpy change, ΔH, has a negative value, there is more energy released by the formation of the bonds in the products than is absorbed by the breaking of the bonds in the reactants. Therefore, thermal energy is released to the surroundings, and the reaction is exothermic. 4. Solution: Step 1: Write the equation that represents the complete combustion of propane. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) Step 2: Determine the number of moles of reactants and products, the number of moles of bonds broken or formed, and the molar bond energy for each. Organize this information in a table. Substance Number of bonds per mole (nsubstance) Amount of bonds in reaction Bond energy per mole reactants C3H8 8 mol C-H bonds 2 mol C-C bonds 8 mol 2 mol 413 kJ/mol 347 kJ/mol O2 1 mol O=O bonds 5 mol 495 kJ/mol products CO2 2 mol C=O bonds 6 mol 799 kJ/mol H2O 2 mol O-H bonds 8 mol 467 kJ/mol Step 3: Calculate the enthalpy change, ΔH, of the reaction. ()
bonds brokenbonds formedCHCC OO
COHO8 mol2 mol5 mol
6 mol8 mol
8 mo l
HnDnD DDD DD413 kJ
mol×2 mo l
347 kJ
mol×5 mo l
495 kJ
mol6 mo l
799 kJ
mol×8 mo l
467 kJ
mol3304 kJ694 kJ2475 k J(4794 kJ3736 kJ)
2057 kJH
Statement: The quantity of energy released for the complete combustion of 1 mol of propane is 2057 kJ.
Copyright © 2012 Nelson Education Ltd. Chapter 5: Thermochemistry 5.3-4 Section 5.3 Questions, page 313 1. (a) Solution: Step 1: Use the balanced chemical equation to determine the bonding of each substance. H2(g) + Cl2(g) → 2 HCl(g) Step 2: Determine the number of moles of reactants and products, the number of moles of bonds broken or formed, and the molar bond energy for each. Organize this information in a table. Substance Number of bonds per mole (nsubstance) Amount of bonds in reaction Bond energy per mole reactants H2 1 mol H-H bonds 1 mol 432 kJ/mol Cl2 1 mol Cl-Cl bonds 1 mol 239 kJ/mol products HCl 1 mol H-Cl bonds 2 mol 427 kJ/mol Step 3: Calculate the enthalpy change, ΔH, of the reaction. bonds brokenbonds formed
HHCl ClHCl
1 mol1 mol2 mo l
1 mo l
HnDnD DDD432 kJ
mol×1 mo l
239 kJ
mol×2 mo l
427 kJ
mol432 kJ239 kJ854 k J
183 kJH
Statement: The ΔH for the reaction is -183 kJ. (b) Solution: Step 1: Use the balanced chemical equation to determine the bonding of each substance. N2(g) + 3H2(g) → 2 NH3(g) Step 2: Determine the number of moles of reactants and products, the number of moles of bonds broken or formed, and the molar bond energy for each. Organize this information in a table. Substance Number of bonds per mole (nsubstance) Amount of bonds in reaction Bond energy per mole reactants N2 1 mol N≡N bonds 1 mol 941 kJ/mol H2 1 mol H-H bonds 3 mol 432 kJ/mol products NH3 3 mol N-H bonds 6 mol 391 kJ/mol Step 3: Calculate the enthalpy change, ΔH, of the reaction. bonds brokenbonds formed
NNHH NH
1 mol1 mol2 mo l
1 mo l
HnDnD DDD941 kJ
mol×3 mo l
432 kJ
mol×6 mo l
391 kJ
mol941 kJ1296 kJ2346 kJ
109 kJH
Statement: The ΔH for the reaction is -109 kJ. 2. (a) Solution: Step 1: Use the balanced chemical equation to determine the bonding of each substance. HCN(g) + 2 H2(g) → CH3NH2(g)
Copyright © 2012 Nelson Education Ltd. Chapter 5: Thermochemistry 5.3-5 Step 2: Determine the number of moles of reactants and products, the number of moles of bonds broken or formed, and the molar bond energy for each. Organize this information in a table. Substance Number of bonds per mole (nsubstance) Amount of bonds in reaction Bond energy per mole reactants HCN 1 mol C-H bonds 1 mol C≡N bonds 1 mol 1 mol 413 kJ/mol 891 kJ/mol H2 1 mol H-H bonds 2 mol 432 kJ/mol products CH3NH2 1 mol C-H bonds 2 mol C-N bonds 2 mol N-H bonds 3 mol 1 mol 2 mol 413 kJ/mol 305 kJ/mol 391 kJ/mol Step 3: Calculate the enthalpy change, ΔH, of the reaction. ()
bonds brokenbonds formedCHCN HH
CHCN NH
1 mol1 mol2 mo l
3 mol1 mol2 m ol
1 mo l
HnDnD DDD DDD413 kJ
mol×1 mo l
891 kJ
mol×2 mo l
432 kJ
mol3 mo l
413 kJ
mol×1 mo l
305 kJ
mol×2 mo l
391 kJ
mol413 kJ891 kJ864 k J(1239 kJ305 kJ782 kJ)
158 kJH
Statement: The ΔH for the reaction is -158 kJ. (b) Solution: Step 1: Use the balanced chemical equation to determine the bonding of each substance. N2H4(g) + 2 F2(g) → N2(g) + 4 HF(g) Step 2: Determine the number of moles of reactants and products, the number of moles of bonds broken or formed, and the molar bond energy for each. Organize this information in a table. Substance Number of bonds per mole (nsubstance) Amount of bonds in reaction Bond energy per mole reactants N2H4 4 mol N-H bonds 1 mol N-N bonds 4 mol 1 mol 391 kJ/mol 160 kJ/mol F2 2 mol F-F bonds 2 mol 154 kJ/mol products N2 1 mol N≡N bonds 1 mol 941 kJ/mol HF 1 mol H-F bonds 4 mol 565 kJ/mol
Copyright © 2012 Nelson Education Ltd. Chapter 5: Thermochemistry 5.3-6 Step 3: Calculate the enthalpy change, ΔH, of the reaction. ()
bonds brokenbonds formedNHNN FF
NNHF4 mol1 mol2 mol
1 mol4 mol
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