Chem12 SM Ch05 Section5 2 final ok revised PDF

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

Chem 12 SM Ch5 Review final new ok revised

Δ = −. Statement: The enthalpy change ΔH

Chem12 SM Ch05 Section5 3 final ok revised

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) +

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

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:.

Chem12 SM Ch05 Section5 2 final ok revised

HO(l)

6.0 mL=V

; initial

25 °C=T

; final

75 °C=T

; 2

HO( l)

4.18 J/(g°C)=⋅c

; 2

HO(l )

1.00 g/mL=d

Required: thermal energy required, q Analysis: =ΔqmcT Solution: Step 1: Determine the mass of water, m. 6.0 mL mVd

1.00 g

1 mL

6.0 g=m

Step 2: Determine the change in temperature, ΔT . finalinitial

75 °C25 °C

50 °C

TTT T Step 3: Calculate the quantity of thermal energy, q. 4.18 J (6.0 g) gC qmcT (50 °C

1.3 kJ=q

Statement: The quantity of thermal energy required is 1.3 kJ. 2. Given: ethylene glycol

4.0 kgm=

; ethylene glycol

3.5 J/(g°C)c=⋅

; 250 kJq=

Required: change in temperature, ΔT

Analysis: =ΔqmcT

Solution: Step 1: Convert quantities to the appropriate units. 3

4.0 kg4.010 gm==×

q=250 kJ=250!10 3 J

Step 2: Rearrange =ΔqmcT

to solve for ΔT . 3 3

25010 J

50.0 mLV=

; NaOH(aq)

75.0 mLV=

; initial

20.2 °CT=

; final

25.6 °CT=

Required: quantity of energy transferred, q Analysis: =ΔqmcT Solution: Step 1: Determine the total volume of HCl(aq) and NaOH(aq), V . HCl(aq)NaOH(aq)

50.0 mL75.0 mL

125.0 mL

VVV V

Step 2: Determine the total mass of two solutions , m. Since the solution contains dilute HCl(aq) and NaOH(aq), the density and heat capacity are assumed to be the same as those for water. 125.0 mL

mVd=

1.00 g

1 mL

125.0 gm=

Step 3: Determine the change in temperature, ΔT . finalinitial

25.6 °C20.2 °C

5.4 °C

TTT T Step 4: Calculate the quantity of energy transferred, q. 4.18 J (125.0 g) gC qmcT=Δ (5.4 °C

2800 Jq=

Statement: The quantity of energy transferred is 2800 J, or 2.8 kJ. Since the temperature of the surroundings (liquid water in the calorimeter) increased, the reaction was exothermic. Mini Investigation: Thermal Energy Transfer in a Coffee-Cup Calorimeter, page 297 Answers may vary. Sample answers: mass of water in calorimeter = 100.00 g mass of aluminum block = 89.80 g temperature of aluminum block = temperature of hot water bath = 80.0 °C final temperature of water in calorimeter = 27.8 °C initial temperature of tap water = 19.8 °C A. Given: 2

HO(l)

100.00 gm=

; initial

19.8 °CT=

; final

27.8 °CT=

Required: quantity of thermal energy gained by the water, q Analysis: =ΔqmcT

. finalinitial

27.8 °C19.8 °C

8.0 °C

TTT T Step 2: Calculate the quantity of thermal energy gained by the water, q. 4.18 J (100.00 g) gC qmcT=Δ (8.0 °C

3300 Jq=

Statement: The quantity of thermal energy gained by the water is 3300 J, or 3.3 kJ. B. Given: Al(s)

89.80 gm=

; initial

80.0 °CT=

; final

27.8 °CT=

; Al(s)

0.900 J/(g°C)c=⋅

Required: quantity of thermal energy transferred to or from aluminum, q Analysis: =ΔqmcT Solution: Step 1: Determine the change in temperature, ΔT . finalinitial

27.8 °C80.0 °C

52.2 °C

TTT T Step 2: Calculate the quantity of thermal energy transferred to or from aluminum, q. 0.900 J (89.80 g) gC qmcT=Δ (52.2 °C

4220 Jq=-

Statement: Since the value of q is negative, energy is transferred from aluminum to water, and the quantity transferred is 4220 J, or 4.22 kJ. C. Answers may vary. Sample answer: There is a difference of about 0.9 kJ in the answers to A and B. The difference could be due to assumptions or experimental errors. D. Answers may vary. Sample answer: Experimental errors include thermal energy lost to the tongs, to the air on transfer from hot to cold water, to the thermometer, to the calorimeter, or to the air above the water-aluminum mixture. The investigation could be improved by taking into account the temperature changes and thus energy transferred to the thermometer, tongs, and calorimeter. Other improvements would be increasing the insulation of calorimeter and faster transfer from hot to cold water.

HO(l )

50.0 gm=

; vap

44.0 kJ/molHΔ=

Required: enthalpy change, HΔ

Analysis: vap

HnHΔ=Δ

Solution: Step 1: Calculate the amount of water in 50.0 g, 2 HO(l) n . 2 HO(l)

18.02 g/molM=

2 2 2 2 HO(l) HO(l) HO(l) HO(l)

50.0 g

18.02 g/mol

2.7747 mol (2 extra digits carried)

m n Mquotesdbs_dbs2.pdfusesText_2

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[PDF] Chem12 SM Ch05 Section5 2 final ok revised

6.0 mL=V

25 °C=T

75 °C=T

HO( l)

4.18 J/(g°C)=⋅c

HO(l )

1.00 g/mL=d

1.00 g

6.0 g=m

75 °C25 °C

50 °C

1.3 kJ=q

4.0 kgm=

3.5 J/(g°C)c=⋅

Required: change in temperature, ΔT

Analysis: =ΔqmcT

4.0 kg4.010 gm==×

Step 2: Rearrange =ΔqmcT

25010 J

50.0 mLV=

75.0 mLV=

20.2 °CT=

25.6 °CT=

50.0 mL75.0 mL

125.0 mL

1.00 g

125.0 gm=

25.6 °C20.2 °C

5.4 °C

2800 Jq=

100.00 gm=

19.8 °CT=

27.8 °CT=

27.8 °C19.8 °C

8.0 °C

3300 Jq=

89.80 gm=

80.0 °CT=

27.8 °CT=

0.900 J/(g°C)c=⋅

27.8 °C80.0 °C

52.2 °C

4220 Jq=-

HO(l )

50.0 gm=

44.0 kJ/molHΔ=

Required: enthalpy change, HΔ

Analysis: vap

HnHΔ=Δ

18.02 g/molM=

50.0 g

18.02 g/mol

2.7747 mol (2 extra digits carried)