Unit 2 Conclusion
Completion requirements
Unit B
Energy and Matter in Chemical Change
Conclusion
By organizing and classifying matter, the study of chemical substances and their changes is starting to make sense.

B13.10 Writing a chemical reaction
In this unit, you started out learning about the underlying structure of matter. You looked at the historical development of atomic theory, including the work of Dalton, Thomson, Rutherford, and Bohr, focusing on the experiments they performed that
allowed them to develop their atomic theories. You learned about the subatomic particles of atoms and how this information is organized on the periodic table.
Once you finished learning about atoms, you learned about how atoms combine to form compounds. You learned how to identify ionic and molecular compounds and how to write chemical formulas and names for those compounds.
The next topic you studied was chemical reactions. You learned how to classify, balance, and predict products for the five main types of chemical reactions: formation, decomposition, single replacement, double replacement, and hydrocarbon combustion.
Finally, you studied chemical amounts by looking at the concept of the mole.
Once you finished learning about atoms, you learned about how atoms combine to form compounds. You learned how to identify ionic and molecular compounds and how to write chemical formulas and names for those compounds.
The next topic you studied was chemical reactions. You learned how to classify, balance, and predict products for the five main types of chemical reactions: formation, decomposition, single replacement, double replacement, and hydrocarbon combustion.
Finally, you studied chemical amounts by looking at the concept of the mole.
Review Questions
Complete the following practice questions to check your understanding of the concepts you have learned. Make sure you write complete answers to the practice
questions in your notes. After you have checked your answers, make corrections to your responses (where necessary) to study from.- A substance in a chemistry lab was labeled with the following WHMIS symbol:
It means the substance can cause health hazards such as cancer, respiratory problems, or reproductive problems. The safety precautions should be to use only in a well-ventilated area and to use personal protective equipment.
- Which scientist determined that the atom was mostly empty space? What evidence did he collect to support this idea?
Rutherford determined the atom was mostly empty space. He shot alpha particles at gold foil; most of the alpha particles passed straight through without being deflected at all. - Identify the number of protons, electrons, and neutrons in an atom of selenium-74
protons = 34
electrons = 34
neutrons = 40 - Draw a modified Bohr diagram for sulfide.
- How many electrons would an ion of cesium contain?
An atom of cesium would contain 55 electrons. An ion of cesium has a charge of 1+, so it must have lost one electron; therefore, an ion of cesium will have 54 electrons. - What happens to electrons in an ionic bond?
Electrons are transferred. The metal atom will lose electrons, and the non-metal ion will gain electrons. - Identify each of the following compounds as ionic or molecular and provide the correct IUPAC name.
Formula Classification IUPAC Name CaF2 PBr5 Li2HPO4 Cr2O3 Ca(NO3)2•4H2O
Formula Classification IUPAC Name CaF2 ionic calcium flouride PBr5 molecular phosphorus pentabromide Li2HPO4 ionic lithium hydrogen phosphate Cr2O3 ionic chromium(III) oxide Ca(NO3)2•4H2O ionic calcium nitrate-water(1/4) - Identify each of the following compounds as ionic or molecular and provide the correct formula.
IUPAC Name Classification Formula potassium sulfide diiodine tetraoxide aluminium sulfite tin(IV) carbonate iron(III) chloride-water(1/6)
IUPAC Name Classification Formula potassium sulfide ionic K2S diiodine tetraoxide molecular I2O4 aluminium sulfite ionic Al2(SO3)3 tin(IV) carbonate ionic Sn(CO3)2 iron(III) chloride-water(1/6) ionic FeCl3•6H2O - Write the formula for sulfuric acid.
First, write the IUPAC name; as the acid name ends in “ic” and does not start with “hydro,” the IUPAC name must end with “ate”: hydrogen sulfate. Next, balance the charge for hydrogen and sulfate: H2SO4 - Four solutions—hydrochloric acid, sodium hydroxide, glucose, and sodium chloride—are tested with red and blue litmus paper and a conductivity metre. What are the expected results?
Substance Classification Red Litmus Blue Litmus Conductivity HCl NaOH C6H12O6 NaCl
Substance Classification Red Litmus Blue Litmus Conductivity HCl acidic red red yes NaOH basic blue blue yes C6H12O6 molecular red blue no NaCl ionic red blue yes - Predict the solubility for the following substances.
Substance Solubility Ag2SO4 LiOH CuCO3 AuF3
Substance Solubility Ag2SO4 slightly soluble LiOH very soluble CuCO3 slightly soluble AuF3 very soluble - Predict the state for the following substances when they are placed in a beaker of water.
Substance State CrSO4 Ba(IO3)2 K3PO4 PbCL2
Substance State CrSO4 (aq) Ba(IO3)2 (s) K3PO4 (aq) PbCL2 (s) - For each of the following reactions, identify the reaction type and balance the equation.
- __KF(l) → __K(s) + __F2(g)
Type: _______________
2KF(l) → 2K(s) + F2(g)
Type: decomposition - __C5H12(l) + __O2(g) → __CO2(g) + __H2O(g)
Type: _______________
C5H12(l) + 8O2(g) → 5CO2(g) + 6H2O(g)
Type: hydrocarbon combustion - __Mg(s) + __HCl(aq) → __H2(g) + __MgCl2(aq)
Type: _______________
Mg(s) + 2HCl(aq) → H2(g) + MgCl2(aq)
Type: single replacement - __BaCl2(aq) + __Li3PO4(aq) → __LiCl(aq) + __Ba3(PO4)2(s)
Type: _______________
3BaCl2(aq) + 2Li3PO4(aq) → 6LiCl(aq) + Ba3(PO4)2(s)
Type: double replacement - __P4(s) + __Cl2(g) →__PCl3(s)
Type: _______________
P4(s) + 6Cl2(g) → 4PCl3(s)
Type: formation14.
- __KF(l) → __K(s) + __F2(g)
- For the balanced equations from a and c of question 13, represent them using a ball and stick model.
- Predict the products, write a word equation, and write a balanced chemical reaction equation for each of the following.
- Lithium bromide decomposes.
lithium bromide → lithium + bromine
2LiBr(s) → 2Li(s) + Br2(l) - Aqueous aluminium chloride reacts with aqueous potassium phosphate.
aluminium chloride + potassium phosphate → potassium chloride + aluminium phosphate
AlCl3(aq) + K3PO4(aq) → 3KCl(aq) + AlPO4(s) - Chlorine gas is bubbled through a solution of nickel(III) bromide.
chlorine + nickel(III) bromide → bromine + nickel(III) chloride
3Cl2(g) + 2NiBr3(aq) → 3Br2(l) + 2NiCl3(aq) - Acetylene (C2H2(g)) undergoes complete combustion.
acetylene + oxygen → carbon dioxide + water vapour
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g) - Zinc reacts with oxygen gas.
zinc + oxygen → zinc oxide
2Zn(s) + O2(g) → 2ZnO(s)
- Lithium bromide decomposes.
- What is the amount in moles of a 25 g sample of potassium oxide?
Write the chemical formula for the substance and determine its molar mass.
«math»«msub»«mi»M«/mi»«mrow»«msub»«mi»K«/mi»«mn»2«/mn»«/msub»«mi»O«/mi»«/mrow»«/msub»«mo»=«/mo»«mfenced»«mrow»«mn»2«/mn»«mo»§#215;«/mo»«mn»39«/mn»«mo».«/mo»«mn»10«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfenced»«mo»+«/mo»«mfenced»«mrow»«mn»1«/mn»«mo»§#215;«/mo»«mn»16«/mn»«mo».«/mo»«mn»00«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfenced»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»M«/mi»«mrow»«msub»«mi»K«/mi»«mn»2«/mn»«/msub»«mi»O«/mi»«/mrow»«/msub»«mo»=«/mo»«mn»94«/mn»«mo».«/mo»«mn»20«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/math»List the variables.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«msub»«mi»K«/mi»«mn»2«/mn»«/msub»«mi»O«/mi»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»m«/mi»«mrow»«msub»«mi»K«/mi»«mn»2«/mn»«/msub»«mi»O«/mi»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»25«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«msub»«mi»K«/mi»«mn»2«/mn»«/msub»«mi»O«/mi»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»94«/mn»«mo».«/mo»«mn»20«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»Identify the correct formula and rearrange if necessary.
«math»«msub»«mi»n«/mi»«mrow»«msub»«mi»K«/mi»«mn»2«/mn»«/msub»«mi»O«/mi»«/mrow»«/msub»«mo»=«/mo»«mfrac»«mi»m«/mi»«mi»M«/mi»«/mfrac»«/math»Substitute the values into the formula.
«math»«msub»«mi»n«/mi»«mrow»«msub»«mi»K«/mi»«mn»2«/mn»«/msub»«mi»O«/mi»«/mrow»«/msub»«mo»=«/mo»«mfrac»«mi»m«/mi»«mi»M«/mi»«/mfrac»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»n«/mi»«mrow»«msub»«mi»K«/mi»«mn»2«/mn»«/msub»«mi»O«/mi»«/mrow»«/msub»«mo»=«/mo»«mfrac»«mrow»«mn»25«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«mrow»«mn»94«/mn»«mo».«/mo»«mn»20«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfrac»«/math»Calculate the answer.
«math»«msub»«mi»n«/mi»«mrow»«msub»«mi»K«/mi»«mn»2«/mn»«/msub»«mi»O«/mi»«/mrow»«/msub»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»265«/mn»«mo»§#8230;«/mo»«mi»mol«/mi»«mo»=«/mo»«mn»0«/mn»«mo».«/mo»«mn»27«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«mo»§#160;«/mo»«mo»(«/mo»«mi»to«/mi»«mo»§#160;«/mo»«mi»two«/mi»«mo»§#160;«/mo»«mi»significant«/mi»«mo»§#160;«/mo»«mi»digits«/mi»«mo»)«/mo»«/math» - What is the mass in grams of a 3.25 mol sample of ammonia, NH3?
Write the chemical formula for the substance and determine its molar mass.
«math»«msub»«mi»NH«/mi»«mn»3«/mn»«/msub»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»M«/mi»«mrow»«mi»N«/mi»«msub»«mi»H«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mfenced»«mrow»«mn»1«/mn»«mo»§#215;«/mo»«mn»14«/mn»«mo».«/mo»«mn»01«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfenced»«mo»+«/mo»«mfenced»«mrow»«mn»3«/mn»«mo»§#215;«/mo»«mn»1«/mn»«mo».«/mo»«mn»01«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mrow»«/mfenced»«mspace linebreak=¨newline¨»«/mspace»«msub»«mi»M«/mi»«mrow»«mi»N«/mi»«msub»«mi»H«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»17«/mn»«mo».«/mo»«mn»04«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/math»List the variables.
«math»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»m«/mi»«mrow»«mi»N«/mi»«msub»«mi»H«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»n«/mi»«mrow»«mi»N«/mi»«msub»«mi»H«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»3«/mn»«mo».«/mo»«mn»25«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»M«/mi»«mrow»«mi»N«/mi»«msub»«mi»H«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»17«/mn»«mo».«/mo»«mn»04«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/mtd»«/mtr»«/mtable»«/math»Identify the correct formula and rearrange if necessary.
«math»«msub»«mi»m«/mi»«mrow»«mi»N«/mi»«msub»«mi»H«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mi»n«/mi»«mo»§#215;«/mo»«mi»M«/mi»«/math»Substitute the values into the formula.
«math»«msub»«mi»m«/mi»«mrow»«mi»N«/mi»«msub»«mi»H«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»3«/mn»«mo».«/mo»«mn»25«/mn»«mo»§#160;«/mo»«mi»mol«/mi»«mo»§#215;«/mo»«mn»17«/mn»«mo».«/mo»«mn»04«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»/«/mo»«mi»mol«/mi»«/math»Calculate the answer.
«math»«msub»«mi»m«/mi»«mrow»«mi»N«/mi»«msub»«mi»H«/mi»«mn»3«/mn»«/msub»«/mrow»«/msub»«mo»=«/mo»«mn»55«/mn»«mo».«/mo»«mn»38«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»=«/mo»«mn»55«/mn»«mo».«/mo»«mn»4«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«mo»§#160;«/mo»«mo»(«/mo»«mi»to«/mi»«mo»§#160;«/mo»«mi»three«/mi»«mo»§#160;«/mo»«mi»significant«/mi»«mo»§#160;«/mo»«mi»digits«/mi»«mo»)«/mo»«/math»