ACS General Chemistry Final Exam Questions With Complete Solutions

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ACS General Chemistry Final Exam Questions With Complete Solutions

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Institution ACS General Chemistry

Course ACS General Chemistry

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ACS General Chemistry Final Exam Questions With Complete Solutions Law of Definite Proportions - Answer-all samples of a given compound have the same proportions of their constituent element /.Mass Number - Answer-A: protons + neutrons /.Atomic Number - Answer-Z: Number of Protons /.Density= - Answer-Mass/Volume /.Frequency (v)= - Answer-Speed of light (c)/wavelength /.Electron Groups= 2 Bonding Groups= 2 Lone Pairs= 0 - Answer-EG= linear MG= linear Bond Angle= 180 /.Electron Groups= 3 Bonding Groups= 3 Lone Pairs= 0 - Answer-EG= trigonal planar MG= trigonal planar Bond Angle= 120 /.Electron Groups= 3 Bonding Groups= 2 Lone Pairs= 1 - Answer-EG= trigonal planar MG= bent Bond Angle= 120 /.Electron Groups= 4 Bonding Groups= 4 Lone Pairs= 0 - Answer-EG= tetrahedral MG= tetrahedral Bond Angle= 109.5 /.Electron Groups= 4 Bonding Groups= 3 Lone Pairs=1 - Answer-EG= tetrahedral MG=trigonal planar Bond Angle= 109.5 /.Electron Groups= 4 Bonding Groups= 2 Lone Pairs= 2 - Answer-EG= tetrahedral MG= Bent Bond Angle= 109.5 /.Electron Groups= 5 Bonding Groups= 5 Lone Pairs= 0 - Answer-EG= trigonal bipyramidal MG= trigonal bipyramidal Bond Angle= 120 (equatorial) 90 (axial) /.Electron Groups= 5 Bonding Groups= 4 Lone Pairs= 1 - Answer-EG= trigonal bipyramidal MG= seesaw Bond Angle= 120 (equatorial) 90 (axial) /.Electron Groups= 5 Bonding Grops= 3 Lone Pairs= 2 - Answer-EG= trigonal bipyramidal MG= t-shaped Bond Angle= 90 /.Electron Groups= 5 Bonding Groups= 2 Lone Pairs= 3 - Answer-EG= trigonal bipyramidal MG= Linear Bond Angle= 180 /.Electron Groups= 6 Bonding Groups= 6 Lone Pairs= 0 - Answer-EG= octahedral MG= octahedral Bond Angle= 90 /.Electron Groups= 6 Bonding Groups= 5 Lone Pairs= 1 - Answer-EG= octahedral MG= square pyramidal Bond Angle= 90 /.Hess' Law - Answer-ΔHrxn= Σ ΔHf (products)- Σ ΔHf (reactants) *ΔS and ΔG can be calculated

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ACS General Chemistry Final Exam
Questions With Complete Solutions
Law of Definite Proportions - Answer-all samples of a given compound have the same
proportions of their constituent element

/.Mass Number - Answer-A: protons + neutrons

/.Atomic Number - Answer-Z: Number of Protons

/.Density= - Answer-Mass/Volume

/.Frequency (v)= - Answer-Speed of light (c)/wavelength

/.Electron Groups= 2
Bonding Groups= 2
Lone Pairs= 0 - Answer-EG= linear
MG= linear
Bond Angle= 180

/.Electron Groups= 3
Bonding Groups= 3
Lone Pairs= 0 - Answer-EG= trigonal planar
MG= trigonal planar
Bond Angle= 120

/.Electron Groups= 3
Bonding Groups= 2
Lone Pairs= 1 - Answer-EG= trigonal planar
MG= bent
Bond Angle= <120

/.Electron Groups= 4
Bonding Groups= 4
Lone Pairs= 0 - Answer-EG= tetrahedral
MG= tetrahedral
Bond Angle= 109.5

/.Electron Groups= 4
Bonding Groups= 3
Lone Pairs=1 - Answer-EG= tetrahedral
MG=trigonal planar
Bond Angle= <109.5

, /.Electron Groups= 4
Bonding Groups= 2
Lone Pairs= 2 - Answer-EG= tetrahedral
MG= Bent
Bond Angle= <109.5

/.Electron Groups= 5
Bonding Groups= 5
Lone Pairs= 0 - Answer-EG= trigonal bipyramidal
MG= trigonal bipyramidal
Bond Angle= 120 (equatorial) 90 (axial)

/.Electron Groups= 5
Bonding Groups= 4
Lone Pairs= 1 - Answer-EG= trigonal bipyramidal
MG= seesaw
Bond Angle= <120 (equatorial) <90 (axial)

/.Electron Groups= 5
Bonding Grops= 3
Lone Pairs= 2 - Answer-EG= trigonal bipyramidal
MG= t-shaped
Bond Angle= <90

/.Electron Groups= 5
Bonding Groups= 2
Lone Pairs= 3 - Answer-EG= trigonal bipyramidal
MG= Linear
Bond Angle= 180

/.Electron Groups= 6
Bonding Groups= 6
Lone Pairs= 0 - Answer-EG= octahedral
MG= octahedral
Bond Angle= 90

/.Electron Groups= 6
Bonding Groups= 5
Lone Pairs= 1 - Answer-EG= octahedral
MG= square pyramidal
Bond Angle= <90

/.Hess' Law - Answer-ΔHrxn= Σ ΔHf (products)- Σ ΔHf (reactants)

*ΔS and ΔG can be calculated the same way*

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Uploaded on: January 17, 2025
Number of pages: 10
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