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Chemistry 100

by: Heidi Archer

Chemistry 100 CHEM 100 002

Heidi Archer
GPA 3.7

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About this Document

End of week Study guide for test
Chemistry and Society
Study Guide
50 ?




Popular in Chemistry and Society

Popular in Chemistry

This 2 page Study Guide was uploaded by Heidi Archer on Wednesday March 2, 2016. The Study Guide belongs to CHEM 100 002 at Indiana State University taught by Jeeewandara in Winter 2016. Since its upload, it has received 45 views. For similar materials see Chemistry and Society in Chemistry at Indiana State University.


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Date Created: 03/02/16
End of Week Study Guide Molecules determine the behavior of ____ ? Matter Break apart H O 2 2hydrogen atoms bounded to 1 oxygen atom H 2 is water . . . H 2 2  hydrogen peroxide. Why did the element change? We changed the subscript Determine the number of each type of atom in glucose, which has the molecular formula  C 6 O12 6 Carbon atoms=6   Hydrogen atoms=12    Oxygen Atoms=6 Determine the number of each type of atom in Pb(C H O ) 2 3 2 2 Pb=2 C=4 H=6 O=4 Elements with unstable electron configurations form compounds with other elements to  ____?   Gain Stability Metals ___ electrons, Nonmetals ___ electrons Lose, gain Compounds formed between a metal and at least one nonmetal are _____.  Compounds  formed between nonmetals are _____  Ionic, Molecular SO  is (Molecular/Ionic)? 2 S and O (nonmetals) so molecular Formula mass of a compound is analogous to the _______ of an element  Atomic mass The formula mass of a compound in amu is numerically equivalent to its molar mass in __?  (g/mol) If an element occurs in only one compound, balance that element ____?               First Change only occurs in ________ never subscripts             Coefficients Propane, C H , 3s o8ten used in place of methane, CH , as a fuel for ho4e heating and  cooking. Much like methane, if burns in (reacts with) oxygen to produce carbon dioxide  (CO ) 2nd water (H O). Bal2nce the reaction for the burning propane in oxygen. The unbalanced equation: C H   +  O3  8   CO   2  H O 2 2 First, balance the elements present in only one compound on each side of the equation.   C H   +  O   →   3CO   +  H O    3 C’s on each side 3 8 2 2 2 C 3  8+  O   →2  3CO   +  42 O 2    8 H’s on each side Balance the element present as a free element last. C 3  8+  5O   →2  3CO   +  4H2O 2    10 O’s on each side Final balanced equation: C 3  8+  5O   →2  3CO   +  42 O 2 Note:  The absence of coefficient in front of a reactant or product implies a “1”.


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