Gen Chem Discussion 1.docx

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Department
Chemistry
Course
CAS CH 131
Professor
Mark Grinstaff
Semester
Fall

Description
2C4H10 + 1302 ­ ▯8C02 + 10H20 10g           =  .14 mole 58g/mol Given: 10g of C4H10 Question: How much C02 formed? 1/4  ratio. Under same constant pressure and temperature, volume to mole ratio will be constant.  Periodicity­ # of neutrons is mass – atomic number Noble gases don’t react as 8 valence electrons CORE ELECTRONS= ATOMIC NUMBER – VALENCE ELECTRONS (what column it  is in) 99.9% of space in atom is empty…orange in a hockey rink (example) diatomic molecules – almost never exist by themselves C4H10: mass percent = molecular mass= 8 grams, 48 grams of carbon, 1 mole of C4H10 Mass  percent of carbon = 48/58 Mass of molecule/mass of total molecules Limiting combustion: What limits the combustion switch to moles and compare ratio…larger ratio is the  limiting agent 5 moles of C4H10  ▯ ­
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