Grams to Moles to Atoms

CHEM 001 7 CHAPTERS Ex , MIDTERM 2 8 GRAMS TO PRACTICE PROBLEMS 9 , ( 10 , W 6 ( SECTION molar H2O of mass ) P1 VI grams H2o ( Iom } ) JIT I mot H2O × 18.02 PRACTICE 4 grams mot Mgsoy Q5P2 : : If H2O , ( °m% I' ) - 7) , W 8 PI 16199048ms × 18.02 ) = I moi grams H2O z3mofetgmef-zo-1.la x 1024 atoms t I Im ) -14/7%981=139 ( :9% ?m7g§o , ' 0.5317 ("mo93mg7§q ) 53.17% is of ⇒ ) ( 212.07g ) the from z 0.5317 5 W 7 P2 34 in ? g) = ( QUIZZES , COMPOSITION PROBLEMS of mass I ( molecules H2O × PERCENT molar " 6.022×10 W7P3 , -11 , zz 70 ? I 2 : - ATOMS TO 11 grams . 7 WORKSHEETS MOLES : Blinky Guide = 112.8g of oxygen mass of oxygen . Mg 504 MOLECULAR PRACTICE EMPIRICAL & PROBLEMS Q5P3 : , FORMULAS Wb ( SECTION VI ) P2 W7P4 , W7P5 , $8000002.03 000063.57 DRAFT .lk# At 151.15g 30 31.76g 63.57g glmolc 4.67g molar mass of C : 12.01g 11 : miiiiiisissoiti.is?oioo:iiiiii ( ) mot I 11.985 C 01/46.00 glmol 4/42.01 0/1.985 5.293 ( ) O HI / ( ( H 1.01 ) = 4.624 1.985 = mot mot H = empirical molar mass 302.3 of glmol 151.15g Imd Cotta = 2 : ⇒ ① mold 8( fifty ) ' molecular : : -1 11.985 Coo 7 2.329×3=6.987 2.666 = = 1 x 3 x 3 7.998 = 3 = → 7 H7O3 ( Iom If! ) -13/7%980 ) , Cho Hey 06 C → → = 8 = 1 mot Coo # 703 H O REACTIVITY SERIES practice YOU PROBLEMS REMEMBER EACH ANY , My 2 IS N Oz EACH o NEED NOT DO IS 1 S THE MEMORIZE TO NEW My ( NO 3) z ⇒ - SERIES MUST FORMED COMPOUNDS + REACTION practice W8P3 : . BE ELECTRICALLY NEUTRAL NEUTRAL IS TYPES problems W8P2 : REACTANT PRODUCT TO pmtamcepropsu.ms Q6P1,w8P5.w9P1 - - : 4 molar molar mass mass 37.9g 2 of Sify of Site , HF x : t : 'm9f¥ ) 14%99%1+417 I mot ( Iom ,?¥ ) ' Sify 104.09g + I ( Omg I' ) 're = ' - 4 mot HF × × I moi sie , = IT !%i¥ , 3997¥ 21%7 It = 29.1g HE . LIMITING PRACTICE THE OF REACTANT PROBLEMS WITH DIFFERENCE REACTANT THE W8P4 , ONE JUST OF W8( , , WE SO all : BALANCE by 2 2 LIMITING decimal remove 8 13 molar 2 : ( moggy If ) 10 32.00g . of Cy Hao mass 16.05g 32.00g 3 molar : FIND STEP = 1 mot Oz : 4 'T ) ( Ying !'t ) ( zoom -110 Cy Hyo Oz x t x 04h10 8 mol × 58.14g t mot Oz 2 8 mot × 32.00g of CO2 mass mot : 13 moi 0.6154 mot CO2 × mo coz , = Cy Hyo coz 1.104 0.6154 = Oz Yo YIELD ( 27.08g CO2 )( mot I'm ?Y&µo mot 4140mg , = a 27.08g 0.850 ) = 23.0g CO2 CO2 CO2 CO2 ?! ) -11/12 2116in .mg#)=I4m.ofzgfz : APPLY = , GRAMS PRODUCT 4 LIMITING THE 5 REACTANT OF QUANTITIES : FIND STEP FIND TO HAVE GIVEN ARE WE HERE 4 to 2 of Oz molar mass STEP YIELD . multiply I % & W9P2 , THAT IS 7.5 STEP ) ALA PROBLEM PREVIOUS INSTEAD REACTANTS TWO QGPZ : PRODUCT TO tIf¥aI¥f VALENCE ELECTRONS PRACTICE S PROBLEMS Q7P5 : 15252106353104 WIOPI , : SULFUR HAS 6 VALENCE ELECTRONS . You do not need to know how to find the valence electrons for transition metals! ELECTRON CONFIGURATION PRACTICE PROBLEMS Q7P1 : Q7P3 , W10P4 , WTOP , 6 Keetoowah : P ( add electrons 3 P ) ' ORBITAL DIAGRAM : 15252106353106 15252,5353ps 15252136353ps PRACTICE : QFPI , W10P3 , W10P4 : P P PROBLEMS -1 tf ftt : 15 25 It I 2ps 35 I 3ps I , W10P5