ࡱ> ?A>E@ z)bjbj $. BBBBBBBV^^^^$4V(HM"u'$)RT+f'9Bm'BB'}}}:BB}}2}BB -jS^f$'0(+H+VVBBBB+B0}''VVDgVVChemistry 12 Some Review of Chem 11 - Mole Stuff Part 1 - Changing Moles to Grams and Grams to Moles The two conversion factors to remember are: MM grams & 1 mole  1 mole MM grams Where MM stands for the Molar Mass The Molar Mass is calculated by adding up atomic masses from underneath the symbol and the name on the periodic table. eg. The molar mass of Na2SO4 is calculated as follows: 2(23.0) + 32.1 + 4(16.0) = 142.1 grams/mole Here are some examples of converting using the conversion factors: 1. 2.60 moles of Na2SO4 = _____ grams 2.60 moles x 142.1 grams = 369.46 grams  1 mole NOTE: In Chemistry 12 calculations, we must consider significant digits The 2.60 has 3 SDs and the 142.1 has 4 SDs. When multiplying, the answer must be rounded of to the least # of SDs in the numbers being multiplied. So this answer must to rounded to 3 SDs. So the answer is 369 grams ANOTHER NOTE: If a calculation is just one step in a series of calculations, DONT round of the answer. If possible, leave it in your calculator the way it is and go from there. 2. 1053.24 grams of K2Se = _____ moles Solution: The molar mass of K2Se is 2(39.1) + 79.0 = 157.2 g/mole 1053.24 grams of K2Se x 1 mole = 6.700 moles  157.2 grams NOTE: The reason for the two 0s on the end of 6.700 is because the lowest # of SDs in the numbers divided is 4SDs (The 157.2) so the answer must have 4 SDs Now some for you to do Work each of the following out showing the work and the units in the work and in the answer! These will be marked and counted as homework marks. (2 marks each) 1. 833.4 grams of H2O = _____ moles Answer _______________ 2. 2.3 x 10-3 moles of H2SO4 = _____grams Answer _______________ 3. 3.84 grams of (NH4)2CO3 = _____ moles Answer _______________ 4. 2.45 x 10-2 moles of Al(OH)3 = _____grams Answer _______________ 5. 0.3558 grams of nitrogen dioxide = _____ moles Answer _______________ Unit 1 of Chemistry 12 deals with RATES of reactions. Rates are always expressed as a change in amount (grams, moles, litres etc.) per change in time (seconds, min. etc.) Rate = D amount  D time Here s an example of how the grams/mole conversions are used in rate expressions: Change a rate of 0.035 grams H2 per second to moles of H2 per second Solution: 0.035 g H2 x 1 mole H2 = 0.0175 mol H2/s-> rounding to correct SDs-> 0.018 mol H2/s  1 s 2.0 g H2 Notice how the gs cancel out and you are left with the units of mol H2/s Here are some of these for you to do: 6. 2.6 x 10-2 moles of Zn/second = _____grams of Zn/second Answer _______________ 7. 0.1962 grams of Zn/second = _____moles of Zn/second Answer _______________ 8. 0.014 moles of CO2/s = _____grams of CO2/s Answer _______________ 9. 3.718 grams of CO213g \ ] _ `   K L u v Pf#$eܵܥܥܥܥܵܙܙܥܥܥ܅ueh,h{g]H*OJQJmH sH h`5mh{g]5OJQJmH sH &jh,OJQJUmHnHsH uh`5mOJQJmH sH h`5mh{g]H*OJQJmH sH &jh`5mOJQJUmHnHsH u$h`5mh{g]CJOJQJaJmH sH h`5mh{g]OJQJmH sH 'h`5mh{g]>*CJ$OJQJaJ$mH sH ! 123gh  C D q r % ,^,gd`5mgd`5m$a$gd`5mz)% l   W 7 8 u /OPgh67gd`5m789:YZ,-./0OPgd`5megrsuw ]HLNPBDvx-.n h`5mh{g]CJOJQJmH sH "h,h{g]6H*OJQJmH sH h`5mh{g]6OJQJmH sH &jh,OJQJUmHnHsH uh,h{g]OJQJmH sH h,OJQJmH sH h,h{g]H*OJQJmH sH h`5mh{g]OJQJmH sH h,h{g]H*OJQJmH sH +TU^`1Z[ gd`5m.13PVWXY23(((R(S(k(l(p(²wkkiUhOiOJQJmH sH h,h{g]H*OJQJmH sH h`5mhOiOJQJmH sH h{g]OJQJmH sH hOih{g]H*OJQJmH sH h,h,H*OJQJmH sH h`5mh,OJQJmH sH h,OJQJmH sH &jh,OJQJUmHnHsH uh`5mh{g]OJQJmH sH "24nopqrst(((( (!("(gd`5m/s = _____moles of CO2/s Answer _______________ 10. 1.12 L of CO2/s = _____moles of CO2/s (at Standard Temp. and Pressure) HINT: Recall that for gases at STP there are 22.4 L/ 1 mole so conversion factors could be: 22.4 L or 1 mole  1 mole 22.4 L Answer _______________ "(A(B(((((!)U)V)W)X)Y)Z)[)z)gd`5mp(((!)#)U)y)z)ιޮh`5mh{g]OJQJ)jhOi6OJQJUmHnHsH uh`5mh{g]6OJQJmH sH h`5mh{g]OJQJmH sH #h`5mh{g]6CJOJQJmH sH ":p`5m/ =!"8#$%L@L Normal5$7$8$9DH$CJ_HmH sH tH DA@D Default Paragraph FontVi@V  Table Normal :V 44 la (k(No List . 123ghCDqr%lW78u/OPgh6789:YZ,-./0OPT U j ( ) 3 4   . / m n o p q r s % & ' ( ) * + J K | } ~  < P 0(0p0p0p0p0p0p0000p00p00p0 0 00p0 0 0p0 000 0 00p0p00 00 00(00 000p0p000p0p00p0p0p00p0p0000000000000 00(000p0 00(00 00 0 0800p0p0p0p0p0p0p00p0p00000000 000p000 0p000p00 00000p0 00000p0p0p0p0p000p0p00p0p00p0p00p0p00p0p00p0p0000p0p000 00 0 000 :0:00e.p(z) % 7"(z) z) 8  @b (  HB  C DHB  C DHB  C DHB  C DHB  C DHB  C DHB  C DHB   C DHB   C DB S  ?uj  vt7 t [t vtn tr% %tt Vt %^ t  % Zc~ms JQ36"(qylvV ] O T  . / 2 R V   e i > E 333333333333333333333333333rr88uvU U ] ` j k 4 4 - . : : < < n n o r - .  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