1 ST ROUND, SOLUTIONS

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1 ST ROUND, SOLUTIONS Problem (Lithuni) Self destructing pper ( points) Solution ( ) ( ) ( ) [Al HO OH ) ph pk lg [Al H O ( ) ( ) [Al H O OH [Al ( ).9 [Al H O.9.47 [Al ( ) ( ) ( ) [Al H O OH.9 pk ph lg. lg 4.9 [Al H O.47 b) Al (SO ) H O [Al(H O) [Al(H O) H O H c [Al(H O) [Al(H O) 4 OH - K From () :[OH [H From (4) :[Al(H O) [Al(H O) [Al(H O) [Al(H O) (OH) (OH) (OH) (OH) SO [H [H [SO c [Al(H O) 4 K K [Al(HO) (OH) [H [Al(H O) - [H [OH [OH [SO - 4 c () () () (4) () () (7) Using (,,7) in [Al(H O) [Al(H O) [H c [Al(H O) () : (c [Al(H O) 4c [Al(H O) K [H K ) [H c [H K [H c [H (8)

2 Using (4) in () : K K [Al(H O) K [Al(H O) [Al(H O) [Al(H O) c[h [H [Al(H O) [H [Al(H O) ([H K ) c[h c[h [H K (c [Al(HO) )[H [Al(H O) c[h (9) Using (9) in (8) :[H c[h c [H K [H K [H c[h ck [H c[h K [H K K [H K [H (ck K )[H K K K ([H K [H ) [H [H. [H. [H. [H. -7 M ph. [OH M [SO M [Al(H O). -8 M [Al(H O) (OH) M 9 Problem (Ltvi) Serching for secrets of Supermn ( points). Rdioctive decys re st order rections, so: C ln t C If t t(hlf life) yers, then C, C. ln ln. points.77 yers t/. If C, C, then: C ln ln.77 t t 848 8, yers. points,c npu : nt : nxe : npm : ndilium : nhg : nh : : : : : : :.998 :.:. :.97 :.9 :.4. :.9 :.77 : 8.4 :. :: 8.8 : :: 8. : :: 9 Pu T Xe Pm 7 (Al ) Hg H 8 Sum

3 It is not possible minerl formul becuse it does not contin proper mount of nions (such s oxygen which is not in this minerl)... point 4. Fluoride ions cn replce hydroxide ions in LiNSiB O 7 (OH). Similr compounds in nture re ptite nd fluoroptite. [ points. Volume of monoclinic cell is equl to re of fce between nd b powers height of cell. V * b * c * sin β.74 *.847 * 7.89 * sin Å. points Or you cn use freewre inplotr to clculte cell volumes (it lso wors for triclinic cells) nd mny other dt: 8. Å 8. * m 8. * * cm.8 * cm If we count correctly green colored lithium toms nd blue colored sodium toms, then we cn find tht Z (number of molecules in unit cell) is 4. M[LiNSiB O 7 (OH) *.8 8 * g/mol Mss of one molecule : M 9.4 m.4 g N. A m 4 m 4.4 density. 9 g cm point V V.8. Brgg eqution: where: n is n integer determined by the order given (in this problem it is ), λ is the wvelength of X rys, nd moving electrons, protons nd neutrons (λ.48 nm.48 Å), d is the spcing between the plnes in the tomic lttice, nd θ is the ngle between the incident ry nd the scttering plnes. So from this eqution: θ λ sin d λ θ rcsin d

4 d spcing (Å) Reltive intensity thet, o 4. 9,.7 9, ,. 74 8,.7 4 9,.94,4. 8 4,4. points Jdrite PXRD pttern (found somewhere in literture; Bruer AXS presenttion, Rig, 9) nd it corresponds to results clculted. In ll clculted positions we cn find diffrction signls. 7. None of those diffrction ptterns refers to jdrite. (. points) U Th He α decy Th P e β decy P U e β decy U Th He α decy Th R He α decy R Rn He α decy. points Firstly we hve to clculte mount of wter in given hydrte. Cu(UO ) (PO 4 ) (8 )H O M ( H O) x w( H O). M ( nhydrous compound) M ( H O) 8x. ( 79. 8x) 4

5 Molr mss clcultions: 8x.47.78x 4.9x.47 M r n M r * n sum 79. x.4 So molr mss of hydrted compound Cu(UO ) (PO 4 ) (.4)H O is: M * g/mol m. n[cu(uo ) (PO 4 ) (.4)H O. mol. mol M N N A * n. * *. *. * toms (corrected lter) point. Formtion of 4 Th nd 4 P cn be ssigned s negligible nd it cn be ssumed tht 8 U decys to form 4 U with hlf life of 4.48 * 9 yers. 8 8 dn( dt 4 4 d( dt 8U Th) d( Th) dt R) d( R) dt Rn) d( Rn) dt At the beginning: 4U 8U Th R ln 4.48 ln.8 yers 4 ln 9.9 yers 78 ln.47 yers. yers U where is s smll s possible. N( 8 t t N( 8 4U Th R N ( 8 U).9974 *. *.99 * toms N ( 4 U).4 *. *. * 7 toms N ( Th) N ( R) N ( Rn) t t 8

6 for rest prticles similr equtions cn be obtined. point Inputting them into MS Excel worsheet we cn obtin results tht re shown below. MS Excel function mx(..) is used to find mximum vlue of Th in million yers. It hs been found tht fter one million yers it is.89 * toms (mximum vlue for tht time) nd vlue of.474* yers cn be chieved in.974 million yers. Correct:.8* 7 toms in.974 million yers.. See nswer between grph.. points. Function extremes such s mximum points cn be determined from function derivtive. It must be equl to zero. t t N( o ( ) N( so o

7 e ln e t t.9 t t t t t ( ) t (9.9.8 ) t (.7 ) t (.7 ) t (.7 t (.7 ).9 ln. ).8..8 t 849 yers.7. points. This big difference cn be explined by fct tht in clcultion of t mx formtion of 4 U (from 8 U) hs not ten into ccount, but in Excel file it is included. If we exclude this from Excel file (simply set constnt ), then we obtin sme result, see figure bellow.. points Problem (Lithuni) Push those rrows (8 points). Genertion of free crboction Me Al Me Al Alyltion Me Me H H Me 7

8 . Genertion of free crboction Acyltion O O H. Regents O Mechnism of lyltion s in prt. Side crboction formtion: These crboctions re ble to further rect with the bezene thus producing side products. Plese note, tht there ws n error in this question. Those students, who provided even slightly logicl solutions, where wrded with mrs. e re sorry for the inconvenience

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10 Problem 4 (Estoni) Some nme for problem ( points) Solution Originl compound is CH COCH CH CHCH CHCH CH COOCH CH Mechnisms: ) )

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