Takmičenje iz hemije srednja škola

Posto ovde nema resenja, ja cu da uradim ove zadatke pa cu moje da ispisem, mozda ce nekome trebati.

1996-okr:
1. 66,9%
2. a) K<Na<Ca<Mg;  b) Ca<Mg<K<Na
3. a) 1; b) 1
4. 6,15
5. 5,78
6. b
7. 116,67
8. [HCN]=5,34*10^-3M; [CN^-]=5,23*10^-3M
9. a,b,d,e
10. 45,6 dm³ H₂ ; 22,8 dm³ O₂
11. 86,8%
12. 1,2,3,4 (K,L,M,N)
13. 5,4*10¹²
14. 4,32*10^-1
15. a) 0,15 min; b) 1465 min
16. +283
17. a) B₂O₃; b) P₂O₅ (P₄O₁₀); c) Mn₂O₇
18. a) SnCl₂+2HgCl₂-->SnCl₄+Hg₂Cl₂
b) 5Br^- + BrO₃^- + 6H⁺ --> 3Br₂ + 3H₂O
19. 20,2 g Cu₂O; 4,8 g CuO
20. 0,104

"A new scientific truth does not triumph by convincing opponents and making them see the light, but rather because its opponents eventually die, and a new generation grows up that is familiar with it" Max Planck

Kako si u 12. zadatku zakljucio da ne moze 4. perioda? Zar ne moze n=4, l=2(d)?

Da li se 4d elementi nalaze u 4. periodi? :)

"A new scientific truth does not triumph by convincing opponents and making them see the light, but rather because its opponents eventually die, and a new generation grows up that is familiar with it" Max Planck

Tacno. Nezgodan zadatak.

Ustvari, zasto ne bi bio neki 3d element, oni se nalaze u cetvrtoj periodi a n+l=4+2=6? U 7. periodi se nalaze 6d i 5f elementi, ali ni kod njih zbir nije 6, pa be trebalo da bude i 7. perioda u resenju.

Nisam te bas lepo razumeo za ovo prvo, a 7. periodu namerno nisam stavio u resenje. U zadatku ne pise da zbir glavnog i orbitalnog kvantnog broja za valentne elektrone bude 6, nego samo da njihov zbir bude 6. Svi elementi u 7. periodi imaju elektrone ciji je n+l=6.

"A new scientific truth does not triumph by convincing opponents and making them see the light, but rather because its opponents eventually die, and a new generation grows up that is familiar with it" Max Planck

Ok, hvala na postavljanju resenja.