... - " " ..,.., , -
|
|
- George Clarke
- 5 years ago
- Views:
Transcription
1 " "
2
3 ( ) 60 = 98 K
4 10 11 " 40" " " " " 4
5 CuSO 4 CuSO 4( ) Cu + ( ) + SO 4 ( ) ( ) CuSO 4 5H O ( ) 50 CuSO 4 5H O CuSO 4 + 5H O CuSO 4 5H O [Cu(H O) 4 ]SO 4 H O [Fe(H O) 6 ]SO 4 H O FeSO 4 7H O CuSO 4 H O CuSO 4 5H O AlCl 3 6H O Na CO 3 10H O Na CO 3 10H O Na CO H O ( 10 ); K SO 3 H O K SO 3 + H O ( 00 ); CaCl 6H O CaCl + 6H O ( 50 ) CuSO 4 5H O ( ) 105 CuSO 4 H O ( ) CuSO 4 H O ( ) 50 5
6 ( ) Al (SO 4 ) 3 18H O Al (SO 4 ) H O ( 40 ); oso 4 7H O CoSO 4 + 7H O ( 410 ); (FeCl 3 6H O) = Fe O 3 + 6HCl + 9H O ( 50 ); (AlCl 3 6H O) = Al O 3 + 6HCl + 9H O ( ) 1? ( )? 3 K 391%? 4 644? acl H O 10? % 7 10% (II) 5%? 6
7 " " 1 (m 0 ) 5 10 (m 1 ) (m ) 30 (m 3 ) m = m m 7
8 * * * ( ) ( ) (U) (q) (W) U q W q = U + W (1) (1) ( = const) q r = U + p V = (U + pv ) (U 1 + pv 1 ) = H H 1 = H () H W = 0 (1) q v = U (3) =const H H 1 H ( ) +? ( ) ( ) rh (98) = 86 / rh (1 ) = const VT = const 500 1) Ca ( ) +?O ( ) ao ( ) fh (CaO; T) = 6369 / 8
9 ) ( ) + ( ) + ( ) ( ) ( ) fh (Ca(OH) ; T) = 9880 / 3) ( ) +? ( ) ( ) fh ( ; T) = 850 / 4) ( ) + ( ) ( ) ( ) rh =? fh ( ( H 0) ( H 0) ) (4) (1) (3) (4) = () (1) (3) ( ) + ( ) + ( ) + ao ( ) + ( ) ( ) ( ) + Ca ( ) +?O ( ) + ( ) +? ( ) ( ) + ( ) ( ) ( ) rh 4 = rh r H 1 r H 3 rh 4 = 9880 ( 6369) ( 850) = 661 / 0 n = /56 = = ?? ? 7? 8 fh (98) rh (98)? 9 9
10 / 86 / K lo 3( ) KCl ( ) + 3/O ( ) rh (98)= 494 ; K lo 4( ) KCl ( ) + O ( ) rh (98)= 33 4K lo 3( ) 3K lo 4( ) + KCl ( ) 98 K ( ) (IV) / 4 98 K ZnS Zn + S rh 1 = 005 ; ZnS + 3 ZnO + SO rh = 8935 ; SO + SO 3 rh 3 = 198 ; ZnSO 4 ZnO + SO 3 rh 4 = * 13 ( ) ? " " ( )
11 1 NaOH ( ) Na + ( ) + OH ( ) rh 1 1) 001 ) 00 3) 1 4) 5) NaOH ( ) + H + ( ) + Cl ( ) H O + Na + ( ) + Cl ( ) rh 1) HCl 3 Na + ( ) + OH ( ) + H + ( ) + Cl ( ) H O + Na + ( ) + Cl ( ) rh 3 1) ) ) 4) 1) ; ) ; 3) c ; 4) ; 5) N OH ; 6) 1 NaOH; 7) rh r H 1 + rh 3 ; 8) 11
12 " " ( ) = C/ t (4) aa + bb + cc + dd ( ) = k[a] na [B] nb (5) 1
13 n A n B n A + n B k (5) ( ) ( ) /10 (6) / 1 = k = Ae E/RT (7) R + K K K + + K 1? 3 ) Cl Cl (k 1 ); ) Cl Cl (k 1 ); ) l + Cl l + Cl (k ) 4? 13
14 5? 6?? 1 * + / [A] 0 M [B] 0 M 0 M c (IV) 80 (IV)? ? 5 + = + 88 K /( ) 9796 / 313 K? 6 75 / ? 14
15 3 " " (Na S O 3 ) Na S O 3 Na S O 3 H SO 4 /( ) ??? 15
16 16 ( ) ( ) ma + nb pc + qd 1 = k 1 [A] m [B] n = k 1 [C] p [D] q (8) 1 = K = k 1 /k 1 = [C] p [D] q / [A] m [D] n (9) [A] [B] [C] [D] (9) V = const p = const K ( ) ( ) (
17 ) D 1 [A] 0 = 3 [B] 0 = M [A] M [B] M [C] M [D] M K = [C] [D] / [A] [B] = (x x) / (3 x) ( x) = 1 M [A] = 3 1 = 18 M; [B] = 1 = 08 M; [C] = [D] = 1 M N O 4( ) NO ( ) rh o = 580 ) N O 4 ; ) NO ; ) ; ) ; )? ) N O 4 ; ) NO ; ) ; ) ( ) ; ) 17
18 1?? 3 3( ) ( ) + ( ) 4 K ) S ( ) + 4H ( ) H 4( ) + H S ( ) ; ) 4NH 3( ) + 5O ( ) 4NO ( ) + 6H O ( ) ; ) 4NH 3( ) + 3O ( ) N ( ) + 6H O ( ) 5 NO + O NO NO [NO] = 0056 [O ] = 008 [NO ] = 0044 / D 1 [A] = 3 [B] = / 7 N + 3H NH CrO 4 ( ) + + ( ) Cr O 7 ( ) + H O ( ) ) ?? ) 1 1 1? 18
19 Cr O 7 ( ) + ( ) CrO 4 ( ) + + ( ) +??? BaCrO 4( ) Ba + (p ) + ro 4 ( ) ) ) 4 5? 3) 4 5? CrO 4 ( ) + + ( ) Cr O 7 ( ) + H O ( ) Cr O 7 ( ) + ( ) CrO 4 ( ) + + ( )? ( ) ( ) ( ) 19
20 1 W = m / (m 1 + m ) (10) W ; m ; m 1 1 = n / V (11) / ; n ; V = n / (n + n 1 ) (1) X ; n 1 n 1 S (II)? (II) 00 n = V C n(cuso 4 ) = 0 01= 00 n(cuso 4 5H O) = n(cuso 4 ) M(CuSO 4 5H O) = = 50 / m = = 5 5 0
21 38% HCl 119 / 1? 1614? n(hcl) = 1 = HCl m(hcl) = 365 = 73 m( HCl) = 73 / 038 = 191 V( HCl) = 191 / 119 = % 100 m(c H 5 OH) = = 96 ; m(h O) = = 4 ; M(C H 5 OH) = = 46 / ; (H O) = = 18 / n(c H 5 OH) = 96 / 46 = 09 ; n(h O) = 4/18 = 0 ; (C H 5 OH) = 09 / (09 + 0) = 0905 (H O) = = ? K SO W = 14 / 114 = 0176 m(k SO 4 ) = = 7051 K SO 4 0 (400 ) W(K SO 4 ) = (7051 x)/(400 x) = 111 / 1111 = 01 = (II) % FeCl + xh O FeCl xh O 1
22 = = 990 FeCl = = 155 FeCl (FeCl ) = 17 / (FeCl xh O) = ( ) / W(FeCl ) = 155 / 43 = (17 / ) = 4 FeCl 4H O % %? 30% ( = 1 / )? 3 %? ? % HNO 3 ( = 15 / ) 30 48% HNO 3 ( = 13 / ) 145 / NH 4 Cl 50 " "
23 Na S O 3 5H O CrK(SO 4 ) 1H O CoSO 4 7H O K 4 [Fe(CN) 6 ] 3H O CuSO 4 5H O CuNO 3 3H O NiSO 4 7H O CoCl 6H O CrCl 3 6H O 1) ) 001 3) 4) 100 5% CuSO 4 1) ) 5% CuSO ( 001; 0015; 000; 0030 ) 3) 4) 3
24 ( ) K + K K + + K p [ K ][ A [ KA] ] K p K n n n n K K ( ~ 1) ( < 1) ( << 1) 4 1
25 1???? ) ) (II) ) (IV) ) )? 6 85% ( = 104 / 3 ) ( 945%) 7 HNO ? ( %) " " 6 ( ) AgNO 3 NO
26 N Cl + AgNO 3 AgCl ( ) + N NO 3 SO 4 Cl + Ag + AgCl Cl CO 3 PO 4 3 AsO 4 3 AgCl ( ) + N 4 [Ag(N 3 ) ]Cl + Cl Cl ( ) [Ag(N 3 ) ]Cl + N 3 AgCl ( ) + N 4 NO 3 7 " " SO 4 Cl ( ) NO 3 ( ) CO 3 6 ( ) ( ) 4 3 ( ) NO 3 " " (II) 5 6 [Fe(NO)SO 4 ]
27 CO 3 6FeSO 4 + NaNO 3 + 4H SO 4 3(Fe (SO 4 ) 3 + NO ( ) Na SO 4 FeSO 4 + NO ( ) [Fe(NO)SO 4 ] (II) Fe + Fe 3+ 3 NO NO ( ) + 1 3Fe + + NO Fe 3+ + NO ( ) Na CO 3 + CH 3 COOOH CH 3 COONa + H O + CO ( ) CO CO 3 H O + CO ( ) SO Na SO 4 + Cl SO 4( ) + NaCl SO SO 4( ) 7
28 4 3 NH 4 OH MgCl NH 4 Cl Na 4 + MgCl + NH 4 Cl Mg NH 4 4( ) +NaCl + HCl Mg + + NH 4 + MgNH 4 4( ) MgNH 4 4( ) + 3HCl MgCl + NH 4 Cl " " 8 Fe 3+ (III) 1 (II) K 4 [Fe(CN) 6 ] ( ) 4 5 (III) 4 5 (II) (III) (" ") " " 4Fe [Fe(CN) 6 ] 4 Fe 4 [Fe(CN) 6 ] 3 " " (NH 4 CNS) (K 4 CNS) 4 5 (III) 45 (III) 8
29 Fe 3+ + ncns [Fe(CNS) n ] (3n) ( ) (III) (III) Cu + NH (II) 3 CuSO 4 + NH 4 [Cu( )] SO 4( ) + (NH 4 ) SO 4 Cu + + NH 4 [Cu( )] + + NH 4 + [Cu( )] SO 4( ) + 6NH 4 + (NH 4 ) SO 4 [Cu(NH 3 ) 4 ] SO Cu + + 4NH 4 [Cu(NH 3 ) 4 ] ZnCl + N Zn( ) ( ) + N Cl 8 10 N Zn( ) ( ) + N N [Zn( ) 4 ] 9
30 Zn( ) ( ) + Cl ZnCl + (NH 4 ) [Hg(CNS) 4 ] ( ) Zn + + [Hg(CNS) 4 ] Zn[Hg(CNS) 4 ] ( ) K + 9 " ( " 1 N 3 [ (N ) 6 ] 4 5 K + + N 3 [ (N ) 6 ] K N [ (N ) 6 ] ( ) + N K + + Pb + + Cu + + 6N K Pb[ u(n ) 6 ] ( ) 30
31 NH N NH N NH 3( ) + N + + K [ gi 4 ] ( ) ( ) NH NH 3( ) + NH 3( ) + K [ gi 4 ] + 3K [NH Hg O]I ( ) + 7KI + N + Zn( 3 ) U ( 3 ) N + + Zn + +3U N Zn(U ) 3 ( 3 )
32 10 " " ( ) ( ) 3 Na + H O NaOH + CH 3 COOH 3 + H O CH 3 COOH NH 4 Cl + H O NH 4 OH + HCl NH H O NH 4 OH + + 3
33 4 3 NH 4 + H O CH 3 COOH + NH 4 OH 3 + H O CH 3 COOH + NH H O NH 4 OH + + ( ) = lg [ + ] <7 7< 14 = 7 ( ) < 7 > (III)? NaCl BaCl FeCl 3? 3 33
34 11 " " HCl 1 CH 3 COOH NaOH 3 NH 4 OH 4 CH 3 COONa 5 NH 4 NO 3 6 NaNO
35 * ( ) 1) ; ) ; 3) 1 H S S H S H SO 4 H SO 4 S H SO 4 S 3 ( ) Fe +3 Fe(III) Fe( ) 3 4 ( ) 1 FeCl 3 + KI Fe +3 Fe + ; I 35
36 I FeCl 3 + KI FeCl + KCl + I Fe +3 + e Fe + I e I Fe +3 +I Fe + + I FeCl 3 + KI FeCl + KCl + I + ( ) ( ) ; (II) FeS Fe 3+ (Fe(NO 3 ) Fe NO 3 ) SO 4 (H SO 4 SO ) FeS Fe 3+ +SO ( ) FeS +8H Fe 3+ + SO FeS 15 FeS +8H 15 Fe 3+ + SO NO 3 NO 1 + ( ) NO NO + 36
37 ( +1) NO NO + NO NO + 15 FeS +8 H 15 Fe 3+ + SO FeS +8H + 15NO Fe 3+ + SO NO +15 8H 16 + FeS + 15NO Fe 3+ + SO NO +7 NO FeS + 18 NO 3( ) Fe(NO 3 ) 3 +H SO NO +7 1 Cu + H SO 4 ( ) Fe + H SO 4 ( ) Fe + H SO 4 ( ) Zn + H SO 4 ( ) Cu S + O CuO + SO Ca 3 (PO 4 ) + C + SiO CaSiO 3 + P + CO 3 H S + HNO 3 H SO 4 + NO + H O KI + H SO 4 ( ) I + S + K SO 4 + H O 4 * MnO + KBr + H SO 4 Br + KMnO 4 + SO + H O K Cr O 7 +K S + H SO 4 S + CrCl 3 + NaClO + NaOH NaNO + Cl + NaOH 37
38 1 " " 000 M % 005 5% (365%) 10% % 3? 005? %
39 * * * 1 Cl e Cl K Na + + e Na o = 714 NaCl Na + Cl ( ) ( ) ( ) ( ) ( ) ( F ) F 39
40 Na SO 4 c = 13 SO 4 S O 8 = 01 SO 4 Cl e Cl = 1360 K + + = 088 NaCl + Cl + H + NaOH ( ) (II) Cu(NO 3 ) ( ) Cu + + NO 3 Cu o e Cu + K Cu + + e Cu o = 19 K Cu + + e Cu o = 0337 Cu(NO 3 ) + Cu + 4HNO 3 + O 1) ) 3) m = I t / n F (13) 40
41 m ; / ; n ; I ; t ; F / % ( ) NaNO 3( ) A + + K H O 4e 4H + + O H O H + O m(nano 3 ) = = 5 m(h O) = 50 5 = 5 W (NaNO 3 ) = 0 0 = 5 / (50 ) = 15 n = 15 / 18 = / = 347 ( ) V(H ) = = 1556 V(O ) = = ( = 117 / ) 540 AgNO 3( ) A Ag + + e Ag o 4 K H O 4e 4H + + O 4AgNO 3 + H O = 4Ag o + 4HNO 3 + O n(agno 3 ) = = 014 n(ag) = 54 / 107 = 005 n(hno 3 ) = 005 m(hno 3 ) = = 315 n(o ) = 0015 m(o ) = 04 n(agno 3 ) = = 85 41
42 m(h O) = = 045 n(h O) = = = = 111 W(HNO 3 ) =315 / 111 = % W(AgNO 3 ) = ( ) 170 / 111 = % W(AgNO 3 ) = 1130% W(HNO 3 ) = 83% % NaOH * 400 8% (II) 05 1 C H SO 4 CuCl AgNO 3 3 ) ; ) 4 ) ; ) 5? 6 7 (II) ) Sn e Sn + ; ) Cl e Cl ; ) H O 4e 4H + + O? ) KCl; ) CuCl ; ) Cu(NO 3 )? 4
43 8 (II) H O 4e 4H + + O? " " 13 1 U 05 5% (II) U ? 7 [OH ] 8 01 (III) 9 10 U
44 I 1 CaSO 4? M r CaSO 4 M r (CaSO 4 ) = H SO 4? m ( H) 1 m( S) 3 m( O) 16 m H m( S) 4m( O) m (H) m (S) m (O) = CaSO 4? M r CaSO %
45 % W O % CaSO 4 Ca 09; S 03; O Na 706%; N 1647%; O 5747%? Na x N y O z ( %) x y z x y z NaNO %; 143% 05 14? CH 1 ( ) x x 1 3 x (05 14 ) ( 1 ) /
46 x = 8 / /14 / = ( ) 4 x 1 x + x = 8 x = C H 4 C H 4 ( ) 36 4 ( ); 16? 1 D M( ) ( ) 16 M 3 / 36 1 / n( HO) 0 18 / 4 n( N) / n( H) n( HO) 0 4 n N) n( N ) 0 ( 3 n ( ) n( N) (N ) x x 1 x (14 + ) = 3 x = N H 4 (H N NH ; ) 46
47 II 1 1) ; ) ; 3)? 700 )? 1 m( N m( H m( ) ) m( He) ) / 8 / DH N 1400 / 8 / D N / 8 / D He N / ( 1 ) 15 / 1) 1400; ) 096; 3) (IV) ( M CO m( CO ) 1 44 / / 44 47
48 (IV) 4 V( CO) (IV) ( ) (IV) ( )? n ( NO ) V V m (IV) M NO / m V M V m n ( NO ) / m M 56 (IV) 115 / ? m( H m( 10 D H ( ) ) 4 (05 4 / ) (05 ) x / )
49 V(H ) = 0 V(O ) = 03 ( ) ( H )% % = 0 40 % (O ) = (100 40)% = 60% ( ) 40% 60% III 1 100% %? (m ) (m) m = m W = = 50 ( ) m 1 = m m = = % % * 50 10% 96% ( = 184 / 3 )? 1 96% x m = (x 096) m 1 = (x 096 x) 50 10% m = = 5 m 1 = 50 5 = 45 3 x x 096 = 5 x = % 50 10% V = m / = 51 / 184 / = 83 V = 45 (x 096 x) = % % 49
50 * 40% % 60%? (VI) % m(so 3 ) = = = 90 (VI) 90 m = / / 98 / = 735 (VI) = % 40 (VI) 40 m = / / 98 / = 365 (VI) % m = / / 98 / = 49 10% % x (VI) 365 x/100 60% y (VI) 49 y/ x y x = 11 x 49 y % ( = 184 / )? 96% m / 9 H SO 4 m = =
51 m' = = W 06 6% 14 6% IV 100 C 0 C 50? ( 100 ) 0 C C C 0 C 100 C C C = V ( ) 50% 161 Zn H SO4 ZnSO4 H n ZnSO ) 161 /161 / 0 1 ( 4 n(znso 4 ) = n(h ) = n(h SO 4 ) V( H ) Vm n 4 51
52 m( H SO 4 n M 9 8 m( ) = 98 / 05 = % ; 4 ( ) 31 95% Ca ( PO ) C 3SiO 3CaSiO P 5CO n / n Ca3 ( PO4 ) 05n( P) n(p) = 0 m(p) = 0 31 / = 6 m'(p) = = % 6? Ca ( PO ) C 3SiO 3CaSiO P 5CO n(p) = 6 /31 / = 0 n Ca3 ( PO4 ) 05n( P) n Ca PO ) 0 1 3( 4 3 ( PO4 ) m Ca / 31 30% m = 31 / 07 = % 5
53 KCl 10 85% 1 AgNO3 KCl KNO3 AgCl 745 n(kcl) / n( AgNO ) / n( AgCl) 0 05 m(agcl) = / = % ( = 105 / )? 1 Mg + H SO 4 = MgSO 4 + H n(mg) = n(mgso 4 ) = n(h ) = m(mg) / M(Mg) = 48 / 4 = 00 n(h SO 4 ) = / 98 = 06 m(mgso 4 ) = n(mgso 4 ) M(MgSO 4 ) = 0 10 = 4 m(h ) = n(h ) M(H ) = 0 = 04 3) m 1 ( ) = V(H SO 4 ) = = 10 m ( ) = m 1 ( ) + m(mg) m(h ) = =
54 4) MgSO 4 W(MgSO 4 ) = m(mgso 4 ) / m ( ) = 4 / 144 = % 11% % ( = 111 / ) 300 0% ( = 110 / ) 8%? 1) H lo 4 + NaOH = NaClO 4 + H O ) n(naoh) = / 40 = 165 n(h lo 4 ) = / 1005 = 033 3) n(naclo 4 ) = n(h lo 4 ) = 033 m(naclo 4 ) = = 404 4) x = x V(H O) = 64 VI V(H O) = 64 ) ( 1 1 S O SO ( r H ( T) 97 / ) rh ( T)' ( 97 / 1000 ) / 3 /
55 ( ) 1970? O O r H ( T) 394 / O O r H ( T) 394 / x 1970 x = ( )/( 394 ) = 5 ( ) ( ) V Vm n 5 4 / VII 1 ( ) ( )? 1) n(h ) V 179 n ( H ) 008 V m 4 / ) x x Fe + H SO 4 = FeSO 4 + H y y Zn + H SO 4 = ZnSO 4 + H 3) x y x 65y 466 x = 008 y 56(008 y) + 65y = y + 65y =
56 018 = 9y y = 00 ( ) x = 006 ( ) n( Fe) 006( ); n( Zn) 00( ) m( Fe) n M ( ) 4) m( Zn) ( ) m(fe) = 336 ; m(zn) = ( )? 1 Fe Cu 3Cl Cl 56 FeCl CuCl x y m Fe 56 x ; m Cu 64 y 56x 64y 3 (1) () 1 5x y x 64y 15x y 005 x 0005 m( Fe) 56 / y m( Cu) 64 / ; 7 ( ) 1 04? 1 m(cu) = x x n( Cu) ( )
57 x 64 CuSO4 Fe FeSO4 x 64 Cu 1 1 n( Fe) n( Cu) x ( 64 ) 3 m(fe) m( Fe) x x( ) 4 m(cu) x 0875x = 04x; x = (II) 0% 10 (II) 1 m( ) = = m ( ) = 50 = 48 3 m(cuso 4 ) m( CuSO4 ) W1 ( CuSO4 ) m1 ( ) x x x CuSO4 Fe FeSO4 1 Cu 1 n ( Cu) n( Fe) ( ) 5 x 6 n CuSO ) ( 4 n( CuSO4 ) n( Cu) 7 n (Cu) x + 64x = 10; x = 05 8 m(cuso 4 ) m (CuSO 4 ) = = 40 9 m (CuSO 4 ) m (CuSO 4 ) = = W ( CuSO4 ) % W CuSO ) 4% ( 4 57
58 1 11 ( ) ( ) ( ) ( ) ( ) /
59 59
60 ( ) 60
61 = 98 K E E Li e Li 3045 D e?d K e K 95 H e?h 0000 Rb e Rb 95 Ge e Ge +001 Cs e Cs 9 Cu e Cu 0337 Ba e Ba 906 Cu e Cu 051 Sr e Sr 888 Te 4 4e Te 0560 Ca e Ca 870 Rh e Rh 0600 Na e Na 713 Ag e Ag 0799 Mg e Mg 380?Hg e Hg 0798 Be e Be 1847 Hg e Hg 0854 U 3 3e U 1800 Pd e Pd 0987 Al 3 3e Al 1660 Pt e Pt 100 Ti e Ti 168 H + +?O e H O +18 Zr 4 4e Zr 159 Au 3 3e Au 1498 V e V 1186 Au e Au 1691 Mn e Mn 1180 H O e?h + OH 0879 Cr 3 3e Cr 0913 S e S 0480 Zn e Zn 0763 H O?O e OH 0401 Cr e Cr 0744?J e J 0536 Ga 3 3e Ga 059?Br e Br 1066 Fe e Fe 0440?Cl e Cl 1358 Cd e Cd 040 H e H 00 Tl e Tl 0335?F e F 870 Co e Co 077 Ca(OH) e Ca OH 300 Ni e Ni 050 Sr(OH) e Sr OH 880 Mo 3 e Mo 000 Ba(OH) e Ba OH 810 Sn e Sn 0140 Mg(OH) e Mg OH 690 Pb e Pb 016 Al(OH) 3 3e Al 3OH 300 Fe 3 3e Fe 0036 MnCO 3 e Mn CO
62 ( ) Cr(OH) 3 3e Cr 3OH 1480 V 3 e V 055 Zn(OH) e Zn OH 145 GeO 4H 4e Ge H O 0150 CdS e Cd S 1175 Ti 4 e Ti ZnCO 3 e Zn CO Sn 4 e Sn 0150 PbS e Pb S 0930 Sb O 3 6H 6e Sb 3H O 015 Cd(OH) e Cd OH 0809 Cu e Cu 0153 HgS e Hg S 0690 UO 4H e U 4 H O 0334 PbO H O e Pb OH Fe(CN) 6 e Fe(CN) Ag CrO 4 e Ag CrO H 3 AsO 4 H e H 3 AsO 3 H O 0559 PbSO 4 e Pb SO MnO 4 e MnO PbBr e Pb Br 084 UO 4H e U 4 H O 060 PbCl e Pb Cl 068 O H e H O 068 AgJ e Ag J 015 C 6 H 4 O H e C 6 H 4 (OH) 0699?Hg J e Hg J 0040 Fe 3 e Fe 0771 AgCN e Ag CN 0017 Hg e Hg AgBr e Ag Br 0071 Pu 4 e Pu HgO H O e Hg OH 0098 JO 3 6H 5e?J 3H O 1195?Hg Br e Hg Br MnO 4H + + e Mn + + H O 130 AgCl e Ag Cl 0 Tl 3 e Tl 150?Hg Cl e Hg Cl 068 Cr O 7 14H + + 6e Cr 3 + 7H O AgJO 3 e Ag JO PbO 4H e Pb H O 1456 Hg SO 4 e Hg SO MnO 4 8H 5e Mn 4H O 150 Ag C H 3 O e Ag C H 3 O Ce 4 e Ce Ag SO 4 e Ag SO PbO 4H SO 4 PbSO 4 H O e H O 6e W 8OH 105 MnO 4 4H + + 3e MnO H O Cr 3 e Cr 0410 Co 3 e Co 1840 Ti 3 e Ti 0369 WO 4 6
63 " " 1 * 63
64 1 ( )
65
66 This document was created with WinPDF available at http//wwwdaneprairiecom The unregistered version of WinPDF is for evaluation or noncommercial use only
Chapter 4: 11, 15, 18, 20, 22, 24, 30, 33, 36, 40, 42, 43, 46, 48, 50, 56, 60, 62, 64, 75, 82, 94 + C 2 H 3 O 2
Chapter : 11, 15, 18, 0,,, 0,, 6, 0,,, 6, 8, 50, 56, 60, 6, 6, 5, 8, 9 11. a. NaBr Na Br b. MgCl Mg Cl c. Al(NO ) Al NO d. (NH ) SO NH SO e. NaOH Na OH f. FeSO Fe SO g. KMnO K MnO h. HClO H ClO i. NH C
More informationChapter 4: 11, 16, 18, 20, 22, 24, 30, 34, 36, 40, 42, 44, 46, 48, 50, 56, 60, 62, 64, 76, 82, 94 + C 2 H 3 O 2
Chapter : 11, 16, 18, 0,,, 0,, 6, 0,,, 6, 8, 50, 56, 60, 6, 6, 76, 8, 9 11. a. NaBr Na Br b. MgCl Mg Cl c. Al(NO ) Al NO d. (NH ) SO NH SO e. HI H I f. FeSO Fe SO g. KMnO K MnO h. HClO H ClO i. NH C H
More informationPractice Worksheet - Answer Key. Solubility #1 (KEY)
Practice Worksheet - Answer Key Solubility #1 (KEY) 1 Indicate whether the following compounds are ionic or covalent a) NaCl ionic f) Sr(OH) 2 ionic b) CaBr 2 ionic g) MgCO 3 ionic c) SO 2 covalent h)
More informationCHEM 12 Unit 3 Review package (solubility)
CHEM 12 Unit 3 Review package (solubility) 1. Which of the following combinations would form an ionic solid? A. Metalloid - metal B. Metal non-metal C. Metalloid metalloid D. Non-metal non-metal 2. Which
More informationUnit 5: Chemical Equations and Reactions & Stoichiometry
pg. 10 Unit 5: Chemical Equations and Reactions & Stoichiometry Chapter 8: Chemical Equations and Reactions 8.1: Describing Chemical Reactions Selected Chemistry Assignment Answers (Section Review on pg.
More informationHW 7 KEY!! Chap. 7, #'s 11, 12, odd, 31, 33, 35, 39, 40, 53, 59, 67, 70, all, 77, 82, 84, 88, 89 (plus a couple of unassigned ones)
HW 7 KEY!! Chap. 7, #'s 11, 12, 15-21 odd, 31, 33, 35, 39, 40, 53, 59, 67, 70, 72-75 all, 77, 82, 84, 88, 89 (plus a couple of unassigned ones) 11) NOTE: I used the solubility rules that I have provided
More informationReaction Writing Sheet #1 Key
Reaction Writing Sheet #1 Key Write and balance each of the following reactions and indicate the reaction type(s) present: 1. zinc + sulfur zinc sulfide 8 Zn (s) + S 8 (s) 8 ZnS (s) synthesis 2. potassium
More informationSolubility Multiple Choice. January Which of the following units could be used to describe solubility? A. g/s B. g/l C. M/L D.
Solubility Multiple Choice January 1999 14. Which of the following units could be used to describe solubility? A. g/s B. g/l C. M/L D. mol/s 15. Consider the following anions: When 10.0mL of 0.20M Pb(NO3)
More informationWRITING CHEMICAL EQUATIONS 2002, 1989 by David A. Katz. All rights reserved. Permission for classroom used provided original copyright is included.
WRITING CHEMICAL EQUATIONS 2002, 1989 by David A. Katz. All rights reserved. Permission for classroom used provided original copyright is included. David A. Katz Chemist, Educator, Science Communicator,
More information1) What is the volume of a tank that can hold Kg of methanol whose density is 0.788g/cm 3?
1) Convert the following 1) 125 g to Kg 6) 26.9 dm 3 to cm 3 11) 1.8µL to cm 3 16) 4.8 lb to Kg 21) 23 F to K 2) 21.3 Km to cm 7) 18.2 ml to cm 3 12) 2.45 L to µm 3 17) 1.2 m to inches 22) 180 ºC to K
More informationFall 2011 CHEM Test 4, Form A
Fall 2011 CHEM 1110.40413 Test 4, Form A Part I. Multiple Choice: Clearly circle the best answer. (60 pts) Name: 1. The common constituent in all acid solutions is A) H 2 SO 4 B) H 2 C) H + D) OH 2. Which
More informationTASK 1 Writing formulas of ionic compounds. TASK 2 Writing formulas 1. TASK 3 Writing formulas 2. TASK 4 Writing balanced equations 1
TASK 1 Writing formulas of ionic compounds 1 AgBr 2 Na 2CO 3 3 K 2O 4 Fe 2O 3 5 CrCl 3 6 Ca(OH) 2 7 Al(NO 3) 3 8 Na 2SO 4 9 PbO 10 Na 3PO 4 11 Zn(HCO 3) 2 12 (NH 4) 2SO 4 13 Ga(OH) 3 14 SrSe 15 RaSO 4
More informationDirections: Use the rules for Assigning Oxidation numbers to determine the oxidation number assigned to each element in each of the given formulas.
Oxidation Numbers #00 Directions: Use the rules for Assigning Oxidation numbers to determine the oxidation number assigned to each element in each of the given formulas. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.
More informationAMOUNT OF SUBSTANCE. TASK 1 Writing formulas of ionic compounds. TASK 2 Writing formulas 1. TASK 3 Writing formulas 2
AMOUNT OF SUBSTANCE TASK 1 Writing formulas of ionic compounds 1 AgBr 2 Na 2CO 3 3 K 2O 4 Fe 2O 3 5 CrCl 3 6 Ca(OH) 2 7 Al(NO 3) 3 8 Na 2SO 4 9 PbO 10 Na 3PO 4 11 Zn(HCO 3) 2 12 (NH 4) 2SO 4 13 Ga(OH)
More informationCHEMICAL REACTIONS WORDS, SYMBOLS AND ABBREVIATIONS
CHEMICAL REACTIONS All chemical reactions have two parts: (1) A substance that undergoes a reaction is called a. In other words, reactants are the substances you start with. (2) When reactants undergo
More information[ ]:543.4(075.8) 35.20: ,..,..,.., : /... ;. 2-. ISBN , - [ ]:543.4(075.8) 35.20:34.
.. - 2-2009 [661.87.+661.88]:543.4(075.8) 35.20:34.2373-60..,..,..,..,.. -60 : /... ;. 2-. : -, 2008. 134. ISBN 5-98298-299-7 -., -,,. - «,, -, -», - 550800,, 240600 «-», -. [661.87.+661.88]:543.4(075.8)
More informationClosed Book Exam - No books or notes allowed. All work must be shown for full credit. You may use a calculator.
Chem 110 Name Exam #4 April 1, 2017 Closed Book Exam - No books or notes allowed. All work must be shown for full credit. You may use a calculator. Question Part I (40) Part II (60) Credit TOTAL Part I
More informationFaculty of Natural and Agricultural Sciences Chemistry Department. Semester Test 1. Analytical Chemistry CMY 283. Time: 120 min Marks: 100 Pages: 6
Faculty of Natural and Agricultural Sciences Chemistry Department Semester Test 1 Analytical Chemistry CMY 283 Date: 5 September 2016 Lecturers : Prof P Forbes, Dr Laurens, Mr SA Nsibande Time: 120 min
More informationBALANCING EQUATIONS NOTES
BALANCING EQUATIONS NOTES WHY DO WE NEED TO BALANCE CHEMICAL EQUATIONS? The LAW OF CONSERVATION OF MASS says that matter cannot be created or destroyed. In other words, you cannot end up with any more
More informationChem 12 Practice Solubility Test
Chem 12 Practice Solubility Test 1. Which combination of factors will affect the rate of the following reaction? Zn (s) + 2HCl ZnCl 2 + H 2(g) A. Temperature and surface area only B. Temperature and concentration
More informationBalancing Equations Notes
. Unit 6 Chemical Equations and Reactions What is a Chemical Equation? A Chemical Equation is a written representation of the process that occurs in a chemical reaction. A chemical equation is written
More informationFaculty of Natural and Agricultural Sciences Chemistry Department. Semester Test 1 MEMO. Analytical Chemistry CMY 283
Faculty of Natural and Agricultural Sciences Chemistry Department Semester Test 1 MEMO Analytical Chemistry CMY 283 Date: 5 September 2016 Lecturers : Prof P Forbes, Dr Laurens, Mr SA Nsibande Time: 90
More informationCHAPTER Describing Chemical Reactions Reactants Products. New substances produced The arrow means yields TYPES OF EQUATIONS.
CHAPTER 11 Chemical Reactions 11.1 Describing Chemical Reactions Reactants Products New substances produced The arrow means yields Where do Chemical Reactions occur? Everywhere!!! In living organisms In
More informationUnit Learning Targets (L.T.):
Unit 9: Chemical Equations and Reactions Chapters 8 and 19 Name Block Unit Learning Targets (L.T.): By the end of the unit, students will be able to: Chapter 8: 1. Correctly write and balance chemical
More informationTYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY
CHAPTER FOUR TYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY Questions 9. "Slightly soluble" refers to substances that dissolve only to a small extent. A slightly soluble salt may still dissociate
More informationGeneral Chemistry Multiple Choice Questions Chapter 8
1 Write the skeleton chemical equation for the following word equation: Hydrochloric acid plus magnesium yields magnesium chloride and hydrogen gas. a HClO 4 + Mg --> MgClO 4 + H 2 b HClO 4 + Mg --> MgClO
More informationHomework #3 Chapter 4 Types of Chemical Reactions and Solution Stoichiometry
Homework #3 Chapter 4 Types of Chemical Reactions and Solution Stoichiometry 13. Determine the concentrations of the solutions Solution A 4 particles 1.0 L Solution B 6 paticles 4.0 L Solution C 4 particles
More informationStoichiometry Problem Set 1 page g MgO g Al g Sr 3 (PO 4 ) M NaOH 5. a g Ni 5. b.
Stoichiometry (Baverstock/Kailley) 2006 1 Stoichiometry Problem Set 1 page 1 1. 4.15 g MgO 2. 26 g Al 3. 1.70 g Sr 3 (PO 4 ) 2 4. 0.628 M NaOH 5. a. 91.4 g Ni 5. b. 106 g Ni 3 N 2 6. a. 12.3 L O 2 6. b.
More information] after equilibrium has been established?
Chemistry 1 Solubility Equilibrium onster Review 1. A saturated solution forms when a 0. 10 mol of salt is added to 10. L of water. The salt is A. Li S B. CuBr C. Zn( OH) ( ) D. NH CO 4. Consider the following
More information(please print) (1) (18) H IIA IIIA IVA VA VIA VIIA He (2) (13) (14) (15) (16) (17)
CHEM 10113, Quiz 3 September 28, 2011 Name (please print) All equations must be balanced and show phases for full credit. Significant figures count, show charges as appropriate, and please box your answers!
More informationWriting Chemical formula with polyatomic groups
Writing Chemical formula with polyatomic groups 1. Use the Periodic table to determine the combining powers of single elements. Eg. Magnesium is in Group 2 and has a combining power of 2. 2. Use Table
More information7.01 Chemical Reactions
7.01 Chemical Reactions The Law of Conservation of Mass Dr. Fred Omega Garces Chemistry 100 Miramar College 1 Chemical Reactions Making Substances Chemical Reactions; the heart of chemistry is the chemical
More informationCHAPTER 1 QUANTITATIVE CHEMISTRY
Page 4 Ex 4 (a) element; (b) mixture; (c) compound; (d) element; (e) compound 5. (a) mixture; (b) compound; (c) mixture; (d) element; (e) compound. Page 5 Ex 1.1 3. C 4. D 5. C 6. D 7. a) 0.20 b) 1.2 10
More informationE5 Lewis Acids and Bases: lab 2. Session two lab Parts 2B, 3, and 4. Session one lab Parts 1and 2A. Aquo Complex Ions
E5 Lewis Acids and Bases: lab 2 Session one lab Parts 1and 2A Session two lab Parts 2B, 3, and 4 Part 2B. Complexation, Structure and Periodicity Compare the reactivity of aquo complex ions containing
More informationTYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY
CHAPTER TYPES OF CHEMICA REACTIONS AND SOUTION STOICHIOMETRY Questions 17. a. Polarity is a term applied to covalent compounds. Polar covalent compounds have an unequal sharing of electrons in bonds that
More informationChemistry COPYRIGHT SASTA 2018 WORKBOOK TOPICS. SACE STAGE 1 Australian Curriculum. SECOND EDITION Rhys Lewis. > Materials and their atoms
WORKBOOK SACE STAGE 1 Australian Curriculum Chemistry TOPICS > Materials and their atoms > Combining atoms > Molecules > Mixtures and solutions > Acids and bases > Redox reactions Supporting Teachers of
More informationOxidation numbers are used to identify the path of electrons in redox reactions. Each element in the compound must be assigned an oxidation number.
Packet 9: Oxidation-Reduction Reactions Many reactions are oxidation-reduction reactions A.k.a redox Reaction where one atom loses electrons and another atom gains electrons Atoms that lose electrons are
More informationName: 1. Law of Conservation of Mass atoms going into the reaction (reactants) must equal atoms coming out of the reaction (products)
Unit 8: Balancing and Identifying Chemical Reactions Packet Name: 1. Law of Conservation of Mass atoms going into the reaction (reactants) must equal atoms coming out of the reaction (products) C O 2 CO
More informationE5 Lewis Acids and Bases: lab 2. Session two lab Parts 2B, 3, and 4. Session one lab Parts 1and 2A. Aquo Complex Ions. Aquo Complex Ion Reactions
E5 Lewis Acids and Bases: lab 2 Session one lab Parts 1and 2A Part 2B. Complexation, Structure and Periodicity Compare the reactivity of aquo complex ions containing pre-transition, transition, and post-transition
More informationReview of Chemistry 11
Review of Chemistry 11 HCl C 3 H 8 SO 2 NH 4 Cl KOH H 2 SO 4 H 2 O AgNO 3 PbSO 4 H 3 PO 4 Ca(OH) 2 Al(OH) 3 P 2 O 5 Ba(OH) 2 CH 3 COOH 1. Classify the above as ionic or covalent by making two lists. Describe
More informationCHEM 107 (Spring-2004) Exam 2 (100 pts)
CHEM 107 (Spring-2004) Exam 2 (100 pts) Name: ------------------------------------------------------------------------, SSN -------------------------------- LAST NAME, First (Circle the alphabet segment
More informationBroughton High School
Name: Section: Chapter 21 Chemical Reactions Vocabulary Words 1. Balanced Chemical Reactions 2. Catalyst Broughton High School 3. Chemical Equation 4. Chemical Reaction 5. Coefficient 6. Combustion Reaction
More information7.01 Chemical Reactions
7.01 Chemical Reactions The Law of Conservation of Mass Dr. Fred Omega Garces Chemistry 152 Miramar College 1 Chemical Reactions Making Substances Chemical Reactions; the heart of chemistry is the chemical
More informationCHEMICAL REACTION. Engr. Yvonne Ligaya F. Musico 1
CHEMICAL REACTION Engr. Yvonne Ligaya F. Musico 1 Chemical Reaction Engr. Yvonne Ligaya F. Musico 2 Introduction Chemical reactions occur when bonds between the outermost parts of atoms are formed or broken
More informationTypes of Reactions. There are five types of chemical reactions we observed in the lab:
Chemical Reactions Acids and Bases Acids: Form hydrogen ions (H + ) when dissolved in water. HCl (aq) H + (aq) + Cl - (aq) Examples: HCl (hydrochloric acid), HNO 3 (nitric acid), H 2 SO 4 (sulfuric acid),
More informationTRU Chemistry Contest Chemistry 12 May 21, 2003 Time: 90 minutes
TRU Chemistry Contest Chemistry 12 May 21, 2003 Time: 90 minutes Last Name First name School Teacher Please follow the instructions below. We will send your teacher a report on your performance. Top performers
More informationUnit 1 - Foundations of Chemistry
Unit 1 - Foundations of Chemistry Chapter 2 - Chemical Reactions Unit 1 - Foundations of Chemistry 1 / 42 2.1 - Chemical Equations Physical and Chemical Changes Physical change: A substance changes its
More informationCHEMISTRY Midterm #2 October 26, Pb(NO 3 ) 2 + Na 2 SO 4 PbSO 4 + 2NaNO 3
CHEMISTRY 123-02 Midterm #2 October 26, 2004 The total number of points in this exam is 100. The total exam time is 50 min. Good luck! PART I: MULTIPLE CHOICE (Each multiple choice question has a 2-point
More informationChapter 19. Solubility and Simultaneous Equilibria p
Chapter 19 Solubility and Simultaneous Equilibria p. 832 857 Solubility Product ) The product of molar concentrations of the constituent ions, each raised ot the power of its stoichiometric coefficients
More informationO 2. Cl 2. SbCl 3. NaBr. NaCl
Name: Date: Chemistry ~ Ms. Hart Class: Anions or Cations 6.6 The Mole 1. Mg + O 2 à MgO Mg, O 2, and MgO are there? Mg: MgO? Mg O 2 MgO O 2:Mg? 2. Sb + Cl 2 à SbCl 3 Sb Cl 2 SbCl 3 Sb: Cl 2? 3. NaBr +
More informationPetr Vanýsek. As + 3 H e AsH H e 2 As + 3 H 2. AsO 4
Electrochemical Series Petr Vanýsek There are three tables for this electrochemical series. Each table lists standard reduction potentials, E values, at 298.15 K (25 C), and at a pressure of 101.325 kpa
More informationCircle the letters only. NO ANSWERS in the Columns! (3 points each)
Chemistry 1304.001 Name (please print) Exam 4 (100 points) April 12, 2017 On my honor, I have neither given nor received unauthorized aid on this exam. Signed Date Circle the letters only. NO ANSWERS in
More informationChemistry 12 Provincial Exam Workbook Unit 03: Solubility Equilibrium. Multiple Choice Questions
R. Janssen, MSEC Chemistry 1 Provincial Workbook (Unit 0), P 1 / 7 Chemistry 1 Provincial Exam Workbook Unit 0: Solubility Equilibrium Multiple Choice Questions 1. Which of the following would be true
More informationUNIT III: SOLUBILITY EQUILIBRIUM YEAR END REVIEW (Chemistry 12)
I. Multiple Choice UNIT III: SOLUBILITY EQUILIBRIUM YEAR END REVIEW (Chemistry 12) 1) Which one of the following would form an ionic solution when dissolved in water? A. I 2 C. Ca(NO 3 ) 2 B. CH 3 OH D.
More informationChemistry. Bridging the Gap Summer Homework. Name..
Chemistry Bridging the Gap Summer Homework Name.. Standard Form Number Number in standard form 0.008 8 x 10-3 0.07 7 x 10-2 0.55 5.5 x 10-1 0.000052 0.048 0.0086 0.00086 0.000086 0.0000000001 0.000455
More informationINSTRUCTIONS: Exam III. November 10, 1999 Lab Section
CHEM 1215 Exam III John III. Gelder November 10, 1999 Name TA's Name Lab Section INSTRUCTIONS: 1. This examination consists of a total of 7 different pages. The last page includes a periodic table and
More informationChapter 8. Chemical Equations. Flames and sparks result when aluminum foil is dropped Into liquid bromine.
Chapter 8 Chemical Equations Flames and sparks result when aluminum foil is dropped Into liquid bromine. Introduction to General, Organic, and Biochemistry 10e John Wiley & Sons, Inc Morris Hein, Scott
More informationChapter 4 Suggested end-of-chapter problems with solutions
Chapter 4 Suggested end-of-chapter problems with solutions a. 5.6 g NaHCO 1 mol NaHCO 84.01 g NaHCO = 6.69 10 mol NaHCO M = 6.69 10 mol 50.0 m 1000 m = 0.677 M NaHCO b. 0.1846 g K Cr O 7 1 mol K 94.0 g
More information4.02 Chemical Reactions
4.02 Chemical Reactions The Law of Conservation of Mass Dr. Fred Omega Garces Chemistry 111 Miramar College 1 Chemical Reactions Making Substances Chemical Reactions; the heart of chemistry is the chemical
More information7. Relax and do well.
CHEM 1215 Exam II John II. Gelder October 7, 1998 Name TA's Name Lab Section INSTRUCTIONS: 1. This examination consists of a total of 5 different pages. The last page includes a periodic table and a solubility
More informationChapter 1 -- Introduction to Qualitative Analysis. NAME: Lab Section: Date: Sign-Off:
Chapter 1 -- Introduction to Qualitative Analysis NAME: Lab Section: Date: Sign-Off: Chapter 1 -- Introduction to Qualitative Analysis Introduction Qualitative analysis is a way inorganic chemists have
More informationExperiment VI.4 - Complete Analyse
Experiment VI.4 - Complete Analyse The Ions found I the unknown substance were: NH 4 +, K +, Sr 2+, Cu 2+ ;B(OH) 4 -, Br -, Cl -, CrO 4. The testing process is described below. Cation analyse In a first
More information7. Relax and do well.
CHEM 1215 Exam II John II. Gelder October 7, 1998 Name TA's Name Lab Section INSTRUCTIONS: 1. This examination consists of a total of 5 different pages. The last page includes a periodic table and a solubility
More informationSCHOOL YEAR CH- 19 OXIDATION-REDUCTION REACTIONS SUBJECT: CHEMISTRY GRADE: 12
SCHOOL YEAR 2017-18 NAME: CH- 19 OXIDATION-REDUCTION REACTIONS SUBJECT: CHEMISTRY GRADE: 12 TEST A Choose the best answer from the options that follow each question. 1. During oxidation, one or more electrons
More informationPractice Final CH142, Spring 2012
Practice Final CH142, Spring 2012 First here are a group of practice problems on Latimer Diagrams: 1. The Latimer diagram for nitrogen oxides in given below. Is NO stable with respect to disproportionation
More informationChem 101 Practice Exam 3 Fall 2012 You will have a Solubility Table and Periodic Table
Chem 101 Practice Exam Fall 01 You will have a Solubility Table and Periodic Table 1. A 1.6-mol sample of KClO was decomposed according to the equation KClO (s) KCl(s) O (g) How many moles of O are formed
More informationChemistry Standard level Paper 1
M15/4/CHEMI/SPM/ENG/TZ1/XX Chemistry Standard level Paper 1 Thursday 14 May 2015 (afternoon) 45 minutes Instructions to candidates Do not open this examination paper until instructed to do so. Answer all
More informationChemical Reactions. Chemical Reactions 5 signs/evidence of chemical reactions:
Chemical Reactions Chemical Reactions 5 signs/evidence of chemical reactions: Chemical Reaction: a process in which one or more substances are converted into new substances with different chemical and
More information7. Relax and do well.
CHEM 1215 Exam II John II. Gelder October 13, 1999 Name TA's Name Lab Section INSTRUCTIONS: 1. This examination consists of a total of 5 different pages. The last page includes a periodic table and a solubility
More informationSOLUBILITY REVIEW QUESTIONS
Solubility Problem Set 1 SOLUBILITY REVIEW QUESTIONS 1. What is the solubility of calcium sulphate in M, g/l, and g/100 ml? 2. What is the solubility of silver chromate? In a saturated solution of silver
More informationCircle the letters only. NO ANSWERS in the Columns!
Chemistry 1304.001 Name (please print) Exam 5 (100 points) April 18, 2018 On my honor, I have neither given nor received unauthorized aid on this exam. Signed Date Circle the letters only. NO ANSWERS in
More informationUNIT 9 - STOICHIOMETRY
General Stoichiometry Notes STOICHIOMETRY: tells relative amts of reactants & products in a chemical reaction Given an amount of a substance involved in a chemical reaction, we can figure out the amount
More informationSolution Stoichiometry
Chapter 8 Solution Stoichiometry Note to teacher: You will notice that there are two different formats for the Sample Problems in the student textbook. Where appropriate, the Sample Problem contains the
More informationAssignment #1: Redox Reaction Skill Drills
Assignment #1: Redox Reaction Skill Drills Skill #1 Assigning Oxidation Numbers (Text Reference: p. 639 641) All elements have an oxidation number of 0. In compounds, oxidation numbers add up to 0. o Group
More informationTYPES OF CHEMICAL REACTIONS AND SOLUTION STOICHIOMETRY
CHAPTER TYPES OF CHEMICA REACTIONS AND SOUTION STOICHIOMETRY Questions 1. a. Polarity is a term applied to covalent compounds. Polar covalent compounds have an unequal sharing of electrons in bonds that
More informationOctober 19, 1999 Page 1. Chapter 4 Practice Worksheet Dr. Palmer Graves, Instructor MULTIPLE CHOICE
October 19, 1999 Page 1 MULTIPLE CHOICE Section 4.1 Some Ways that Chemical Reactions Occur 1. The reaction of HNO (aq) + KOH(aq) KNO (aq) + H O(l) is best classified as a(n) a) acid-base neutralization
More informationAqueous Equilibria: Part II- Solubility Product
Aqueous Equilibria: Part II- Solubility Product PSI AP Chemistry Name-------------------------- I Solubility Product, K sp MC #63-103 a) Writing K sp expression b) Solving for K sp c) Solving for (molar)
More information8. Relax and do well.
CHEM 1225 Exam I John I. Gelder February 4, 1999 Name KEY TA's Name Lab Section Please sign your name below to give permission to post your course scores on homework, laboratories and exams. If you do
More informationLast 4 Digits of USC ID:
Chemistry 05 B Practice Exam Dr. Jessica Parr First Letter of last Name PLEASE PRINT YOUR NAME IN BLOCK LETTERS Name: Last 4 Digits of USC ID: Lab TA s Name: Question Points Score Grader 8 2 4 3 9 4 0
More informationQ (S) + ML (AQ) L (S) + QM (AQ) Is this reaction exo or endothermic? 2Al 2 O 3(S) + heat 4Al (S) + 3O 2(G) CaCrO 4(AQ) + K (S)
CH 4(G) + 2O 2(G) 2H 2 O (G) + CO 2(G) set up C 5 H 12 + O 2 PQ P + Q 2Mg (S) + O 2(G) 2MgO (S) + heat H 2 O 2 O 2 + H 2 O PF 5(S) P (S) + F 2(G) RZ (AQ) + QP (AQ) RP (AQ) + QZ (S) Fe 2 S 3(S) + heat 2Fe
More informationChemical Reactions. G.10 Science Chemistry Shaftesbury High School
G.10 Science Chemistry Chemical Reactions 2240 Grant Avenue Winnipeg, Manitoba R3P 0P7 Phone: 204-888-5898 Fax: 204-896-5492 http://www.pembinatrails.ca/shaftesbury/ G.10 Science WHMIS and MSDS Reading
More informationChapter 4. Page 162, Practice Problems. 1. Atom Reactant Side Product Side Na 8 8 O H 8 8 Cr 4 4. Page 159, Quick Check
Chapter 4 Page 157, Quick Check 1. Color change (red to silver), evolution of gas (O 2) and increase in solid mass. 2. CaO (s) + H 2O (l) 2 Ca(OH) 2(s) 3. Carbonic Acid decomposes to carbon dioxide gas
More informationStation #1: Classifying Reactions. Station #1: Classifying Reactions
Station #1: Classifying Reactions 1. PbCl2 + AgNO3 Pb(NO3)2 + AgCl 2. NH3 + HCl NH4Cl 3. AlCl3 + Na2SO4 Al2(SO4)3 + NaCl 4. Zn + S ZnS 5. Al2(SO4)3 + BaCl2 BaSO4 + AlCl3 6. Al2S3 Al + S 7. H2SO4 + Fe H2
More informationChemistry 1A3 Instructor: R.S. Dumont Supplementary Problem Solutions Types of Reactions
Chemistry 1A3 Instructor: R.S. Dumont Supplementary Problem Solutions Types of Reactions 1. (a) 2 Ca(s) + O 2 (g) 2CaO(s) (b) 2 Na(s) + Cl 2 (g) 2NaCl(s) (c) 16 Ga(s) + 3 S 8 (s) 8Ga 2 S 3 (s) (d) 3 Ba(s)
More information**The partially (-) oxygen pulls apart and surrounds the (+) cation. The partially (+) hydrogen pulls apart and surrounds the (-) anion.
#19 Notes Unit 3: Reactions in Solutions Ch. Reactions in Solutions I. Solvation -the act of dissolving (solute (salt) dissolves in the solvent (water)) Hydration: dissolving in water, the universal solvent.
More informationBalancing CHEMICAL EQUATIONS
Balancing CHEMICAL EQUATIONS CHEMICAL REACTIONS involves a chemical change in the identity of one or more chemical species Ex. Rusting of iron (Fe): chemical rxn btwn water and iron involve the breaking
More informationChapter 4. Properties of Aqueous Solutions. Electrolytes in Aqueous Solutions. Strong, weak, or nonelectrolyte. Electrolytic Properties
Chapter 4 Reactions in Aqueous Solution Observing and Predicting Reactions How do we know whether a reaction occurs? What observations indicate a reaction has occurred? In your groups, make a list of changes
More information! b. Calculate the ph of the saturated solution. (Hint: How many OH ions form for every Zn(OH) 2 that dissolves? Calculate poh, then ph.)! (8.
AP Chem Worksheet: Solubility Product, K sp Page 1 Write your chemical equations for dissolving the solid and the K sp expression before trying to solve the problems!! 1. The molar solubility of copper(i)
More informationInstructions. 1. Do not open the exam until you are told to start.
Name: Lab Day and Time: Instructions 1. Do not open the exam until you are told to start. 2. This exam is closed note and closed book. You are not allowed to use any outside material while taking this
More informationCHAPTER 4 THREE MAJOR CLASSES OF CHEMICAL REACTIONS
CHAPTER 4 THREE MAJOR CLASSES OF CHEMICAL REACTIONS END OF CHAPTER PROBLEMS 4.1 Plan: Review the discussion on the polar nature of water. Water is polar because the distribution of its bonding electrons
More informationName: Date: Period: Page: Balancing Equations
Name: Date: Period: Page: Balancing Equations In a chemical reaction, one or more reactants change into one or more products. Chemists use chemical equations as a quick shorthand notation to convey as
More information1. Which one of the following would form an ionic solution when dissolved in water? A. I2 C. Ca(NO3)2 B. CH3OH D. C12H22O11
Chemistry 12 Solubility Equilibrium Review Package Name: Date: Block: I. Multiple Choice 1. Which one of the following would form an ionic solution when dissolved in water? A. I2 C. Ca(NO3)2 B. CH3OH D.
More informationUNIT 7 Assignment Electrochemistry (Chap 5-pg & Chap 19-pg ) ANSWERS. 2 MnO 4 2 NO 3
UNIT 7 Assignment Electrochemistry (Chap 5pg 193229 & Chap 19pg 845899) ANSWERS 7.1. Assign Oxidation numbers to the elements in bold print. 1) Mg 0 2) P 4 0 3) Cl I 4) HCl I 5) CaO +II 6) Na 2 O +I 7)
More informationPractice Packet Unit 12: Electrochemistry
Regents Chemistry: PRACTICE PACKET: ELECTROCHEMISTRY Practice Packet Unit 12: Electrochemistry Redox and Batteries? Ain t nobody got time for that!!! 1 For each word, provide a short but specific definition
More informationPart 2. Multiple choice (use answer card). 90 pts. total. 3 pts. each.
1 Exam I CHEM 1303.001 Name (print legibly) Seat no. On my honor, I have neither given nor received unauthorized aid on this exam. Signed Date Part 1. Nomenclature. 10 pts. total. 2 pts. each. Fill in
More informationPDF created with pdffactory trial version A) mol Answer: moles FeS 2 8 mol SO 2 /4 mol FeS 2 = mol SO 2.
Part A. [2 points each] For each question, circle the letter of the one correct answer and enter the answer on the TEST SCORING SHEET in pencil only. The TEST SCORING ANSWER SHEET will be considered final.
More informationChapter 8 Chemical Reactions
Chemistry/ PEP Name: Date: Chapter 8 Chemical Reactions Chapter 8: 1 7, 9 18, 20, 21, 24 26, 29 31, 46, 55, 69 Practice Problems 1. Write a skeleton equation for each chemical reaction. Include the appropriate
More informationCHEM 108 (Spring-2008) Exam. 3 (105 pts)
CHEM 08 (Spring-008) Exam. (05 pts) Name: --------------------------------------------------------------------------, CLID # -------------------------------- LAST NAME, First (Circle the alphabet segment
More informationSCH 3UI Unit 5 Outline Chemical Reactions Homework Questions and Assignments complete handouts: Balancing Equations #1, #2, #3, #4
Lesson Topics Covered 1 Note: Chemical Reactions and Chemical Equations definition of chemical reaction four signs of chemical change the Law of Conservation of Mass balancing chemical equations SCH 3UI
More informationChemical Reactions Chapter 17
Chemical Reactions Chapter 17 I. Physical Change Changes which affect the size or shape of the substance BUT NOT its chemical properties or formulas. e.g.; ice melting. Water is still water II. Chemical
More information