Give the products from the reaction of but-l-yne with:?
469836
(a) 1,2-dibromobut-1-ene\n(b) 2,2-Dichlorobutane (based on Markovnikovs rule)\n(c) don't know\n(d) don't know.  Na is often used to reduce multiple bonds in a free-radical manner, however you need a source of H.\n(e) KMnO4 is a very strong oxidising agent and is commonly used as tests for unsaturation.  The side product is MnO2, showing that two atoms of oxygen are transferred to the product.  The product formed I think is 1-hydroxy-butan-2-one.
860351
The right chemical reaction is as follow:\n      6CO2+12H2O+energy=C6H12O6+6O2+6H2O\nPlz do not get confused with this reaction\n     6CO2 + 6H2O + light (energy) --> C6H12O6 + 6O2\nBecause 1st one is latest one and you can't get in some conventional books.Later one is wrong.
1121888
Well you got C3H3Br + CH3S,  I'd guess something like ethane and bromosulfide.. C2H6 + BrS..
216732
This might not be the simplest way possible, but here goes:\n\n(Starting with propanone)\n1. H3O+/I2 (to form 1-iodopropanone)\n2. (Ph)3P (triphenylphosphine forms the ylide)\n3. butanone* (Wittig reaction, now you should have hepta-3-ene-2,6-one)\n4. H2/Pd/C (hydrogenate the C=C; hopefully this'll leave the C=O's alone)\n5. H3O+ (intramolecular aldol condensation to close the ring, with dehydration to form the C=C double bond)\n\n*can be made from propylaldehyde and CH3MgBr\n\nNote: No guarantees this'll work as desired. :)
664298
molecular: \nHCl + K2CrO4 + SnCl2 --> CrCl3 + SnCl4 + KCl + H2O\n\nnet ionic: H+ + CrO42- + Sn2+ --> Cr3+ + H2O + Sn4+\n\nreduction half reaction: \n3 e- + 3 H+ + 5 HCl + K2CrO4 --> CrCl3 + 2 KCl + 4 H2O\noxidation half reaction:\n2 HCl + SnCl2 --> SnCl4 + 2 H+ + 2 e-\n\nBalancing them:\n6 e- + 6 H+ + 10 HCl + 2 K2CrO4 --> 2 CrCl3 + 4 KCl + 8 H2O\n6 HCl + 3 SnCl2 --> 3 SnCl4 + 6 H+ + 6 e-\n-----------------------------------------------------------\n6 H+ + 16 HCl + 2 K2CrO4 + 3 SnCl2 --> 2 CrCl3 + 4 KCl + 8 H2O + 3 SnCl4
892172
44 A) 2AgO --> 4Ag + O2\n     B)NiCO3 --> NiO + CO2\n     C)NH4NO3 --> N2O + 2H2O\n48)A)2HCl + Ca(OH)2 --> CaCl2 + 2H2O\n     B)3Ag2SO4 + 2AlCl3--> 6AgCl + Al2(SO4)3\n     C)&D) I've no idea, sry\n50)A)C4H8 + 6O2 --> 4CO2 + 4H2O\n     B)2C8H18 + 25O2 --> 16CO2 + 18H2O\n     C)2C3H8O3 + 7O2 --> 6CO2 + 8H2O\n     D)2C3H6O + 5O2 --> 6CO2 + 6H2O\n52)A)2Al + 3H2SO4 --> Al2(SO4)3 + 3H2\n     B)2HCl + Ba(OH)2--> BaCl2 + 2H2O\n     C)Au + HCl--> No Reaction(gold is below hydrogen in the metal reactivity series)\n54)A)2Cl2 + 2NaI --> 2NaCl2 + I2\n     B)2NH3--> N2 + 3H2\n     C)4Na + O2--> 2Na2O\n     D)Mg + H2SO4--> MgSO4 + H2\n     E)MgCl2 + Ca(OH)2--> Mg(OH)2 + CaCl2\n     F)H2 + Cl2--> 2HCl
821017
(CH3)2 N2H2+2N2O4-->3N2+4H2O+2CO2
618652
.........CH3\n..........|\nCH3 - C  - CH2 - CH2 - CH2 - CH3\n.........      |\n........      CH3\n\nPlease ignore the full stops in the above image (which i had to insert to get the correct spacing).\n\nI hope that this helps you.
788892
OF2 (~104), NF3(~107), CF4 (~109.5),  BF3 (120), BeF2(~180)
861726
You will have to first find the standard enthalpy values of the reactants and products. They're given in the appendices of any standard GenChem or PhysChem books.\n\nThen, you subtract the sum of enthalpies of reactants from the sum of enthalpies of the product to get the enthalpy of the reaction.
871220
The second one:\nAF + H --> AFM + TM\n\nActivation E: +10KJ/mol is moderate\nHeat of Rx= +5.2 KJ/mol is positive, making this an endothermic reaction (requires the input of heat or energy to accomplish).\n\nSee the site below:
813911
HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)
382396
reactants ----> products
222097
OH-(aq) + HC2H3O2(aq)---> H2O(l) + C2H3O2-(aq)
781970
2Al + 3F2 ----> 2AlF3\nAl + 3OH- ----> Al(OH)3
956142
2Rb(s)+Cl2(g)--2RbCl(s)
732503
U CAN MAKE A PROJ ON REACTIONS ON THE FOLLOWING PTS  .\n INTRO OF REACTIONS\nDEFINITION OF A REACTION\n TYPES OF REACTIONS\n THEN, DESCRIBE THEM SEPARATELY