Please, please help me with these questions about chemistry!?
959309
#3:   113.6g ammonia is the maxium that can be produced, i think.\n\nBecause:   when you set up the problem you do:   5 moles N2 x 2 moles of NH3 (coming from the balanced equation) x 17.04g NH3 (formula mass) equaling.....170.4g NH3\n\nthen....\n\nyou have to do it again with the other number(10 moles of H2)\n\n10 moles H2 x 2 moles NH3 / 3 moles H2 x 17.04 equaling....113.6.   \n\nyou pick the lowest number becuase that is the maxium you can have.   so your answer is 113.6g NH3.\n\n\n\n#4:   15.33g NO2 is the maxium that can be produced \n\nbecause: 20g of O2 to grams of NO2.  To do that, you do...20g O2 / 32g O2 (Formula mass of O2) x 46.0g NO2 (formula mass of NO2)\nwhich equals......28.76\n\nthen do:\n\n10gNO / 30.0g NO (formula mass of NO) x 46.01(formula mass of NO2) and that equals....15.33\n\nso your answer is the lowest number which is 15.33.\n\n\ni hope that helps and i hope i'm right...
857176
S(s) + O2(g)------>SO2\n\n2Mg(s) + O2(g) -------> 2MgO(s)\n\nH2(g) + Br2 (g) -------> 2HBr(g)\n\n4Fe(s) + 3O2 -------> 2Fe2O3 (s)\n\nThen there are single replacement \n\nA + BC-------> AC + B\nZn(s) + 2HCl(aq)------->ZnCl2 (aq) + H2 (g)\nwhere one element replaces another\n\nthen there is double replacement \n\nAB+CD -------> AD + BC\n\nIn this one the positive ions in the reacting compounds switch places\n\nBaCl2 (aq) + Na2So4 (aq) --------.BaSo4 (s) + 2NaCl (aq)
832296
A) Ag2O (s) >>> 2 Ag(s) + 1/2 O2 (g)\nB) Nickel (II) Carbonate >>>>>Nickel (II) Oxide + [Carbon Dioxide], NiCO3 -> NiO + CO2\nC) Ammonium Nitrate >>>>> Dinitrogen monoxide + Water,  \nC does not balance as is... but this does:\nNH3NO3 -> N2O +  1.5 H2O + 0.5 O2\n\n48) A) 2 HCl (aq) + Ca (OH)2 (aq) >>>> 2 H2O + CaCl2 (aq)\nB) 3 Ag2 SO4 (aq) + 2 AlCl3 (aq) >>>> 6 AgCl (s) + Al2(SO4)3 (aq) (silver chloride is a precipitate)\nC) H2C2O4 (aq) + 2 KOH (aq)>>>> 2 H2O + K2C2O4 (aq)\nD) CdBr2 (aq) + Na2S (Aq)>>> CdS(s) + 2 NaBr (aq) (Cadium sulfide is a precipitate)\n50 Write a Balanced equation for the complete combustion of each compound:\nA) Butene (C4H8): C4H8 + 6 O2 -> 4 CO2 + 4 H2O\nB) Octane (C8H18): C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O\nC) glycerol (C3H8O3): C3H8O3 + 3.5 O2 -> 3 CO2 + 4 H2O\nD) Acetone (C3H6O): C3H6O + 4 O2 -> 3 CO2 + 3 H2O\n52: Complete these:\nA) Al (s) + H2SO4 (aq)>>>> AlSO4(aq) + H2(g)\nB) 2 HCl (aq) + Ba (OH)2 (aq)>>>> BaCl2 (aq) + 2 H2O\nC) Au (s) + 3 HCl (aq) >>>>> AuCl3(aq) + 1.5 H2 (g)\n54) EACH IS BALANCED BUT "INCORRECT", so correct these:\nA) Cl-2 + NaI(aq) >>> NaCl2(aq) + I-\nB) NH3(aq) >>> N- + H+3\nC) Na (s) + O2(g) >>NaO2(s)\nD) 2Mg (s) + H2SO4(aq) >>Mg2SO4(aq) + H2(g)\nE) MgCl2(aq) + Ca(OH)2(s) >>> Mg(OH)2(s) + CaCl2(aq)\nF) H2 (g) + Cl2 (g) >> H2Cl2 (g)
701443
1 - the reaction is reversed and multiplied by 4, so the enthalpy change is 11508 kJ\n\n2 - The reaction is a combination of the other three (I'll label them A, B and C, in the order you have them listed, for simplicity)\nThe original reaction = 2A - 2B -2C (the first one is just doubled, the next two are doubles and reversed), so 2*321.8 - 2*571.7 - 2*288.6 = -2542.2 kJ\n\n3 - This one is done similarly to question 2.  A - 0.5B + 1.5C\n180.8 - 0.5*(-906.2) + 1.5*(-483.6) = -91.5 kJ\n\nHope this helps
176528
http://www.chem1.com/CQ/ionbunk.html\n\nskim this article.
93462
I hope this helps:\n\nhttp://chemistryformulas.com/
802199
Hi,, hydrogen, oxygen, and nitrogen... \n\n good luck.....
901730
Na2S (ag) + 2 AgNO3 (ag) --> 2 NaNO3 (ag) + Ag2S (cr)\n\nB. Double displacement.\n\nA. Ba(NO 3)2 and H2O\n\nB. If there is a double arrow (pointing in both directions), then it is an equilibrium reaction.\n\nC. Ba(OH)2 and HNO3\n\nD. Barium\n\nE. 6 oxygen atoms. \n\nF. Ba(OH)2
698715
www.chemistrycoach.com or try www.cbseguess.com!\n\nenjoy!
250561
i can help with part of it. i forgot a lot of my chemistry stuff. so ok  first thing is to find the number of mols of each reactant put in. which is  mass over molar mass. and then find the molar mass of 0.566g of H2. so you know how many moles of H2 are left over, so you should lknow how many moles reacted and you should know the number of mols of the other 2 substances too and you know the volume and you can figure out the masses so then   do that. \nand Kc is just the  conc of products over reactants. if youre in AP   i assume you know how to do that\nthen c   and    um partial pressures. i think it has something to do with mole ratios and   the total pressure. im not completely sure, but im sure you can figure it out. good job
781038
I agree with birchardv here...........we need the rest of the problem, and you would learn a lot more if you did the homework yourself.  But do whatever floats your boat.  Add the rest of the problem and I'll get back to you.\n\nOkay Chemist, this problem is really long; if you want a good answer with plenty of steps, I'll need more room.  Email me and I'll send you the solution.
315648
Try Chemistry.com.  It may not directly help you with 'chemistry', but the site is supposed to be geared toward helping you find 'the one' with advanced techniques based on the things that users would define as 'chemistry'.\n\n(Sorry for all the quotes.)
496291
You need to take more care in writing equations when you post your questions. \n\nPart a) is difficult to decipher - you will need to re post it.\n\nPart b) is presumably:\n\nCH3COOH(l) + 2O2(g) --> 2CO2(g) + 2H2O(l)\n\nPart c) is presumably:\n\nFe2O3(s) + 3CO(g) --> 2Fe(s) + 3CO2(g)\n\nI would next use heats of formation to calculate the enthalpy change in the above reactions using the formula:\n\ndelta Ho = sum(heats of formation of products) - sum (heats of formation of reactants).\n\nIf you have a data book or data sheet, these values can be found there (Note: heats of formation of elements are zero.) \n\nIf you want web sites that might help you, try googling "enthalpy change problems" +tutorial.\n\nI hope this helps you.
734957
Na+ Cl-\nC +4 O-2 each\nK+ Br-\nN-3 H+1 each\nMg+2 O-2
1081543
i find  your question, interesting but quite inexplicable .\n this is something new to me . I was not  very attentive in my chemistry studies ..\n\n My memory centre is all clogged up with other facts , just now ; \n this must be the reason for me , to draw a blank here\n \n i will go hang my head in shame .\n \n or hunt up that  answer,  LIT UP for all to see , on Google or some other wondrous search engine .\n\n good luck \n \n.
936280
You are so smart - in fact, so smart that no one in THREE DAYS has even tried to answer your question.  Perhaps you can take this question and apply it to another realm:  \n\nhttp://antoine.frostburg.edu/chem/senese/101/index.shtml\n\nThis way you can get your homework done for you by people who know chemistry on a much larger scale... Answers is not here for homework completion
1053423
Chemistry rocks!  Try these sites...