Basic Chem, need help ...?
921419
First of all let's find how many moles of KClO3 you have and how many moles of O2 are produced\n\nmole KClO3=mass/MW =0.261/122.5 =2.13*10^-3\n\n2KClO3(s) → 2KCl(s) + 3O2(g) \nthus 2mole KClO3 produce 3 mole O2\n2.13*10^-3 mole KClO3 produce x mole O2\n\n=> 2x=3*2.13*10^-3 => x=3.19*10^-3 mole O2\n\nIn order to find the volume you use PV=nRT, but how much is P?\nYou know that the total pressure is 722.7 mm Hg and the pressure comming from water is 17.5 mm Hg. Thus the pressure coming from O2 is 722.7-17.5= 705.2 mm Hg.\nYou have to convert into atm.\n1 atm is 760.0 mm Hg\nThus 705.2 mm Hg is 705.2/760.0 = 0.928 atm. This value of P will go in the equation\n\nPV=nRT=> V=nRT/P= (3.19*10^-3)*0.082*293/0.928= 0.08259 L= 82.59ml
980626
ChemDraw from CambridgeSoft Corporation may be help you. Look at www.cambridgesoft.com
752218
2KOH + Cl2 ----> KCl + KClO + H2O\n6KOH + 3Cl2 ----> 5KCl + KClO3 + 3H2O
796834
try asking this in the chemistry section next time....
880509
Hey, Blond fuss?  Good luck if U can't find help @ this late hour. Keep working on it.  Something from what U learned so far must b in your head - find it. No one else working on this problem awake?\n  I'm tired (12:35 a.m. here @ UMAss Amherst).  Had 1st-yr Chem 100 in 1978, not Chem 1111 !! Barely remember orbitals.  U do have books there, don't U?   Library maybe not open now as ours is.    \n Aces 2 computer. So U have awhile 2 figure it out.  Good luck.
735321
At the top of your screen, 'Search:' type in 'chemistry help'...leads you to many sites offering a wide vaiety of help in chem...browse through them till you find the type of help you need...'let your browser do the walking'  :-))\n\nYou can narrow the search with...chemistry help math...I see some sites there which seem very helpful...good luck.
298587
Look up the site chemfinder.com.  its a good website if you have not heard of it already.
964798
This is a chem, not bio question
747676
2C12H22O11 +24O2 ---> 24CO2 +22H2O\na)F\nb)F\nc)T\nd)F\ne)T\nf)T
799202
for those questions, all you have to do is translate them into the equation\ncheck chemtutor.com for hellp, it explains it much clearer than i can
1081230
yup.  Email me with any other chem questions at\nfortitudinousskeptic@yahoo.com\nI teach chemistry.
245982
Sorry, can't help. You use different units than I do.\n but concentration(mol per dm3) = no. of moles(mol) divided by volume(dm3)
1030709
1 H2 +1 Cl2 ------> 2 HCl\n\n     2 - H - 2\n     2 - Cl - 2
856873
The balanced chemical equation and the molecular weights of the reactants and products.\n\nIn simplest form, assume\n\nXY + AB = XB + AY.\n\nAssume the molecular weight of XY is 50, and of XB 35.\n\nFor every 50 grams (or pounds, tons, or any other measure of mass) of XY you use, you will get 35 grams (or whatever) of XB.\n\nEDIT:\n\nOK. Atomic weights are H=1, S=32, Cl=35.5, Fe=56. Molecular weights are FeS=88, HCl=36.5, H2S=34, FeCl2=127.One gram molecular weight of each is its molecular weight in grams. So if you react 88 grams of FeS with 36.5 x 2 grams of HCl (the balanced equation tells you that you need 2 moles of HCl, so you multiply by 2), you will get 34 grams of H2S and 127 grams of FeCl2. \n\nNotice that the weights balance. You put in 88 + (36.5 X 2) grams, or 161 grams. You got out 127 + 34, or 161 grams.\n\nYou need the mass of any one of those compounds - either reactant or product - to determine all the others. Calculate the moles of the one for which you have the mass, determine the ratio of the moles of that one to the others, multiply by that ratio and then multiply result by molecular weight.
739042
2Al + 6HNO3 ----> 2Al(NO3)3(s) + 3H2(g)
1142668
Problem \ntake D as Delta(change)\nEstimate the change in enthalpy, DH, for the following reaction:\n\nH2(g) + Cl2(g) --> 2 HCl(g) \n\nSolution \n\nTo work this problem, think of the reaction in terms of simple steps:\n\n\nFirst, the reactant molecules, H2 and Cl2, break down into their atoms\nH2(g) --> 2 H(g)\nCl2(g) --> 2 Cl(g)\n\n\nNext, these atoms combine to form HCl molecules\n2 H(g) + 2 Cl(g) --> 2 HCl(g) \n\nIn the first step, the H-H and Cl-Cl bonds are broken. In both cases, one mole of bonds is broken. When we look up the single bond energies for the H-H and Cl-Cl bonds, we find them to be +436 kJ/mol and + 243 kJ/mol, therefore for the first step of the reaction:\n\nDH1 = +(436 kJ + 243 kJ) = +679 kJ\n\nBond breaking requires energy, so we expect the value for DH to be positive for this step.\n\nIn the second step of the reaction, two moles are H-Cl bonds are formed. Bond breaking liberates energy, so we expect the DH for this portion of the reaction to have a negative value. Using the table, the single bond energy for one mole of H-Cl bonds is found to be 431 kJ:\n\nDH2 = -2(431 kJ) = -862 kJ\n\nBy applying Hess's Law, DH = DH1 + DH2\n\nDH = +679 kJ - 862 kJ\n\nDH = -183 kJ\n\n\nAnswer\n\nDH = -183 kJ
813911
HBr ---> H(+) + Br(-)\nH2SO3 ---> H(+) + HSO3(-)\nCH3-COOH ---> H(+) + CH3-COO(-)