Chemistry help!!! --- composition by mass problem?
959337
The molal freezing point depression constant is 1.86 degree/molal\n\nWe observe a 5.5 degree depression in freezing point therefore its concentration is (5.5 degree*1 molal/1.86 degree) = 2.96 molal\n\nWe assume complete ionization of each substance therefore for x moles of NaNO3 there is 2x moles of particles, and for y moles of Mg(NO3)2 there are 3y moles of particles.\n\nLets scale things up to a liter of water, so instead of 6.6 g of solid we use 6.6 g * 20 = 132 g\n\nM.Wt. of NaNO3 is  84.99 g\nM.Wt of Mg(NO3)2 is 148.3 g (both are rounded)\n\n2x + 3y = 2.96\n\n x(84.99g/mole) + y(148.3g/mole) = 132\n\nTo solve the system, multiply row 1 by -42.495 and add to row 2\n\n-127.485 y  + 148.3 y= 132 -125.785\n   20.815 y = 6.215 ----------------> y = 0.299 moles, plug this into equation 1 to solve for x to get x = 1.032 moles\n\nremember we had scaled this up by 20, therefore we will have to scale it back down.\n\n0.299 moles Mg(NO3)2* 148.3 g/mole *(1/20) = 2.217 g\n\n1.032 moles NaNO3 *(85.99g/mole)*(1/20) = 4.437 g\n\nSum to double check our answer 2.217 + 4.437 = 6.654 g, which is close enough to 6.6 since we expect rounding errors.\n\nPlease check for computational errors.
844758
Using the Michaelis-Menten equation.\n\nMore details on teh attached link\n\nhttp://www.le.ac.uk/by/teach/biochemweb/tutorials/michment2print.html\n\nGood luck
749866
From the final mass of the MgO, calculate the number of moles of Mg that are present.  Then use a little algebra.  3 times the number of moles of Mg3N2 plus the number of moles of MgO equals the total number of moles of Mg.  Convert all of the moles to the appropriate mass, and solve for the number of moles of Mg3N2.  I leave the solution as an exercise for the student.
863456
I assume the reaction in question is:\n\nA(g) <-> 2B(g)\n\nThe equilibrium constant, Keq, for this reaction is given by:\n\nKeq = ([B]^2)/[A]\n\nwhere the quantities in square brackets mean "the concentration of" the relevant species at equilibrium (or, in the case of gases, one commonly uses equilibrium partial pressures instead of concentrations).\n\nPlugging in the values we are given for the equilibrium concentrations, we have that:\n\nKeq = (0.154^2)/0.381\n\nKeq = 0.0622
929547
I find 16.4% \nHere we go:\n1) As air is a mixture of O2 and N2 two reactions are taking place:\n2 Mg + O2 -> 2 MgO\n3 Mg + N2 -> Mg3N2 \nThere are 4 parts of N2 for one part of O2, so the second reaction takes place four times as frequently as the first. Effectively we have\n14 Mg + O2 + 4 N2 -> 2 MgO + 4 Mg3N2.\n\n2) The molar mass of MgO is 40.3 gram. In MgO the magnesium makes up 24.3/(24.3+16.0) = 60.3 % of the mass.\nThe molar mass of Mg3N2 is 100.9 gram, the magnesium makes up 3*24.3/(3*24.3 + 2*14.0) =  72.2 % of the mass. \n\n3) Assume you have 1000 gram of reaction product. If it is pure MgO you will find 603 gram of Mg. But in the mixture you will have  \ntwo moles of Mg2N3 (201.8 g) for every mole of MgO (40.3 gram).\nIn 1000 gram of mixture, 833.5 gram of  Mg2N3 and 166.5 gram of MgO. This gives 72.2% * 833.5 + 60.3% * 166.5 = 702 gram of Mg.\n\n4) Finally: the error is (702-603)/603 = 16.4%
880546
you should learn how to do your own homework hun you'll never be a chemist with someone else doing your work.\n\n\nuse the molar mass from the periodic table (units = g/1mol)\n\nCl = 35.5\nH = 1.00\n\nMolar mass, formula weight, molecular weight, formula mass = 36.5 grams HCl/1 mol HCl\n\n\nThis is a simple problem is you can't do this problem maybe you should try changing majors.  I hear history or english is nice.
762286
I agree with Keith P.\n\nYou solve this by finding the number of moles of Sn and F.  You divide the sample (g) by the formula weight of the element.\n1.0g/118.71g = 0.0084\n0.640g/ 19g = 0.0336\n\nNext you divide the smaller of the two by the other to establish a sort of ratio.\n0.0336>0.0084\n\n0.0336/0.0084 = 4\nThus there is one atom of Sn for every 4 atoms of F and the empirical formula is SnF4.
105785
Chemistry.....ouch!!!  When teaching the method for converting percentage composition to an empirical formula, I have devised the following rhyme: \n\nPercent to mass\nMass to mole\nDivide by small\nMultiply 'til whole
763458
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 5.601 g x (12 / 44) = 1.528 g\nhydrogen: 4.586 g x (2 / 18) = 0.510 g\noxygen: 4.078 - 1.528 - .510 = 2.040 g\n\n2) convert to moles.\ncarbon: 1.528 / 12 = 0.127 mol\nhydrogen: 0.510 / 1 = 0.510 mol\noxygen: 2.040 / 16 = 0.128 mol\n\n3) divide each molar amount by the lowest value to obtain small integers\ncarbon: 0.127 / 0.127 = 1\nhydrogen: 0.510  / 0.127 = 4\noxygen: 0.128 / 0.127 = 1\n\nThe empirical formula of the substance is (CH4O)n\nThe atomic masses add up to 32, which is also the molar mass of the compound. Therefore n=1 and we find CH4O
675016
This is a calorimetry problem! There is an equation for it:       \nq = mc(change in temperature)\n\nq = the heat required \nm= mass\nc= specific heat \n\nAlso, write out the reaction and products.
58627
OK. You've posted this (and several other) questions(with exactly the same words and numbers) for the fourth day in a row, and everytime you've asked, one person has responded and we've all got different answers. \n\nI can't tell if you're a student that's gonna keep posting until several people give you the same answer, of if you're just someone that likes making the nerdy people solve chem problems. If you're the latter, you should really find a better use for your time. \n\nIf you're the former, try working the problem yourself and then compare your answers to ours, or compare our approaches, your notes, and textbook and see where they differ. Our different answers can probably be attributed to some math or conversion error. \n\nRemember, when all else fails follow the units and you'll be right more times than not.
36586
Ag+ + SO4(2-) ---> Ag2SO4\nYeah, it's double replacement, and according to my solubility rules, AgSO4 is insoluble.
912712
1) determine the mass of carbon, hydrogen and oxygen.\ncarbon: 4.793 g x (12 / 44) = 1.307 g\nhydrogen: 2.452 g x (2 / 18) = 0.272 g\noxygen: 2.018 - 1.307 - .272 = 0.439 g\n\n2) convert to moles.\ncarbon: 1.569 / 12 = 0.109 mol\nhydrogen: 0.272 / 1 = 0.272 mol\noxygen: 0.439 / 16 = 0.0274 mol\n\n3) divide each molar amount by the lowest value to find small integers:\ncarbon: 0.109 / 0.0274 = 4\nhydrogen: 0.272  / 0.0274 = 10\noxygen: 0.0274 / 0.0274 = 1\n\nThe empirical formula of the substance is (C4H10O)n\nThe atomic masses add up to 74, which is also the molar mass of the compound. Therefore n=1 and we find C4H10O
295875
for each one you are given a ratio i am assuming you are combining them both, make sure you write out a balanced equation for each one to make sure, some may not be a 1 to 1 ratio, next Molarity=mol/L therfore 1M = 1 mol per liter, use your periodic table to figure out the molecular weight for each one, each one of these is a strong acid, strong base reaction, therefore they will create water as a product and a salt, in the first one you are given 1.0 mol of HCl which is reacting w/ NaOH 1 to 1, therefore you only need to convert to # of moles of NaOH i am assuming you can do this hopefully this helps
978323
*your equation is written wrong it should be:\n3Cu(NO3)2 + 2Na3(PO4) -> Cu3(PO4)2 + 6Na(NO3)\n\nyou have to find your limiter and your excess.\n 1.) change grams to moles using moler mass\n2.2 g Cu(NO3)2 times 1mol/188 g = .0117mol\n1.29 Na3(PO4) times 1mol/164 g = .00787 mol\n\n2.) divide the two by the # of moles in the balanced equation:\nCu(NO3)2: .0117 mol/3 = .00390 mol\n Na3(PO4): .00787 mol/2=.00394 mol\n\n3.) So now you know you don't really have a limiter, but we'll go with Cu(NO3)2 b/c there's a little less. \n\nthere are 3 moles of Cu(NO3)2 used for every 1 mol of Cu3(PO4) produced.\n\n.0117 mol Cu(NO3)2 times (1 mol Cu3(PO4) / 3 mol Cu(NO3)2) = .00390 mol Cu3(PO4) produced\n\nconvert that to grams\n\n.00390 mol Cu3(PO4) times 286 g/mol = 1.12 g \n\nthat's your actual\n\nso % yield = .89/1.12 x 100 = 79.5%\n\noh yeah you find moler mass by adding the molecular weights of the atoms in the compound (times the number of the atoms in that compound) for ex. Cu3(PO4) 63.6 x 3 + 31.0 + 16.0 x 4 = 286\n\nDon't forget significant figures!!\n\nhope that helped!
730330
Different chemical composition, one has a reaction the other doesn't..☺
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.