1. Which will have a greater boiling point elevation: 3.00g Ca(NO3)2 (the 3 and the 2 are subscripts) in 60.0g of water or 6.00g Ca(NO3)2 in 30.0g of Water?\n\n3. (multiple choice) The molarity of a 0.50-L solution containing 9.0g NaCl in water is:\na. 18 M\nb. 0.055 M\nc. 0.31 M\nd. 114 M\n\n4. (multiple choice) Which involves a colligative property?\na. heating a solvent\nb. allowing a carbonated beverage to warm to room temperature\nc. adding salt to the water in which pasta is being cooked\nd. pouring a concentrated solution into a dilute solution
138979
Doesn't anyone do their own homework anymore.  No wonder the youth of American are failing.  I dread our future.
529024
I'm not going to solve these for you.  My homework days are far behind me.  I will try to teach you how to solve them, though.\n\n1.  Keep your eye on the moles and don't get confused by the changing volumes.  The final solution is 0.00436 M  and you know the final volume so calculate how much KNO3 is there (=0.00436 M x Vf, in liters).  That will tell you how much was in 25 mL of the second solution.\nSo how much was in the total second solution?  Multiply the moles above by five (25 mL goes into 125 mL five times.)  This is also how many moles was in original 20 mL solution.\nSo, to find the molarity of the original solution, divide the number of moles by the volume, in liters (20 mL = 0.02 L).\n\n2.  Same thing goes for this one.  Multiply 1.66 M by .037 L (remember that M is in terms of liters, not millilieters!) to get the moles in the first solution.  Do the same for the second.  Add the two volumes together to get the resulting volume (in liters!!!)  Divide the sum of the moles in both solutions by the total combined volume and you have your answer.\n\n3.  Same would go for this one if you hadn't left out what volume of solution the Pb++ was in.  Anyway, the way to do this one is to notice that the reaction of Pb+2 ion and SO4-2 ion is a one-to-one proposition.  So, if you figure out how many moles of Na2SO4 you have in 0.009 grams of the stuff, you will also calculate how many moles of Pb(NO3)2 you have.  Divide that by the volume in liters and you have your answer.  [Nice touch by Teach to throw in univalent species to cross you all up.  These are irrelevant to the reaction you're considering.  Still, I like that.  Makes you think.]\n\nGood luck.
309653
CaCO3 ---> CaO + CO2\n\nNo. of moles of CaCO3 = 37.0/(40+12+(16*3)) = 0.37 moles\n\nMole ratio of CaCO3 to CO2 is 1:1, so the number of moles of CO2 = 0.37 moles.\n\nUsing PV = nRT,\n\nP = 1.15 atm = 116523.76 N/m2\nn = 0.37 moles\nR = 8.3144 J/molK\nT = 25+273.15 = 298.15 K\n\nV = 0.37*8.3144*298.15/116523.76 = 0.0078714 m3\nV = 7871.4 cm3
477308
1) 2Al + Fe2O3=Al2O3 + 2Fe\n2) 8.53g BaCl2*(1mole BaCl2/ 208.3G BaCl2)=.0410mole BaCl2\n.0410mole BaCl2*(2moles KCl/1moleBaCl2)=.0820 mole KCl2\n.0820mole KCl2*(110.1gKCl2/1mole KCl2)=9.03gKCl2\n3)i don't know this one\n4) 2NaOH + H2 is reduced\n5) i don't know this one\n6) 5gNa*(1moleNa/23gNa)=.217mole Na\n.217mole Na*(6.02x10^23/1mole Na)=1.31x10^23 atoms Na
54244
the difference in the boiling point of this soln compared to pure water equals Kb times m, where Kb is the boiling point constant(0.512 for water) and m is the molality of the soln.\nso the answer si 100.512 degrees C.
253664
The reaction is as follows:\n\nCaCO3 + 2HCl -----------> CaCl2 + H2O + CO2\n\nThe reaction produces H2O & CO2 as biproducts and CO2 at room temperature will be a gas. So it will escape from the recation media (Contaner)\nAs a result, this reaction is not a reversible reaction (since one byproduct is missing) and hence as time goes water will adds up and HCl will dilute more and more ... rection kinetics slowed down.\n\nCaCl2 is very much  soluble in water.. so its not the reason\n\nIs this answer is enough to deserve 10 pts from you?\n\nHave fun in chemistry always!
400379
a) 2NaCl ---> 2Na + Cl2\n\nb) pretty sure its 9.2793 g of Na
183108
1) H3COCH3 (or C2H6O) + 3O2 (g) --> 2CO2 (g) + 3H2O (g)\n2) HNO2 (aq) + NaOH (aq) <--> NaNO2 (aq) + H2O (l)\n3) 2K+ + Cr2O7(2-) + Fe(NO3)2 + 2H+ <--> CrO4(-) + Fe3+ + 2K+ + 2NO3(-) \n4) Na3PO4 + NH4NO3 = No reaction, all soluble salts.\n5) 2HCl (aq) + Na2S (s) --> 2NaCl (aq) + H2S (g)
362979
The response depends on what you call "not concentrated".  Any saturated solution, as you correctly indicate contains the maximum amount of solute at a specific temperature.  \n\nIf by "not concentrated" you mean that the maximum is low (compared to what?) then you can find sparingly soluble salts that reach saturation at relatively low concentrations (compared, for example, to NaCl).  The absolute value of these concentrations will be very much lower than the saturation level of NaCl.
519157
Iodomethane CH3I has a relative molecular mass of 142 ( C = 12;  +  H3 = 3  +  I  =  127)\nIodomethane is a liquid.\nIt is dense (because of the iodine) 2.28 times heavier than water\nIt has a boiling point of 42C\n\nHalf the volume of calcium hydroxide would be needed because it is dibasic ( has two OH groups)\n\n  Ca(OH)2  +   2HCl  =   CaCl2 +  2H2O     \n  2NaOH  +  2HCl   =  2NaCl +  2H2O  (the reason for all the 2's is to have the same amount of acid in both equations)\n\nThe mass of calcium hydroxide needed to make up a 1 Molar solution is 74g  ( Ca = 40  O = 16 and H = 1)\n\nHowever you wouldn't be able to make a 1 molar solution of calcium hydroxide as it isn't soluble enough!
429913
For the first question , the key words are FINAL MOLAR CONCENTRATION.  This assumes that the reaction has gone to completion and kinetics are no longer involved.  Since the balanced reaction is A + 2B ---> 3C, you can determine that the limited reagent is "B" and that "C" is created at a 2:3 ratio.  At the end of the reaction there will be 0.180M in "C" since is was created in a 2:3 ratio starting with 0.120M "B".\n\nAs for the second question, graphs can be used to determine if the reaction kinetics are first order or second order.  If you make a graph plotting the natural log of the pressure of "A" versus time ( ln(A) vs. t ), and you get a straight line graph, then the reaction will be exhibiting first order kinetics.  If you make a graph plotting the inverse of the pressure of "A" versus time ( 1/(A) vs. t ) and you get a straight line graph, then the reaction will be exhibiting second order kinetics.
473024
The product of a neutrilization reaction is always water and a salt, so use that as a guide.  For part 2, use the units as a guide.  Hint M=molarity=mols/L
29685
You need to check your units.  The specific heat of water is not 4.184 J/g; it is 4.184 J/g*degrees C (or degrees kelvin).  And the heat of solution should not be in J/degree, but in J/g.\n\nTry this equation instead:\n\n[HoS = J/g] = [(Specific Heat = 4.184 J/g*C)*(mass = 100g)*(Temperature change = Initial C - Final C)] / [Mass of CaCl2 = 3 g]\n\nBy subtracting the final from initial you should get the negative value and not need the -1 at the front.\n\nSometimes enthalpy of solution is in J/mol rather than J/g, in which case you modify the above by calculating the mols of CaCl2 in 3 grams and switching that out with the 3g in the above equation:\n\n(3g) / (110.99 g/mol) = 0.027 mol\n\nTheoretically, you should want to know the heat capacity of the CaCl2 because it is also being heated, but this is assumed to be negligible.  Write that into your error analysis and you teacher will love you.  Also, make sure to check your significant figures; i don't know how many sig figs you used in weighing out your 3 grams, but that will be necessary to know how much to round everything.  Good luck.
411260
Take 1.875 L of the .2 molar stock solution of NaBr and add .625 L of water to it and you get 2.5 L of a .15 molar solution of NaBr this just use the c1v1=c2v2 equation.\n\nNote:\nThis is the only way it could work. You do not want to add more NaBr because your concentration is already higher then desired.   If you just added 0.5L of water to the 2.0 L of .2 molar stock solution your molarity would still be to high at .16 meaning you have to dilute a smaller portion of your 2L .2 molar stock solution.
93467
When you add a salt to water and the container feels cold, yes, an endothermic reaction is taking place.  The solution is grabbing heat from the surroundings to get the salt to dissolve.  When you add a salt to a pure solvent (say, water), the freezing point will go down and the boiling point will go up.  That's because the vapor pressure of the pure solvent is lowered (this would be easier to explain with a phase diagram).  Remember, the definition of BP is the temp at which the solution's vapor pressure equals the external pressure.  By lowering the vapor pressure, the whole phase diagram shifts...BP to the right (higher), FP to the left (lower).  You may want to check out a intro level chemistry text or at least draw out the phase diagram to see more clearly what gets lost is just words here.  Did you ever try boiling water without heating it?  If you have the means, hook up a vacuum to a flask of water.  You don't change the vapor pressure of the water, but you reduce the external pressure enough so that the water does boil.\n\nOne more thing...  The change in BP and FP is related to how much salt is added to the solvent:  delta T (change in T) = Km  where K is either what's called the molal freezing-point depression constant or molal boiling-point elevation constant;  m is the molality of the solution defined as moles of solute (salt) / kilograms of solvent. By knowing the constant for the solvent, how much solvent you have and how much solute you add, you can predict by how much (in degrees (C) the BP or FP will change.\n\nI would just like to add a couple of more comments... The reaction that is endothermic in this case is  NaCl(s) --> Na+(aq) + Cl-(aq)  ...  The reason that the freezing point is lowered is as J. Schultz said - the vapor pressure is lowered ....  However there is a simple analogy (that may be useful) which gives a simplified picture ... Freezing involves the water molecules becoming very ordered in crystalline lattice...  However when there are foreign particles such as Na+ and Cl- ions in the water they hinder the formation of the crystalline structure because they interfere with the intermolecular forces between the water molecules.  Also, the freezing point depression and boiling point elevation are called colligative properties...  that is, they are properties that depend only on the number of solute particles and not on their chemical properties...  The third famous colligative property is osmotic pressure.
645855
1)\nSodium Sulfate (Na2SO4) will break into ions in solution,\nNa2SO4 (s) --H2O--> 2 Na+1 (aq) + SO4-1 (aq)\nFor every 1 mole of Na2SO4 dissolved, 3 moles of ions are formed.  If the concentration of Na2SO4 is 1.15 mol/Liter, then the ion concentration must be 3 times as much, 3.45 mol/Liter.\n\n2)\nTlCl3 is an ionic compound which when dissolved in water will break into ions since Chloride compounds are generally soluble,\nTlCl3 (s) --H2O--> Tl+3 (aq) + 3Cl- (aq)\n\nC2H5OH, otherwise known as ethanol, will easily dissolve (easily mincible) in water, but will not break into ions (as it is not an ionic compound).\n\nFe2(CO3)3 (although it is an ionic compound) in considered insoluble in water.  Carbonates in generally are considered insoluble unless they are bonded with a group 1 element or the Ammonium ion.\n\n3)\n.356 mol/Liter HCl means that for every 1 Liter of solution, there are .356 moles of HCl dissolved in it.\nMolarity  = moles / Liter --> moles  = Molarity * Liters\nSo if you originally had a 6.5 M solution with a volume of .0572 Liters, you originally had .3718 moles of HCl.\n\n.356 M = .3718 moles HCl / x Liters\nx Liters = .3718 moles / .1.044 Liters.\n\nSo if then old volume was 57.2 mL and the new volume should be 1044 mL, then 986.8 mL shoudl be added to solution.
70400
1 calarie can take 1 cc (cubic centimeter) of water up by 1 degree C. The density of water is 1 g/cc.\n7,310,000/50.8 = 143898 degrees C [quie huge]