Octadecan-1-ol has a melting point of 60-61 C. Why is the result from the experiment falls outside the range?
549677
This is a chemistry question, the anomaly is due to Hydrogen bonding.
136403
You can calculate the freezing point depression of the solution is you know three things, firstly, the freezing point of the pure solvent, and secondly, the concentration of the solute, and thirdly, the freezing point depression constant of the solvent.\nIn this problem, you know both (One is given, two of them are assumed knowledge).\n\nThe freezing point of pure water is 0 degrees Celcius.\nThe concentration of the solute is 52 percent.  You need to convert this to a molaltiy concentration to use it (or fiind a different freezing point depression constant).\nLook up for your self the k_f value.\n\nthe freezing point depression = k_f * m,\nwhere k_f is the freezing point depression and m is the concentraion of the solute in terms of molality.\n\nmolality = moles of solute / kg is solvent
162986
We typed "how does salt melt ice" into the Yahoo! Search box and were soon traveling the well-worn path to How Stuff Works. \nThere we learned that the answer has to do with melting points. Most people know that water freezes at a temperature of 32°Fahrenheit (0°Celsius). Salt dissolves fairly rapidly in ice, lowering its freezing point in an effect known as freezing point depression, and melting it in the process. \n\nBut what's going on at the molecular level? Basically, salt molecules are bonding with water molecules, making it more difficult for ice to form. The result is a solution with a lower freezing temperature. For example, a 10% salt solution freezes at 20°F (-6°C), while a 20% salt solution freezes at 2°F (-16°C). \n\nYou've also probably noticed the effect of salt water on freezing points if you've tried to make ice cream or if you've wondered why the ocean doesn't freeze at the same time as bodies of fresh water, such as rivers, lakes, and ponds.
874636
There's a problem with your nomenclature.  Octadecane is a straight-chain alkane with the formula, C18H38.  A carboxy group can only be attached at either end, so the name octadecanoic acid (aka, "stearic acid") requires no number to say where the acid group is.  Besides, it couldn't be at the 9 or 10 positions, anyway.  What do the "9" and "10" mean in your question?
48785
Boiling Point: 444.6 °C (717.75 K, 832.28 °F)
716723
ice never melts when it is in an environment of 0C.  any environment above that temperature will eventally raise the temperature of the ice to its melting point...\n\nit is an inherent property of ice to melt at temperatures above 0C.\n\ndry ice is not ice.\nsteel is not ice.\nice is solid water which is only found at or below 0C (at atmospheric pressure).\n\nthis has to do with the gibbs free energy of the transformation \nH2O (solid) ----> H2O (liquid)
684151
The freezing point of pure water is 0°C, but that melting point can be lowered by the adding any solvent such as a salt. Depending on the amount of salt the freezing temperature of a saline solution can range from below 0°C to -21.1°C. The frozen solution is a mixture of separate NaCl·2H2O crystals and ice crystals, not a homogeneous mixture of salt and water. This heterogeneous mixture is called a eutectic mixture. \nThe use of ordinary salt on sidewalks is effective because it requires a colder temperature to freeze water and it is a cheaper alternative to other solvents or methods of ice removal.
718253
Melting point 1811 K\n(1538 °C, 2800 °F)
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.
872765
It won't matter how much you started with or how much higher the solution was heated.  When the solution cools to 20C it will hold 0.22g/100mL of solvent (water).\n\nSo if you start with 1.75g and 0.22g remain in solution, the amount crystallized at 20C will be the difference, 1.53g.
149250
When the temperature of the water in the bottle approaches 32 degrees farehheit (the temperature where water freezes) is expands.  This pre-freeze expansion can rupture most nearly full containers.\n\nThe one good thing about this unusual physical phenomenon is that this is the reason that tce forms on the tops of lake or pond water.  This enables people to skate on the ponds in winter time.  The top layer of water freezes, while the lower water stays in a liquid form.
523531
At the heart of the issue is a phenomenon called Boiling Point Elevation. The boiling point of water is elevated when solutes are dissolved in the water. The boiling point also depends on the atmospheric pressure, but for the this question, I'm going to assume that we are at one atmosphere of pressure so that the boiling point of water is 100 degrees centigrade. \n\nSome people put salt in water when they make pasta, believing that the dissolved salt raises the boiling point of the water, and that the water is actually hotter than 100 degrees centigrade when the pasta is added. I say that they believe this because I'm doubt people add enough salt to result in a significant elevation in the boiling point. For water, the boiling point is elevated by 0.513 degrees centigrade for each mole of dissolved solute-ions per kilogram of solution. A mole of salt (NaCl) is 58.44 grams, and since one mole of NaCl disassociates into two moles of ions (Na+ ions and Cl- ions), 58.44 grams of salt (or 2 ounces) would be required to raise the boiling point of one liter (kilogram) of water (at one atmosphere of pressure) to 101.026 degrees. That would be some pretty salty water. \n\nThere is a related phenomenon known as Freezing Point Depression, in which the freezing point of water is lowered by 1.86 degress centigrade for each mole of solute-ions per kilogram of solution. We exploit the phenomenon of Freezing Point Depression by using salt to melt snow and ice on roads in the winter and to make ice-cream. \n\nNow, how does this relate to your question about the boiling point of blood? Well blood is basically a water solution. The dissolved solutes are more complex than a simple NaCl solution, but the principle of boiling point elevation (and freezing point depression) should still apply. The salt (NaCl) concentration of blood is 0.9%. This is approximately 0.154 moles of NaCl per kilogram (liter), or 0.308 moles of dissolved solute per liter. This should result in a boiling point elevation of a whopping 0.158 degrees centigrade. \n\nNow, there are other molecules dissolved in blood besides Na+ and Cl- ions, so the boiling point may be slightly different, but I think that you can see that it isn't going to be noticeably higher than 100 degrees (at one atmosphere of pressure). \n\nSo the answer to your question is basically, at the same temperature that water does. I hope that isn't too disappointing. If you want more information on Boiling Point Elevation or Freezing Point Depression, you should be able to find it in any college-level chemistry text book.
205901
I don't have to resources in front of me to really compute the answer, but this should give you the answer.\n\n1) you know the freezing point of water is zero degrees Celsius.\n\n2) find the moles of NiSO per kg of water (molality)\n\n3) use the following dimensional analysis (with 'X' being the previous answer): (X mol./kg)*(1.86 degree decrease/1 mole NiSO added) = Y\n\n4)Since there is 100% dissociation, the answer should be  0 - 2*Y (vant Hoff factor).\n\nThis means your answer will be a negative (below zero) number with Celcius as the temperature units
310335
You can never get to -273 C or 0 Kelvin.  But if you could, the pressure would be zero.  It would be difficult to measure since the pressure gauge itself is likely to emit heat (specifically thinking of an ion gauge).  If there was any pressure, then you would not be at 0 K since you would not be at the lowest energy state.  The only exception I can think of is if you only had one molecule in the container then the concept of temperature and pressure becomes ambigious.
515492
2. The temperature that was the most favorite condition for amylase_catalyzed hydrolysis was cold, I tested it at 1.7°C.
382860
Melting point -11°C (262.15 K) \nBoiling point 150.2°C (423.35 K)
175159
Solid\nCopper melts at 1083ºC