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This is not a common phenomenon. Usually it is boiling points that are elevated, and freezing points are depressed. It is possible that a freezing point could be raised (elevated) due to the presence of an impurity with a much higher freezing point.Changes in boiling and freezing points are typically due to impurities in compound.See the Related Questions to the left for more information about freezing point depression and boiling point elevation problems.
You can get depression in the winter called Seasonal Affective Disorder as it is dark most of the day and sleep problems in the summer due to bright light for most of the day.
The climate where you live affects you by determining the clothes you need to wear, the food you eat, and your health condition including allergies, sunburn, and respiratory problems.
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I actually had problems with this question on my crossword puzzle. The answer is actually a caldera. ---------------------------------- A volcanic crater would also be a correct answer.
Common problems associated with freezing point depression include inaccurate measurements of freezing point, difficulty in determining the concentration of solute in a solution, and potential errors in calculations due to variations in the purity of the solvent.
Common molality problems include calculating the molality of a solution, determining the amount of solute needed to achieve a desired molality, and finding the freezing point depression or boiling point elevation of a solution. These problems can be solved effectively by accurately measuring the mass of solute and solvent, using the formula for molality (moles of solute per kilogram of solvent), and applying colligative properties formulas for freezing point depression and boiling point elevation.
Some common challenges students face when solving molecular orbital diagram problems include understanding the concept of molecular orbitals, correctly identifying the atomic orbitals involved, determining the correct electron configurations, and interpreting the energy levels of the molecular orbitals.
For the definition and description of freezing point lower or freezing point depression[http://www.answers.com/freezing+point+depression?cat=technology&gwp=13Click here for the answer from Answers.com]For how to solve a freezing point depression problem with step-by-step instructions click on this question:[http://wiki.answers.com/Q/How_do_you_solve_freezing_point_depression_problemsHow do you solve freezing point depression problems?]
Calculating the freezing point depression of a solvent requires three pieces of information: 1) the molal freezing point depression of that solvent; 2) the molality of the solution; 3) the number of distinct entities produced in solution per molecular formula unit of solute. Hopefully you meant molality instead of molarity (if the solvent is water and it's a dilute solution, assuming that molality = molarity isn't a bad approximation). I suspect you've been told what 1 and 3 are (or been given enough information to figure them out) but simply didn't mention it; otherwise you're pretty much boned.
Some mass spec practice problems that can help improve your understanding of mass spectrometry include identifying the molecular formula of a compound based on its mass spectrum, determining the relative abundance of isotopes in a sample, and calculating the molecular weight of a compound using mass spectrometry data.
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That there were solutions to the problems of the Great Depression.
To solve freezing point depression problems, you need to know the following things:-- the identity of the solute (the thing you are adding to the liquid)-- the identity of the solvent (the liquid)-- whether the solute is molecular or ionic, and if ionic, how many ions it forms-- the amount of solute (in grams or moles)-- the amount of solvent (in kilograms)-- the value of Kf of the solvent (for water, it is -1.858 K·kg/mol (or -1.858 °C molal-1)Then you use the following equation:∆T = i * Kf * mwhere ∆T is the change in the freezing point, "i" is the number of molecules or ions formed upon addition to the solvent, Kf is the freezing point depression constant, and "m" is the molality of the solution.The Kf of water for a freezing point depression is known and is equal to -1.858 °C·kg/mol (or -1.858 °C molal-1)The value of "i" has to do with what you add to the water. If you added sugar, a molecular compound the value of "i" is 1.0. If you add a ionic compound like NaCl, the value of "i" is 2.0 because for every 1 molecule of NaCl, you make 2 ions: one Na+ and one Cl- in water. For MgCl2, the value of "i" is thus 3.0 (for each MgCl2 you get one Mg2+ and two Cl- ions, so a total of 3 ions).To find "m," the molality of a solution you need to know the number of moles of solute and the number of kilograms of solvent (m = moles/kg). If you are given the number of grams of solute, the number of moles is found from the mass and the molar mass of the solute.grams of solute ÷ molar mass of solute = moles of soluteTo find the molality, just divide the moles of solute by the kilograms of solvent moles of solute ÷ kilograms solvent = molality Note: If you are given the volume of the solvent instead of the mass, use the density of the solvent to convert -- the density of water is 1 kilogram per liter)Liters of solvent * density of solvent = kilograms of solventNow just plug all the numbers into the equation at the top of the answer!
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