What is the viscosity of water @ 200 F (93.334 C)?
476488
It is .305785 centipoises at 200 Degs F from the International Critical Tables.
786403
a) a lower freezing point than pure water
635522
It depends on the liquid AND the pressure. 100C is for water at atmospheric pressure.
787311
A. H2O. Molecular water is 2H2O
602506
1. Water boils- 212 degree F/100 degree C\n2. Normal Body temp- 98.6 degree F/37 degree C\n3. Water Freezing temp- 32 degree F/ 0 degree C
857506
Approx. 1200 degrees F.
184315
It's 277K.\nWater density is greatest at 4 degrees Celsius (O degrees Celsius = 273K, so 4 degrees Celsius = 277K).\nWhen the temperature decreases to 0 Celsius, the density of water decreases to 0.999 g/cm^3 and the volume increases about 10%.\nThat's why ice floats on rivers - because at 0 degrees Celsius it has lower density than water!\n\nGreetings from Poland!
1019201
First you have to know the pressure and the exact salinity to answer this question.\nBut generally :\nAt a salinity of 35PPT and a pressure of 1.5 PSI\nthe freezing point would be -1.930 degrees C or 28.5 F
887338
Check the thermometer - I think you will find it is 102 F - at 102C you would cook! The boiling point of water is 212 F or 100C. Your body temperature is normally 98.4 F and so you are in a room about 3.5 F above your normal body temperature - still warn though. the body prefers a lower temperature - so it can control body temperature through the means of perspiration.
177044
use the fmla.... Q=msT\n\nwhere Q is the quantity of heat required, m is mass, s is specific heat capacity of water, T is the temperature difference. i dont hav a calculator with me now, u do the computations plz...
75392
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.
237697
O.k., so you are given the final temperature (-0.430 C). Since the freezing point of water is usually O degrees C, the temperature decrease is 0.43 degrees C. Like in the other problem, let's set up some dimensional analysis.\n\n(0.43 deg.C decr)*(1 mole compound / 1.86 deg.C decr) = X\n\nHere X would equal 0.23 moles compound\n\nNext, you want to find the grams of solute in solution.\n\n(8.02g solute/0.861 kg solvent) = Y\n\nY here is equal to 9.31 g compound/kg solution\n\nNow that you have the mass in grams and the number of moles of the solute, you can divide Y by X, which gives the units for molecular mass (g/ mol.)\n\nThe answer should be around 40.48 g/mol. The solute cannot be determined from this information because no additional information is given concerning what is most likely a compound (not a pure element).
801823
Quantity of energy lost by the water:\n\nQ = (m)(deltaT)(c)\n    = (50.0g)(2.00 C)(4.18 JK-1g-1)\n    = 418 J\n\nSince the water got colder, the reaction was endothermic, and the enthalpy change for the reaction is +418 J.
288276
ρwater ~ 1g/ml\n\nq = 1000g x 4.18J/ºC.g x (37ºC-0ºC) = 155.660kJ (1.6x10^5J to correct Sig Figs)\n\nsecond part. The amount of heat required in part 1 will melt 155.660kJ / 6.01kJ/mole = 25.9 moles of ice or 468g
721040
I get 45.13 degree Celsius.\nReasoning:\nChanging 23 gram ice in water of 0 degree requires 7.66e3 Joule, according to part 1. This amount of energy taken from the 70 degree water decreases the temperature of the latter by 7.66e3 x (70-10) / 2.9e4 = 15.85 degrees (using part 1 and 2).\n\nSo now we have a mixture of 23 gram water at 0 degree and 115 gram (5 times as much!) at 54.15 degrees. This will average out to (5/6) x 54.15 = 45.13 degree Celcius.
778647
By fractional distillation.\nDEG distills at 93 'C, you can collect first and then collect water at 100 'C.
1008978
well this is really a chemistry problem but the formula is q=mC(delta) T, meaning heat= mass times the specific heat capacity times the change in temperature, so to set this up we use, 1.68^10 j = m (4.18 j/g) (80 degrees) becuase 4.18 is C and 80 degrees is the amount u wish to raise it, so then u get 1.68^10 j = 334.4m, divide both sides by 334.4 and you get .53558=m so your mass is .536g of water