A 500milliliter sample of sea water has a mass of 514 grams What is the density of sea water as an unit rate?
413065
density is mass / volume. so 514 grams / 500 mL sample
123586
Density = 13.5 g/cm^3\n13.5 g = about .0297 pounds\n\n4/.0297= about 134.52 cm^3, or 134.52 milliliters.
1018649
Water reaches a density maxima at 4C.
722824
water is denser than wood
588958
the weight will be according 2 density of the material \n\nfor example for water 1  liter wighs 1 kg
747172
I liter of water weighs 1 kg so if the density of diesoline is say .9 then 1kg would weigh .9 kg... it depends on the density!!
96826
By the original definition, the mass of one mL of water is one Gram (temperature dependent, but really, really close at 20 degrees C.
902687
The density of pure water is 1.0\nThe density of ice is .92 (ice expands when it freezes, so the density drops)\nThe density of sea water is 1.025 (sea water is denser then pure water because it is salty).\n\nAn ice cube floating is a glass of water is 92% submerged (compare their densities.)\n\nAn ice berg floating in sea water is 90% submerged (.92/1.025)
603614
The mass of a nickel is a 5.02 g. Its density is 8.90 g/mL!\n\nI hope I'm first!
748523
Saturn has a density less than water.  It's density is 0.6873 g/cm^3.  Water has a density of 1g/cm^3.
1062341
You are given the volume in ml and the density per ml so,in order to get the answer just multiply 220 ml X 7.800 gm/ml.  This is equal to 1,716 grams or 1.716 kilograms.
785219
density of alcohol alone:\n520 - 88 = 432 g\n\n432g/500mL = 0.864 g/mL
255172
Density = mass/volume\nDensity = 615/105 gm/cm3 = 5.86 gm/cm3
374421
mass=density over volume\n\n=)
65002
Density = 5,515 kg/m³
1006390
You need to know the density of whatever it is you want to convert.  1 ml of water at 4 degrees C is has a mass of 1 g or 1000 mg.  To convert any other liquid you must multiply by the specific gravity or density.
71291
Water has its highest density at 4 deg. C.\n\nYou might try a Yahoo! search next time.  That's how I found the info.\n\nTake care,\nDavid Roth\nwww.tutor-homework.com