how do i know the level of liquid in a 50 litre jerrycan is exactly 50 litres or less?
154656
Weigh the can and contents on any accurate scale and subtract the weight of the empty jerrycan. That gives you the weight of the liquid which you can convert to litres by referring to a lookup table of weights vs volumes for different liquids.
185076
Get the dimensions of the rectangular tank into the same units, then get the volume. the unit could be in^3, cm^3, m^3, ft^3, mi^3, mm^3, km^3, etc. Have them be converted to cm^3 using the ff. conversion units:\n\n1 mi = 5280 ft\n1 ft = 12 in\n1 in = 2.54 cm\n1 m = 10 dm = 100 cm = 1000mm = .1 dam = .01 hm = .001 km\n\nNow, 1 cm^3 = 1 mL. Have it converted into liters using 1000 mL = 1 L. To get gallons, use the ff. conversion unit:\n3.8 L = 1 gal\n\nHope this helps ^_^
361907
If the tank is a cylinder, the Volume is πr²h.  If you get the volume in cubic feet, multiply by 7.5 to get gallons.
770705
If the can holds EXACTLY 355 ml, with that many significant figures (3) this means that the contents can be between 354.5 ≤ n < 355.5  Any amount outside that range would be rounded off to either 354 or 356 ml (assuming scientific honesty on the part of the company).  Without a precision tolerance specification (e.g. ±.2ml) that is  the assumed precision.  If the amount was specified as 355.0, then the presumed tolerance would be ±.05ml.
911209
volume is measured in liters or cubic meters
1003456
Depends on the type of gallons as below:\n53 Gallons(British) = 0.241 cubic metres\n53 Gallons(US, dry) = 0.233 cubic metres\n53 Gallons(US, liquid) = 0.2 cubic metres
978082
Well if you know the density of ANY two liquids and the overall weight, you ought to be abler to calculate the proportions:\n\nEG:\n\nWe have X amount of liquid with a SG of 1.2\nWe have Y amount of liquid with an SG of 1.6\nWe also know there is perhaps 3 litres and it weighs 1.4 Kg\n\n1.2 * x + 1.6 * Y = 1.4 * 3\nSimplifies\n\n12x + 16y = 14x3 (42)\n\n12x +16y = 42\n6x +8y = 21\n\nNow do basic algebra to find the proportions of the two.
392516
In the SI system the same units are used to measure any volume irrespective of whether it's liquid, solid, or an enclosed void. There are no messy conversion factors (e.g. cubic inches per gallon etc.)\nThe base unit of length is the metre, so volume is cubic metres. So if your pool is 5 metres long by 4 metres wide by 1.5 metres deep, the volume is 5 x 4 x 1.5 = 30 cubic metres.\nFor a small indoor aquarium, you may prefer to use litres. The litre is a cubic decimetre (the cube of a tenth of a metre i.e. a thousandth of a cubic metre), or 1000 cubic centimetres.\nSo if your fishtank is 30 cm wide by 60 cm long by 40 cm deep, it would be 72,000 cubic centimetres or 72 litres.
1153448
i say its 50 to 1 thats 5ozs to one gallon
492398
normally, liter is used to measure liquids and grammes for measuring weight. however, in the case of water and milk, 2100 ml would amount to 2.1 kg.
67173
About 5 and a half litres, apparently - http://ask.yahoo.com/ask/20000612.html
766934
Assume it is an ideal gas and use PV = nRT\n\nP is in atm 1 atm = 760 torr\n    225 torr / 760 torr = 0.296 atm\nV is in L so 5.7 L\nR is 0.0821 atm*L/mol*K\nT is in K so C + 273 = K\n     27 + 273 = 300 K\n\n(0.296)(5.7) = (0.0821)(300)n\nn = 0.0685 mols
984199
I think this is what you are looking for, you question is difficult to understand.\n\n 7% of 5 liters = 0∙35 Liters.\n\n 2 46/ 47% = 140/ 47%\n 7 = 329 / 47%\n Ratio is 140:329 = 469 parts.\n 2 46/ 47 of 7% = 0∙35 Liters x 140/ 469\n 2 46/ 47 of 7% = 0∙104 477 611 Litres.\n 2 46/ 47 of 7% ≈ 0∙104 Litres.
941831
The point to note is that the mass of the water in the cylinder is continuously varying.\nMass of can = M (constant)\nMass of water = M*(x/H)   -- because it is proportional to the height of the water column.\nCentre of mass of can alone (from the bottom)  = H/2\nCentre of mass of water = x/2\na) Centre of mass of the system = (Mass of can * cm of can+Mass of water * cm of water) / (Mass of can+Mass of water) \n = {(M*H/2) + (M*x/H)*x/2} / (M+M*x/H)\n = 1/2 (H^2+x^2)/(H+x)\n \nb) The value of x for the minimum height of the cm can be found out by putting d(cm)/dx = 0\n This gives, x = H*(sqrt2 - 1). \n At this value of x, the cm turns out to be:\n cm = H*(sqrt2 - 1).
1071364
1.5 liters = 50.7 US fluid ounces
575750
This can be done using the formula M1V1=M2V2.  For example, if you have 2M HCl and wanted 2L of .5M HCl, then putting that into this equation as 2(V1)=(.5)(2).  Solve accordingly.  This is just an example.  This formula can be used in a variety of ways that all.  But, to make it to the desired you, a volumetric flask is needed in which water is added until the etched line on the flask to equal the preset volume of the flask.
783604
volume=pir^2h=1000\nsince there are two variables we need a minimum of two equations