how to find the density of a soluble granular solid?
348154
Since you will get the entrained air as part of any volume measurement, you ideally want to find a solvent that it isn't soluble in. If you want the density of it including the entrained air (between granules).\n\nFinally, you can fill the voids between granules to make a solid in solid 'solution' by taking a known material of much smaller particle size and known density and mixing it all together so that the unknown granules are widely dispersed. \n\nPack the mix to get a volume and weigh it. Now, subtract the known volume and weight of the added material to find the unknown volume and weight.
626920
Trick question! You can figure the VOLUME of dirt you dig up, but without the density of the dirt, you cannot calculate the mass or weight.
143556
Density = Mass / Volume\n\nStep 1) Put the Apple on a scale to get its weight.\nStep 2) Mesuare apple to get its volume (use aproxomation)\nStep 3) Caclulate.
725926
The Density of solid Silver is: 10490 kg/m^3\n10.49 g/mL\nhttp://www.webelements.com/webelements/elements/text/Ag/phys.html\n\nThe Density of solid Gold is: 19300 kg/m^3\n19.30 g/mL\nhttp://www.webelements.com/webelements/elements/text/Au/phys.html
976944
density=mass/volume\n\nd=125.6g/750cm^3=.1674g/cm^3\n\napproximately .167g/cm^3 (dont forget the cubed. if u forget it u'll get it wrong)
991058
mass =concentration\n-------\nvolume
162268
The density of the fluid and the density of the object.
732533
D = M/V\n\nDensity= Mass/Volume
882649
First solve for the force of friction between the block and the table using the formula,\n\nf=uR, where f is the force of friction, u the coefficient of static friction, and R is the force normal to the table. R therefore is the weight of the block equal to 31g.\n\nf=0.425*31*9.8\n =129.1N\n\nThe weight of the sand added to the bucket plus the weight of the bucket itself should be just slightly equal to 129.1N so that the system will begin to move. At that point, however, the block will begin to accelerate because the force of friction acting on it will no longer be static friction but kinetic friction, which is lower in magnitude. Thus there will be a net force equal to 129.1 minus the force of kinetic friction on the block. Thus,\n\nNet force, F=129.1-0.320*31*9.8\n                 =129.1-97.2\n                 =31.9N\n\nF=ma, where F is our net force of 31.9N, m the total mass of block plus mass of sand and bucket, and a is the acceleration of the system.\n\nMass of sand and bucket=129.1/9.8\n                                    =13.2kg\nSubstitute known values:\n\n31.9=(31+13.2)a\na=31.9/44.2\n  =0.72m/s^2
120873
D= M/V\n\nIts a formula used to calulated to density (thinkness) of an object by dividing the Mass of an object by its volume. In this case, you can use the water displacement method that displaces (water) from its orignal volume by adding an object (marble). Using a balancer, measure and record the weight of graduated cylinder(ml), before and after the addition of water. Subtract the weight of the cylinder with water from the cyliner's weight and you will get the mass. You have to pay attention to how much the water displacement after adding the marble. record the volumes before and after (if ness). This will solve for volume. Once this is accomplished, you would have no problem finding the density.
1112775
The density of dry loose gravel regardless of size is on average about 90 to 105 pounds per cubic foot. This assumes a fairly consistent size. If the gravel is a mixture of large and very fine material then it will be some what higher because the fine grains will fill in between the larger pieces. \n\nAny granular material that is evenly graded (all pieces about the same size) will have the same degree of void space between granules. It calculates to around 30% (I don't remember the exact number but you can calculate it you like). Most sand and gravel is made of the same ground up rock, therefore density of the individual grains or pieces is the same. With 30% of any cubic foot the same and the density of the grains the same, sand or gravel will have the same density.
1133778
http://online.unitconverterpro.com/metric-conversion-tables/convert-alpha/density.html
760373
an ore sample weighs 15N (newton) in air. when the sample is suspended by a light cord and totally immersed in water, the tension in the cord is 10.80 N. find the volume and the total density of the sample.. \n\n\nApparent loss in weight=(15 -10.8 )N\nMass of water displaced=(1.50-1.08)kg\n                                        =0.42 kg\nVolume=0.42 l(since cm3= l)\n1 kg= 1 l (litre)\n\nTherefore,volume =0.42 litre\n                             =21/50000 m3\n\n\nsince density equals to mass over volume;\ndensity=0.42 (divide by) 21/50000\n           =100000 kg m-3\n\n\n\n\n\nThis is called Archimedes principle\n\nBelow is a table that will help you to change units from Wikipedia. \n\n\n\nName of unitSymbolDefinitionRelation to SI units\n\ncubic metre (SI unit)\nm³≡ 1 m × 1 m × 1 m\nlitre\nL≡ 1 dm³\nlambda\nλ≡ 1 mm³= 1 μL\ndrop (metric)\n ≡ 1/20 mL= 0.05 mL\nminim (Imperial)\nmin≡ 1/480 fl oz (Imp) = 1/60 fl dr (Imp)≈ 0.059 193 880 208 333 mL\nminim (U.S.)min≡ 1/480 US fl oz = 1/60 US fl dr= 0.061 611 519 921 875 mL\ndrop (U.S.) (alt)gtt≡ 1/456 US fl oz≈ 0.064 854 231 mL\ndrop (Imperial) (alt)gtt≡ 1/1824 gi (Imp)≈ 0.077 886 684 mL\ndrop (U.S.)gtt≡ 1/360 US fl oz= 0.082 148 693 229 1(6) mL\ndrop (medical) ≡ 1/12 mL= 0.08(3) mL\ndrop (Imperial)gtt≡ 1/288 fl oz (Imp)= 0.098 656 467 013 (8) mL\ndash (U.S.) ≡ 1/96 US fl oz = ½ US pinch= 0.308 057 599 609 375 mL\ndash (Imperial) ≡ 1/384 gi (Imp) = ½ pinch (Imp)= 0.369 961 751 302 08(3) mL\npinch (U.S.) ≡ 1/48 US fl oz = 1/8 US tsp= 0.616 115 199 218 75 mL\npinch (Imperial) ≡ 1/192 gi (Imp) = 1/8 tsp (Imp)= 0.739 923 502 604 1(6) mL\nfluid scruple (Imperial)\nfl s≡ 1/24 fl oz (Imp)= 1.183 877 604 1(6) mL\nfluid drachm (Imperial)\nfl dr≡ 1/8 fl oz (Imp)= 3.551 632 812 5 mL\nfluid dram (U.S.); U.S. fluidramfl dr≡ 1/8 US fl oz= 3.696 691 195 312 5 mL\nteaspoon (Canadian)\ntsp≡ 1/6 fl oz (Imp)≈ 4.735 510 416 667 mL\nteaspoon (U.S.)tsp≡ 1/6 US fl oz= 4.928 921 595 mL\nteaspoon (metric) ≡ 5 mL\nteaspoon (Imperial)tsp≡ 1/24 gi (Imp)= 5.919 388 020 8(3) mL\ndessertspoon (Imperial)\n ≡ 1/12 gi (Imp)= 11.838 776 041 (6) mL\ntablespoon (Canadian)\ntbsp≡ ½ fl oz (Imp)= 14.206 531 25 mL\ntablespoon (U.S.)tbsp≡ ½ US fl oz= 14.786 764 782 5 mL\ntablespoon (metric) ≡ 15 mL\ncubic inch\ncu in≡ 1 in³= 16.387 064 mL\ntablespoon (Imperial)tbsp≡ 5/8 fl oz (Imp)= 17.758 164 062 5 mL\npony ≡ 3/4 US fl oz= 22.180 147 171 875 mL\nfluid ounce (Imperial)\nfl oz (Imp)≡ 1/160 gal (Imp)= 28.413 062 5 mL\nfluid ounce (U.S.)fl oz (US)≡ 1/128 gal (US)= 29.573 529 562 5 mL\nshot ≡ 1 US fl oz= 29.573 529 562 5 mL\njigger ≡ 1 ½ US fl oz= 44.360 294 343 75 mL\ngill (U.S.)\ngi (US)≡ 4 US fl oz= 118.294 118 25 mL\ngill (Imperial); Noggingi (Imp); nog≡ 5 fl oz (Imp)= 142.065 312 5 mL\ncup (Canadian)\nc (CA)≡ 8 fl oz (Imp)= 227.3045 mL\ncup (U.S.)c (US)≡8 US fl oz ≡ 1/16 gal (US)= 236.588 236 5 mL\ncup (metric)c≡ 250 mL\nbreakfast cup ≡ 10 fl oz (Imp)= 284.130 625 mL\npint (U.S. fluid)\npt (US fl)≡ 1/8 gal (US)= 473.176 473 mL\npint (U.S. dry)pt (US dry)≡ 1/4 bu (US lvl)= 0.550 610 471 357 5 L\npint (Imperial)pt (Imp)≡ 1/8 gal (Imp)= 0.568 261 25 L\nfifth ≡ 1/5 US gal= 0.757 082 356 8 L\nquart (U.S. fluid)\nqt (US)≡ 1/4 gal (US fl)= 0.946 352 946 L\nquart (U.S. dry)qt (US)≡ 1/32 bu (US lvl) = ¼ gal (US dry)= 1.101 220 942 715 L\nquart (Imperial)qt (Imp)≡ ¼ gal (Imp)= 1.136 522 5 L\npottle; quartern ≡ ½ gal (Imp) = 80 fl oz (Imp)= 2.273 045 L\nboard-foot\nfbm≡ 144 cu in= 2.359 737 216 L\ngallon (U.S. fluid; Wine)\ngal (US)≡ 231 cu in= 3.785 411 784 L\ngallon (U.S. dry)gal (US)≡ 1/8 bu (US lvl)= 4.404 883 770 86 L\ngallon (Imperial)gal (Imp)≡= 4.546 09 L\nbeer gallonbeer gal≡ 282 cu in= 4.621 152 048 L\npeck (U.S. dry)\npk≡ ¼ US lvl bu= 8.809 767 541 72 L\npeck (Imperial)\npk≡ 2 gal (Imp)= 9.092 18 L\nbucket (Imperial)\nbkt≡ 4 gal (Imp)= 18.184 36 L\ntimber foot\n ≡ 1 cu ft= 28.316 846 592
721530
Yes, you can get very many experiments on density. Depends on what you want to do with density. If you want to compare the densities of to different solids in relation to a particular liquid (e.g water) You just measure their dimensions and divide their  mass by their volume to get their density (If you don't know already) . Then you find their Relative density using the equation.
753525
that is not enough information, need to know the density of the soil
58191
Density - 1.151 \nNot easy to find
198092
Use:\n\npressure = density x height x g\nwhere g = 980 cm/s^2\n\nfor each liquid.  the answer (in dynes) will be the sum of the two pressures.