How to find out weight of 1m3 of iron and aluminium with the help of Archimedes's principle?
929019
With such a large quantity of ore you will have a problem using the two-beaker method. But the first two suggestions contain half the answer each. Just combine what they are saying.
819316
CuSO4+Fe = FeSO4+Cu\nCompute the number of atoms of CuSO4 in 7 gms of it, the number of atoms in 2gm of Fe. Choose the smaller number of the two. That is the number of atoms of Cu formed. Compute its weight.
895868
Oh dear.  You should be able to do this one.  You simply need to calculate the mass of the core and divide by the total mass.  You have a density and a radius.\n\nMultiply the density of the core by the volume of the core (be sure you use the same units!)  I will leave it to you to look up/remember the equation for the volume of a sphere.\n\nTake the result and divide by the total mass. Are you sure that is the density, btw?  It seems awfully light for iron at the center of the Earth.  Even water has a mass of 1000 kg/m^3.
389126
You don't have to 'get' equivalent weights of substances, its very simple to calculate them.\n\nFirst of all, you need to calculate the molecular wt or know the atomic wt of the compound or element you're dealing with.\n\nThe general formula for the equivalent wt of a compound is \n\nEquivalent weight = Atomic or molecular wt/ x\n\nx has different definitions in different contexts:\n\nIf for an element, x is the valency\n\nEg- E.W. of magnesium = 24/2 = 12\n\nIf for an ion, x is the number of units of charge\n\nEg- E.W. of Al+3 ion = 27/3 = 9\n\nIf for a compund, x is the total amount of formal charge present on the cation/anion, i.e. the number of ions multiplied by the charge on each radical.\n\nEg- For NaCl, the value of x would be 1\n    For MgSO4, the value of x would be 2\n    For AlCl3, the value of x would be 3\n    For Ba3(PO4)2, it would be 6,...\n\nIf for an acid, x is the basicity(no. of ionizable H+ ions)\n\nEg - 1 for HCl, 2 for H2SO4,..\n\nThe same applies for a base, x is the no. of ionizable OH- ions.\n\nNow for KMnO4 (acidified or not),\n\nThe molecular wt is 39+55+64 = 158\n\nThe value of x is 1\nSo the E.W. is 158/1 = 158\n\nHowever, if you are interested in the E.W. of Mn in KMnO4, which is more relevant in redox reactions(using acidified KMnO4), calculate the oxidation state of Mn first. It comes out to be +7.\n\nSo the E.W. is 55/7
23478
Could you be a little more specific?  Are you asking how the atomic weights in the periodic table derived?
1017844
You need the thickness, or the "height" of the aluminum foil.  Your density equation states that aluminum is 2.70 g/cm^3.  You have all the information except the height.  So plug in the numbers:\n\n2.70 g/cm^3 = 2.04g all over 11.5 cm X 14.0 cm X h\n= 2.04 g / 161 X h cm\n2.70 h= 2.04 / 161\n2.70 h = 0.0127\nh = 0.0047 cm\n\nReplace answer for h and you'll find the answer is correct.
579735
Guess a group 3 metal such as boron or Aluminium
736683
.065kg * 35 * specific heat of Aluminum\n\njust make sure your units check out; you may need to convert this equation into calories as specific heat is normally defined in joules I think.
621321
If the density is 'd' kg/m^3 then the weight is '9.81d' Newton. [remember weight is force]
754266
yes \n\nbasic fdormula = mgh\n\nhere mg= weight of body[mass * acc. due to gravity}\nso it can be\nweight* height\n\nalso weight is a kind of force\nso weight = Force represented by F\nand 'd' = distance\nso it can be F*D where f= force[weight] and d=[distance]
341507
You mean find "g" as if you didn't know the constant? Okay.\n\nPretty simply really. Have the height measured that you're going to drop it from to where it's going to land (this will be your "h"). Make it a high one (one meter is enough). Measure the mass of the object (that's the "w"). Then when you do the experiment, time it. Start it at the instant you let it drop and stop it once it touches where ever it lands. That's enough information to find "g." Just solve for "g" then that would be the answer.\n\nActually, for an experiment, you should solve for "g" as a formula before you start doing anything else. So the formula would be like this:\n\ng = (2h)/[(t^2)(w)]\n\n(t^2 is "t squared")\n\nSo just have that formula then do the experiment: measure height of the drop, the mass of the object, then time it when you drop it (you need good reflexes when you drop it, as it falls fast).\n\nPS Penguins!!!
440758
Pascal (Same as Newton per Meter sqaure)\nmm of Hg (Mercury)
1020680
Divide 57 N by 9.81 m/s/s to get the mass in kilograms.\nThe mass will be the same on the surface of Io as it is on the earth.\nTo find the weight on Io multply the mass by 1.81 m/s/s and that will be in Newtons.
702947
The atomic weight is the mass in grams per mole (atomic mass of iron = 55.845).\n\nA mole is 6.02 x 10^23 atoms, therefore the mass of 500 atoms would be:\n\n500/(6.02x10^23) x 55.845 = 4.64 x 10^-20 grams.\n\nNotice that's 10 to the -20 power.
625685
Aluminium has a valanly of 3 Oxgygen has 2\nSo\n4Al + 3O2   -> 2Al2O3
298529
If you know the density of the material, you can measure it's volume.  D = M / V.  Therefore, the mass equals the density of the object times the volume.  The weight of the material is its mass times gravity.
997326
Use Newton's universal gravitation equation, then sum the forces.\n\nF = G*m1*m2/r^2\n\nF = force (newtons)\nG = gravitational constant (6.6742 x 10^-11)\nm1 = mass of object 1 (kg)\nm2 = mass of object 2 (kg)\nr = distance between objects (meters) ==> don't forget to convert the units\n\nIn this case, you do it twice:\nF1 = G * 390 * 390 / 0.11^2 = 0.84mN to the left\nF2 = G * 390 * 390 / 0.39^2 = 0.067mN to the right \n\nNet Force = 0.84 - 0.067 = 0.772mN to the left