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The rate of transfer of a process is equal to the driving force divided by the resistance.The mass transfer coefficient is the resistance to mass transfer. In mass transfer the driving force is the concentration gradient. The mass transfer coefficient is considered anything that contributes to resistance to mass transfer: thermal and eddy diffusivity, distance, etc.Fick's law of diffusion describes convective mass transfer as:N=-c*D*(ca2-ca1)/(z2-z1)where:-c is some constant multiplier (unitless)-The quantity (z2-z1) is the distance between two points. (length i.e. meters)-D is the mass diffusivity or the diffusion coefficient and is dependent on properties of the substance (such as particle size etc.) and temperature. (units: length2/time i.e. m2/s)-The quantity (ca2-ca1) is the concentration gradient between the same two points (the driving force) (units: amount/length3 i.e. mol/m3)-N is the rate of mass transfer (units: mass/(length2*time) i.e. mol/m2*s) )Putting Fick's law in terms of the mass transfer coefficient kc', yields:N=-kc'*(ca2-ca1)where kc'= -c*D/(z2-z1).You can see that the mass transfer coefficient is in fact a function of the diffusivity.
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2.2 mol water = 2.2 (mol) * 18 (g/mol) water = 39.6 (mol*g/mol) = 40 g18 g/mol = mol mass of H2O = 2*H + 1*O = (2*1 + 16) g/mol
Leo Mol has written: 'Leo Mol'
H: 1 g/mol O: 16 g/mol H x 2 = 2 g/mol O x 1 = 16 g/mol 2 g/mol + 16 g/mol = 18 g/mol
(2.43 mol Al2Br6)*(2 mol Al/ 1 mol Al2Br6)= 4.86 mol Al
CH4 + 4S --> CS2 + 2H2S 120 g CH4 / 16.04 g/mol = 7.48 mol CH4 120 g S / 32.06 g/mol = 3.74 mol S 1 mol CH4 requires 4x1 mol S 7.48 mol CH4 requires 29.92 mol S S is limiting 3.74 mol S will yield 3.74 / 4 or 0.935 mol CS2 0.935 mol CS2 x 76.1 g/mol = 71.2 g CS2 theoretical yield
4,51 moles hydrogen exist.
Molar Mass of Al: 2(27.0g/mol) = 54.0g/mol Molar Mass of O: 3(16.0g/mol) = 48.0g/mol Molar Mass of compound: 102.0g.mol (54.0g/mol / 102.0g/mol) x 100% = 52.9%
12.5 (g CuCO3) = [12.5 (g CuCO3) / 123.555 (g/mol CuCO3)] = 0.1012 (mol CuCO3)0.1012 (mol CuCO3)* [1 (mol CuO) / (mol CuCO3)] = 0.1012 (mol CuO)= [0.1012 (mol CuO) * 79.545 (g/mol CuO)] = 8.047 g CuO = 8.05 g CuO
A) 0.500 mol B) 1.000 mol C) 2.00 mol D) 3.00 mol E) 1/16 mol
molar mass is 318 g/mol (2X27)+3(2X12+4X16)=318g/mol