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The ideal gas equation can be rearranged to give an expression for the molar volume of an ideal gas: = = Hence, for a given temperature and pressure, the molar volume is the same for all ideal gases and is based on the gas constant: R = 8.314 462 618 153 24 m 3 ⋅Pa⋅K −1 ⋅mol −1, or about 8.205 736 608 095 96 × 10 −5 m 3 ⋅atm⋅K ...
The molar mass of a substance depends not only on its molecular formula, but also on the distribution of isotopes of each chemical element present in it. For example, the molar mass of calcium-40 is 39.962 590 98 (22) g/mol, whereas the molar mass of calcium-42 is 41.958 618 01 (27) g/mol, and of calcium with the normal isotopic mix is 40.078(4 ...
Mole ratio: Convert moles of Cu to moles of Ag produced; Mole to mass: Convert moles of Ag to grams of Ag produced; The complete balanced equation would be: Cu + 2 AgNO 3 → Cu(NO 3) 2 + 2 Ag. For the mass to mole step, the mass of copper (16.00 g) would be converted to moles of copper by dividing the mass of copper by its molar mass: 63.55 g/mol.
The molar mass of atoms of an element is given by the relative atomic mass of the element multiplied by the molar mass constant, M u ≈ 1.000 000 × 10 −3 kg/mol ≈ 1 g/mol. For normal samples from Earth with typical isotope composition, the atomic weight can be approximated by the standard atomic weight [ 2 ] or the conventional atomic weight.
It is a dimensionless quantity with dimension of / and dimensionless unit of moles per mole (mol/mol or mol ⋅ mol-1) or simply 1; metric prefixes may also be used (e.g., nmol/mol for 10-9). [5] When expressed in percent , it is known as the mole percent or molar percentage (unit symbol %, sometimes "mol%", equivalent to cmol/mol for 10 -2 ).
Using the number density as a function of spatial coordinates, the total number of objects N in the entire volume V can be calculated as = (,,), where dV = dx dy dz is a volume element. If each object possesses the same mass m 0 , the total mass m of all the objects in the volume V can be expressed as m = ∭ V m 0 n ( x , y , z ) d V ...
For a substance X with a specific volume of 0.657 cm 3 /g and a substance Y with a specific volume 0.374 cm 3 /g, the density of each substance can be found by taking the inverse of the specific volume; therefore, substance X has a density of 1.522 g/cm 3 and substance Y has a density of 2.673 g/cm 3. With this information, the specific ...
Two binary solutions of different compositions or even two pure components can be mixed with various mixing ratios by masses, moles, or volumes. The mass fraction of the resulting solution from mixing solutions with masses m 1 and m 2 and mass fractions w 1 and w 2 is given by: