43.2 cm is .432 m
(pi)r2 = area of circle = 0.586 m2
.000000885kg/923kg/m3 = 9.59e-10 m3 = volume of oil cylinder
volume/area = thickness
1.64e-9 m
Firstly, calculate the molecular volume by dividing the product of molecular density and the molecular weight by the Avogadro number. Then take a third root of the volume, the found value is the atomic radius. Then multiply it by 2 to find the diameter.
There are too many different compounds with an approximate relative density of 0.95 for this question to be answered.
Diameter = 5 cm so volume = 4/3*Ï€*(d/2)3 = 65.45 cm3 Then density = 10/64.45 = 0.1528 grams per cm3
0.135 g/cm3
That would depend on several factors; the velocity of the meteoroid, the mass, density and composition of the meteoroid, and the nature of the surface where it strikes.
5
The density of electrons in a water molecule is highest around/near the oxygen atom.
1.408 g/cm3
this is not a county in west Virginia
If there are more molecules in a given volume of air, the density is greater. If there are fewer molecule, the density decreases. Density=Mass/Volume
Firstly, calculate the molecular volume by dividing the product of molecular density and the molecular weight by the Avogadro number. Then take a third root of the volume, the found value is the atomic radius. Then multiply it by 2 to find the diameter.
Height and diameter will give you the volume, if you know the density you can then calculate weight from that.
There are too many different compounds with an approximate relative density of 0.95 for this question to be answered.
To answer this question, the mass of the spherical protein is needed. the fact that the protein is a sphere and that it has a density of 1gcm3 is not enough information to determine the diameter.
Calculate volume of molecule, calculate mass of molecule, compare to bulk density. Lower bulk density indicates empty space.
The sun has an approximate density of 1.4 g cm-3. It is only an approximation because the sun's density varies.
The density of francium hydroxide was not estimated; an approximate estimation can be made by comparison with the alkali metals hydroxides.