Compare the strength of the bonds that hold the atoms in a molecule together with the forces that exist between different molecu
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Answer:
<em>a)</em> <em>1.392 x 10^6 g/cm^3</em>
<em>b) 8.69 x 10^7 lb/ft^3</em>
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Explanation:
mass of the star m = 2.0 x 10^36 kg
radius of the star (assumed to be spherical) r = 7.0 x 10^5 km = 7.0 x 10^8 m
The density of substance ρ = mass/volume
The volume of the star = volume of a sphere =
==> V = = 1.437 x 10^27 m^3
density of the star ρ = (2.0 x 10^36)/(1.437 x 10^27) = 1.392 x 10^9 kg/m^3
in g/cm^3 = (1.392 x 10^9)/1000 = <em>1.392 x 10^6 g/cm^3</em>
in lb/ft^3 = (1.392 x 10^9)/16.018 = <em>8.69 x 10^7 lb/ft^3</em>
Answer:
Explanation:
Hello!
In this case, we can divide the problem in two steps:
1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:
So we solve for C2:
2. Now, since 111 mL of water is added, we compute the final volume, V3:
So, the final concentration of the 139 mL portion is:
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