I think it’s D. Stars and planets.
Answer:
A. 1.63g/dm^3 (3 s.f.)
B. 0.833g/dm^3 (3 s.f.)
C. 1.92g/dm^3 (3 s.f.)
Explanation:
Please see attached picture for full solution.
I am hamster fun boo poop
The number of chlorine atoms needed would simply be the
ratio of distance and diameter. But first convert 200 pm to mm:
<span>200 pm = 2 E-7 mm </span>
So the number of chlorine atoms needed is:
<span>1.0 mm / (2 E-7 mm) = 5,000,000 Chlorine atoms = 5 E6
atoms</span>
Weaken London dispersion forces and stronger dipole dipole forces