this is a quiz 6. Relate the sizes of the three subatomic particles to more tangible everyday objects. A= 7. What particles contain most of the mass of an atom? A= 8. What is in between the nucleus and the electrons? A= 9. What is the density of the nucleus? A= 10. How does this density compare with the density of water? A= 11. What is the most basic unit of matter? A= 12. Inside the most basic unit of matter, there are three different particles. What are they and how are they different from each other? A= 13. What are the three particles’ charges? A= 14. Which particles make up the nucleus of the atom? A= 15. The atom is made up of mostly . 16. What is the definition of an element? A= 17. What makes one element different from another element? A= 18. How many protons, neutrons, and electrons are in Carbon, Hydrogen, Oxygen, and Neon. A= 19. What is an isotope? A=

Adozen is 1221/12 = 1.751.75 x 12 = 21therefore, 1.75 is the answer.

Theoretical temperature at which molecules stop moving and no more heat can be removed from the object. kelvin: absolute zero degrees. none of the above.

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.