Answer:
19.6 J
Step-by-step explanation:
Before the ball is dropped, it has a <em>potential energy
</em>
PE = mgh
PE = 0.2 × 10 × 9.8
PE = 19.6 J
Just before the ball hits the ground, the potential energy has been converted into kinetic (<em>mechanical</em>) energy.
KE = 19.6 J
Answer: The primary consumers are zooplankton, corals, sponges, Atlantic blue tang, and queen conch.
Explanation:
Answer : The solubility of 
in water is, 
Explanation :
The solubility equilibrium reaction will be:
Let the molar solubility be 's'.
The expression for solubility constant for this reaction will be,
![K_{sp}=[Ag^{+}]^3[PO_4^{3-}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BAg%5E%7B%2B%7D%5D%5E3%5BPO_4%5E%7B3-%7D%5D)
Given:
= 
Now put all the given values in the above expression, we get:
Therefore, the solubility of in water is,
Answer:
Average atomic mass of uranium= 237.98 amu.
Explanation:
Given data:
Abundance of U²³⁴ = 0.01%
Abundance of U²³⁵ = 0.17%
Abundance of U²³⁸ = 99.28%
Average atomic mass = ?
Solution:
Average atomic mass of uranium = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) +(abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass of uranium= (234×0.01)+(235×0.71)+(238×99.28)/100
Average atomic mass of uranium= 2.34 + 166.85 + 23628.64 / 100
Average atomic mass of uranium= 23797.83 / 100
Average atomic mass of uranium= 237.98 amu.
