<u>Answer:</u>
<em>The correct equation for measuring the average microscopic weight for 3 isotopes is multiply the rate of abundance by each weight and add them.</em>
<u>Explanation:</u>
To calculate the average microscopic mass of element using weights and relative abundance we have to follow the following steps.
- Take the correct weight of each isotope (that will be in decimal form)
- Multiply the weight of each isotope by its abundance
- Add each of the results together.
<em>This gives the required average microscopic weight of the three isotopes.</em>
<em>Answer:</em>
<h3><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>True</em></h3>
- <em>Because </em><em>Gravity is the force of attraction between two objects, and Earth's gravity pulls matter downward, toward its center. It pulls precipitation down from clouds and pulls water downhill. Gravity also moves air and ocean water. ... Gravity pulls denser air and water downward, forcing less dense air and water to move upward.</em>
<em>Carryonlearning</em>
Explanation:
Given that,
Charge 1, 
Charge 2, 
Distance between charges, r = 0.0209 m
1. The electric force is given by :


F = -492.95 N
2. Distance between two identical charges, 
Electric force is given by :




Hence, this is the required solution.