Join IDNLearn.com and start exploring the answers to your most pressing questions. Get the information you need quickly and accurately with our reliable and thorough Q&A platform.
Sagot :
Answer
A. Neutron
B. There is no change in the number of protons
Procedure
Consider the following chemical equation
[tex]_{80}^{200}\text{Hg }+\text{ }=_{^80}^{201}\text{Hg}[/tex]A. The reaction can happen if there is a neutron that collides with the isotope with a weight of 200 amu, this is because the charge remains constant, therefore it will not be feasible to preserve a neutral charge in case a proton or an electron collides.
In the equation, this is written
[tex]_{80}^{200}\text{Hg }+_0^1\text{ n }=_{^80}^{201}\text{Hg}[/tex]B. Because the protons remain the same in the nucleus, the number of protons determines the nature of the element, mercury will always have 80 protons in its nucleus, if this changes we will be speaking of another element.
Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. For precise answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.
