Answer:
Atomic number of this isotope = 77
Explanation:
Given that,
Mass number = 193
No of neutrons = 116
We need to find the atomic no of this isotope.
We know that,
Atomic mass = No of protons + No. of neutrons
Also, atomic no = no of protons
So,
Atomic mass = atomic no + No. of neutrons
⇒ Atomic no = Atomic mass - no of neutrons
Atomic no = 193 - 116
Atomic no = 77
Hence, 77 is the atomic no of the isotope.
This can be, for example, halogensubstituted hydrocarbons.
CCl₄, C₂F₆.
Or halides halocarboxylic acids, and other compounds.
O
II
Cl₃C-Cl
Answer:
1-ethyl-2-methyl cyclopropane.
Explanation:
- The structure of the molecule will be as shown in the attached image.
- The molecular formula of the compound is C₆H₁₂.
- It has 3 membered ring with 3 C atoms and two substituents one of them with one C atom (methyl) and the other with 2 C atoms (ethyl).
- The ring consist of 3 C atoms, so its name is cyclo propane.
- We numbering the atoms of the ring that give the ethyl substituent the low no. (1) and then methyl group take no. (2).
- <em>Thus, the name of the compound is 1-ethyl-2-methyl cyclopropane.</em>
Answer:
Hi. im an online tutor and i ca help you with all your assignments . check out our wesite https://toplivewriters.com
Explanation: