Across a period I.E increases progressively from left to right
Explanation:
The trend of the first ionization energy is such that across a period I.E increases from left to right due to the decreasing atomic radii caused by the increasing nuclear charge. This not compensated for by successive electronic shells.
- Ionization energy is a measure of the readiness of an atom to lose an electron.
- The lower the value, the easier it is for an atom to lose an electron.
- Elements in group I tend to lose their electrons more readily whereas the halogens hold most tightly to them.
- The first ionization energy is the energy needed to remove the most loosely bonded electron of an atom in the gaseous phase.
Learn more:
Ionization energy brainly.com/question/6324347
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When an electron absorbs energy, it will move up from a lower energy level to a higher energy level, called the "excited state" of the negatively-charged subatomic particle.<span> However, the absorbed energy is released within a small interval of time and the electron moves down to its "ground state."</span>
Answer: 
for the reaction is -186.75 J/K
Explanation:
Change in entropy () for the given reaction under standard condition is given by-
= ![[3\times S_{rhombic}^{0}_{(s)}]+[2\times S_{H_{2}O}^{0}_{(g)}]-[2\times S_{H_{2}S}^{0}_{(g)}]-[1\times S_{SO_{2}}^{0}_{(g)}]](https://tex.z-dn.net/?f=%5B3%5Ctimes%20S_%7Brhombic%7D%5E%7B0%7D_%7B%28s%29%7D%5D%2B%5B2%5Ctimes%20S_%7BH_%7B2%7DO%7D%5E%7B0%7D_%7B%28g%29%7D%5D-%5B2%5Ctimes%20S_%7BH_%7B2%7DS%7D%5E%7B0%7D_%7B%28g%29%7D%5D-%5B1%5Ctimes%20S_%7BSO_%7B2%7D%7D%5E%7B0%7D_%7B%28g%29%7D%5D)
So = ![[3\times 31.8 J/K.mol]+[2\times 188.825 J/K.mol]-[2\times 205.79 J/K.mol]-[1\times 248.22 J/K.mol]](https://tex.z-dn.net/?f=%5B3%5Ctimes%2031.8%20J%2FK.mol%5D%2B%5B2%5Ctimes%20188.825%20J%2FK.mol%5D-%5B2%5Ctimes%20205.79%20J%2FK.mol%5D-%5B1%5Ctimes%20248.22%20J%2FK.mol%5D)
= -186.75 J/K
All the following are equal to Avogadro's number EXCEPT a. the number of atoms of bromine in 1 mol Br₂.
1 mol Br₂ contains Avogadro’s number of molecules of Br₂.
However, each molecule contains two atoms of Br, so there are
<em>2 × Avogadro’s number of Br atoms </em>in 1 mol Br₂.
