Option (i) would have the highest 2nd Ionization Energy.
Option (i) is Sodium.
Can be Written as 2, 8 , 1
For its 1st Ionization energy... It'd be extremely easy to remove that Electron cos its on the outermost shell.
Now After Removing that Electron...
Sodium's Electronic Configuration Reduces to that of Neon Which is 2, 8.
Neon has a very stable Octet.
It would take an ENORMOUS amount of energy to break its Octet stability... that is... Remove 1 electron from its Octet.
So
Option (i) [Sodium] has the highest 2nd Ionization Energy
In chemistry, a solution is a homogeneous mixture composed of two or more substances. In such a mixture, a solute is a substance dissolved in another substance, known as a solvent.
¹/3 C3H8(g) + ⁵/3 O2(g)
Explanation:
The coefficient before every molecule is representative of the number of moles. We can represent it in ration form so as to calculate the question;
C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(l) means;
For every 1 mole of C₃H₈(g) and 5 moles of O₂(g) produces 3 moles of CO₂(g) and 4 moles of H₂O(l).
Therefore to produce 1.00 mole of CO₂(g);
We represent it in ratio;
C₃H₈(g) : CO₂(g)
1 : 3
What about ;
? (x) : 1
We cross multiply;
3x = 1 * 1
X = 1/3
We evaluate the same for O₂;
O₂(g) : CO₂(g)
5 : 3
What about
? (x) : 1
3x = 5 * 1
x = 5/3
Learn More:
For more on evaluating moles in chemical reactions check out;
brainly.com/question/13967925
brainly.com/question/13969737
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Answer:
repel
Explanation:
When it comes to electrical forces, "opposites charges attract" while "like charges repel."
There are primarily two types of charges: positive charge and negative charge. The forces they exert upon each other will depend on their charges. The<u> positive charge has an </u><em><u>attractive force</u></em><u> to a negative charge.</u> On the contrary,<u> it has a</u><em><u> repulsive force</u></em><u> to the same positive charge</u>. Thus, it will repel each other.
So this means that <em>opposite charges will draw closer together</em> while<em> like charges will move apart from each other.</em>