Answer:
It's because removal of electron from an atom, reduces the size of an atom.
Explanation:
When an electron is removed from an atom, it becomes an ion and in this case it will become a postive ion.
When an electron is removed from an atom, the charge balance of an atom is disturbed and positive charge increases in comparison to the negative charge. This results in increase nuclear (positive) charge which exerts greater attraction on the remaining electrons and as a result the remaining electrons are more strongly attracted by the nucleus and in this way the atomic size is decreased. Due to this increased nuclear attraction and reduced atomic size, it bcomes difficult to remove more electeon from the positively charged ion of reduced size. This is the reason that each successive ionization of electron requires a greater amount of energy.
The ionization energy has inverse relation with the size or radius of an atom. This also justifies the reason that why each successive ionization of an electron requires greater amount of energy.
Answer:
The answer would be at 120°C
<em><u>Question</u></em>
<em><u>What </u></em><em><u>does </u></em><em><u>it </u></em><em><u>mean </u></em><em><u>to </u></em><em><u>optimize</u></em><em><u> </u></em><em><u>a </u></em><em><u>solution?</u></em>
<em><u>To find out best possible solution for a given problem within the given constraint is generally termed as optimization</u></em>
<em><u>How </u></em><em><u>are </u></em><em><u>solution</u></em><em><u> </u></em><em><u>optimize</u></em><em><u> </u></em><em><u>?</u></em>
<em><u>To solve an optimization problem, begin by drawing a picture and introducing variables. Find an equation relating the variables. Find a function of one variable to describe the quantity that is to be minimized or maximized. Look for critical points to locate local extrema.</u></em>
setup 1 : to the right
setup 2 : equilibrium
setup 3 : to the left
<h3>Further explanation</h3>
The reaction quotient (Q) : determine a reaction has reached equilibrium
For reaction :
aA+bB⇔cC+dD
![\tt Q=\dfrac{C]^c[D]^d}{[A]^a[B]^b}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7BC%5D%5Ec%5BD%5D%5Ed%7D%7B%5BA%5D%5Ea%5BB%5D%5Eb%7D)
Comparing Q with K( the equilibrium constant) :
K is the product of ions in an equilibrium saturated state
Q is the product of the ion ions from the reacting substance
Q <K = solution has not occurred precipitation, the ratio of the products to reactants is less than the ratio at equilibrium. The reaction moved to the right (products)
Q = Ksp = saturated solution, exactly the precipitate will occur, the system at equilibrium
Q> K = sediment solution, the ratio of the products to reactants is greater than the ratio at equilibrium. The reaction moved to the left (reactants)
Keq = 6.16 x 10⁻³
Q for reaction N₂O₄(0) ⇒ 2NO₂(g)
![\tt Q=\dfrac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=%5Ctt%20Q%3D%5Cdfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
Setup 1 :
Q<K⇒The reaction moved to the right (products)
Setup 2 :
Q=K⇒the system at equilibrium
Setup 3 :
Q>K⇒The reaction moved to the left (reactants)
