The electron configuration that belongs to the atom with the lowest first ionization energy is francium.
<h3>What is ionization energy? </h3>
Ionization energy is defined as the minimum amount of energy required to remove the most loosely electron present in outermost shell.
<h3>Ionization energy across period</h3>
Ionization energy increase as we move from left to right in the period. This can be explained as when we move from left to right along period new electron is added to the same shell which increase the nuclear charge. Hence results int he decrease in size. Due to this decrease in size more energy is required to remove electron from outermost shell.
<h3>Ionization energy along group</h3>
Ionization energy decrease as we move from top to bottom along group. This can be explained as we move from top to bottom new electron is added to new shell. Due to addition of new shell the size of atom increases which results in the decrease in the nuclear charge. Due to this less amount of energy is needed to remove an electron.
Thus, we concluded that the electron configuration that belongs to the atom with the lowest first ionization energy is francium.
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Answer:
0.12M
Explanation:
A balanced equation for the reaction will go a great deal in obtaining our desired result. So, let us write a balanced equation for the reaction
HCl + NaOH —> NaCl + H2O
From the above equation,
nA (mole of the acid) = 1
nB (mole of the base) = 1
Data obtained from the question include:
Vb (volume of the base) = 30mL
Mb (Molarity of the base) = 0.1M
Va (volume of the acid) = 25mL
Ma (Molarity of the acid) =?
The molarity of the acid can be obtained as follow:
MaVa/MbVb = nA/nB
Ma x 25/ 0.1 x 30 = 1
Cross multiply to express in linear form
Ma x 25 = 0.1 x 30
Divide both side by 25
Ma = (0.1 x 30) / 25
Ma = 0.12M
The molarity of the acid is 0.12M
