Answer:
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The theoretical molar yield of lead (II) chloride will be 9 moles.
<h3>Stoichiometric calculation</h3>
First, we need to look at the equation of the reaction:
From the equation, the 1 mole of Pb2+ ion requires 2 moles of Cl- ion in order to produce 1 mole of lead (II) chloride.
Thus, with 18 moles Cl- ions, 9 moles of Pb2+ would be needed, instead of 12 moles. Pb2+ is simply in excess while Cl- can be said to be limiting.
Therefore, Cl- will determine how much of lead (II) chloride that will be produced. The ratio is 2 moles of Cl- to 1 mole of lead (II) chloride.
With 18 moles Cl-, 9 moles of lead (II) chloride will, thus, be produced.
More on mole ratios can be found here: brainly.com/question/14425689
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Answer is: a. Rubidium (Rb) is more reactive than strontium (Sr) because strontium atoms must lose more electrons.
The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
Alkaline metals (group 1), in this example rubidium, have lowest ionizations energy and easy remove valence electrons (one electron), they are most reactive metals.
Earth alkaline metals (group 2), in this example strontium, have higher ionization energy than alkaline metals, because they have two valence electrons, they are less reactive.
Rubidium electron configuration: ₃₇Rb 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s¹; one valence electron is 5s¹ orbital.
Strontium electron configuration: ₃₈Sr 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s²; two valence electrons is 5s² orbital.