For the 52nd electron must pair with one of the electrons in the 5p orbital.
1 answer:
Answer:
This question is incomplete
Explanation:
This question is incomplete, however, the element that has 52 electrons only is Tellurium (Te) and when the electronic configuration of elements with more than 52 electrons are written, the 52nd electron is indicated/paired the same way the 52nd electron of Te is indicated/paired. Hence, while writing the electronic configuration of Te, it is written as
[Kr] 4d¹⁰ 5s² 5p⁴ where [Kr] is the electronic configuration of krypton. Based on this, we can deduce that the 52nd electron will be in the first orbital of the P subshell (as attached in the picture). This is because when indicating the electrons in the subshell, one electron will be spread across each orbital and if any electron is still remaining, it will be added starting from to the first orbital of the subshell, however no two electrons in an orbital in a subshell can have the same spin and hence must face opposite direction based on pauli's exclusion principle (as seen in attached); thus for the 5p-orbital of elements with 52 or more electrons, when one electron each is represented in each box (3 boxes in total) in the 5p-orbital, the remaining electron is paired with the the first electron in the first box of the 5p-orbital
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1) Balance the chemical equation.
2) List the known and unknown quantities.
Reactant 1: Ca(NO3)2.
Amount of substance: 3.40 mol.
Reactant 2: Li3PO4.
Amount of substance: 2.40 mol.
Product: Ca3(PO4)2
Mass: unknown.
3) Which is the limiting reactant?
<em>3.1-How many moles of Li3PO4 do we need to use all of the Ca(NO3)2?</em>
The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.
<em>We need 2.2667 mol Li3PO4 and we have 2.40 mol Li3PO4. We have enough Li3PO4. </em>This is the excess reactant.
<em>3.2-How many moles of Ca(NO3)2 do we need to use all of the Li3PO4?</em>
The molar ratio between Li3PO4 and Ca(NO3)2 is 2 mol Li3PO4: 3 mol Ca(NO3)2.
<em>We need 3.60 mol Ca(NO3)2 and we have 3.40 mol Ca(NO3)2. We do not have enough Ca(NO3)2. </em>This is the limiting reactant.
4) Moles of Ca3(PO4)2 produced from the limiting reactant.
We have 3.40 mol Ca(NO3)2 of the limiting reactant.
The molar ratio between Ca(NO3)2 and Ca3(PO4)2 is 3 mol Ca(NO3)2: 1 mol Ca3(PO4)2.
5) Mass of Ca3(PO4)2 produced.
The molar mass of Ca3(PO4)2 is 310.1767 g/mol.
<em>The mass of Ca3(PO4)2 produced is</em> 351 g Ca3(PO4)2.
Option D.
.
Answer:
Isotopes of an element are atoms of the same element that have different number of neutrons.
