The volume of 5.00 M KBr needed to have 1.00 mol KBr mL
1 answer:
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This set up of a conversion table should show you that if you multiply the grams of BeI2 times .02 moles, it equals <span>5.256 g (your answer) </span>
Answer:
Element:
1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰7p¹
Cations:
1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰
1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶5f¹⁴6d¹⁰
Explanation:
The electron configuration is the distribution of the electrons in the sublevels in order of the crescent energy of them. The crescent energy of the sublevels follows the Linus Pauling's diagram, which is attached below. The sublevel "s" comports until 2 electrons, the sublevel "p" until 6 electrons, the sublevel "d" until 10 electrons, and sublevel "f" until 14 electrons.
So, for the element with an atomic number of 113, the neutral atom will have 113 electrons:
1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰7p¹
Thus the element is at the 7 period (the highest level), and group 13 (most energic sublevel p with 1 electron), the group of the aluminum. It needs to lose 3 electrons to be stable and follow the octet rule, but the subshells of the last shell are too far away in energetic order, thus, it most probably to lose the electron of 7p and form a monovalent cation, and can lose the two electrons of 7s to form a trivalent cation:
1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶7s²5f¹⁴6d¹⁰
1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d¹⁰6p⁶5f¹⁴6d¹⁰
