Chemistry, 07.11.2020 01:40 hcpscyruscm

Indicate the charge the following elements as they achieve the noble gas configuration. Ga O Br P Rb As
S Mg Al Se Li I

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Answer from: 91miketaylor

See explanation

Explanation:

Ga is in group 13 hence it must loose three electrons to form Ga^3+ in order to achieve the noble gas configuration because it has three electrons on its outermost shell.

O is in group 16 hence it must accept two electrons in order to attain the noble gas configuration to form O^2- since oxygen has six electrons on its outermost shell.

Br in group 17 has seven electrons in its outermost shell hence it must form Br^- (gain one electron) in order to attain the noble gas configuration.

P in group 15 must accept three electrons and form P^3- in order to attain the noble gas configuration since it has five electrons on its outermost shell.

S is in group 16 hence it must accept two electrons in order to attain the noble gas configuration to form S^2- since sulphur has six electrons on its outermost shell.

Mg in group 2 has two electrons on its outermost shell and must loose both to attain the noble gas configuration forming Mg^2+.

Al is in group 13 hence it must loose three electrons to form Al^3+ in order to achieve the noble gas configuration because it has three electrons on its outermost shell.

Se is in group 16 hence it must accept two electrons in order to attain the noble gas configuration to form Se^2- since selenium has six electrons on its outermost shell.

Lithium is in group 1 and must loose its only outermost electron in order to attain the noble gas configuration to form Li^+.

Rb is in group 1 and must loose its only outermost electron in order to attain the noble gas configuration to form Rb^+.

As in group 15 must accept three electrons and form As^3- in order to attain the noble gas configuration since it has five electrons on its outermost shell.

I in group 17 has seven electrons in its outermost shell hence it must form I^- (gain one electron) in order to attain the noble gas configuration.

Answer from: Quest

$\frac{239}{94}pu+\frac{1}{0}n ====> \frac{y}{z}x+\frac{134}{54}xe+3\frac{1}{0}$

the catch is that the top numbers are atomic masses the bottom numbers are the numbers of protons and electrons in an uncharged atom. they are the number of that atoms as you would find them on the periodic table. further, the numbers on the left are used to determine the missing numbers on the right

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137 subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54 subtract 54 from both sides.

40 = z

answer

$\frac{103}{40}x=\frac{103}{40}zr$

Answer from: Quest

Ithink d hope that

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Indicate the charge the following elements as they achieve the noble gas configuration. Ga O Br P R...

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