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3 years ago

12

Complete the table for ion charge based upon their losing or gaining electrons in the outer shell. (Use the periodic table as ne

cessary.) Group Most Likely Ionic Charge # of Valence Electrons I +1 II +2 III +3 IV +4 or -4 V -3 VI -2 VII -1 VIII 0

2 answers:

hodyreva [135]3 years ago

5 0

Answer:

Explanation:

Group Most Likely Ionic Charge Number of Valence Electrons

I +1 1

II +2 2

III +3 3

IV +4 or -4 4

V -3 5

VI -2 6

VII -1 7

VIII 0 8

For elements in group IV and above, their ionic charge is (8-number of their valence electrons.)

Degger [83]3 years ago

4 0

<u>Explanation:</u>

Valence electrons are defined as the electrons which are present in the outermost shell of an atom.

The number of valence electrons in a particular group will be equal to the group number.

Number of valence electrons = Group number

The given table follows:

Group Most likely ionic charge No. of valence electrons

I +1 1

II +2 2

III +3 3

IV +4 or -4 4

V -3 5

VI -2 6

VII -1 7

VIII 0 8

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Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

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$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

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