Answer:
.
Explanation:
Electrons are conserved in a chemical equation.
The superscript of 
indicates that each of these ions carries a charge of 
. That corresponds to the shortage of one electron for each 
ion.
Similarly, the superscript 
on each 
ion indicates a shortage of three electrons per such ion.
Assume that the coefficient of (among the reactants) is 
, and that the coefficient of (among the reactants) is 
.
.
There would thus be silver (
) atoms and aluminum (
) atoms on either side of the equation. Hence, the coefficient for 
and 
would be 
and 
, respectively.
.
The ions on the left-hand side of the equation would correspond to the shortage of electrons. On the other hand, the 
ions on the right-hand side of this equation would correspond to the shortage of 
electrons.
Just like atoms, electrons are also conserved in a chemical reaction. Therefore, if the left-hand side has a shortage of electrons, the right-hand side should also be electrons short of being neutral. On the other hand, it is already shown that the right-hand side would have a shortage of electrons. These two expressions should have the same value. Therefore, 
.
The smallest integer and that could satisfy this relation are 
and 
. The equation becomes:
.
Potassium outermost electron occupy "4s" orbital
The volume of CO2 at STP =124.298 L
<h3>Further explanation</h3>
Given
Reaction
4 KMnO4, +4 C3H5(OH)5, -7K2CO3, + 7 Mn2O3, +5 CO2, + 16 H2O
701,52 g of KMnO4
Required
volume of CO2 at STP
Solution
mol KMnO4 (MW=158,034 g/mol) :
mol = mass : MW
mol = 701.52 : 158.034
mol = 4.439
mol CO2 from equation : 5/4 x mol KMnO4 = 5/4 x 4.439 = 5.549
At STP 1 mol = 22.4 L, so for 5.549 moles :
=5.549 x 22.4
=124.298 L
Answer:
Group 8 or Group 0
Explanation:
Group 8 or Group 0 are generally inert gases with Helium as the first member in that group. Their complete duplet (in the case of Helium) and Octet (in the case of Neon) configuration makes them very stable and chemically un-reactive.
