The new concentrations of 
and 
are 0.25M and 19M
Calculation of number of moles of each component,
Molarity of = number of moles/volume in lit = 0. 500 M
Number of moles = molarity of × volume in lit = 0. 500 M× 0.025 L
Number of moles of = 0.0125 mole
Molarity of = number of moles/volume in lit = 0. 38 M
Number of moles = molarity of × volume in lit = 0. 38 M× 0.025 L
Number of moles of = 0.95 mole
Calculation of new concentration at volume 50 ml ( 0.05L)
Molarity of = number of moles/volume in lit = 0.0125 mole/0.05L
Molarity of = 0.25M
Molarity of = number of moles/volume in lit = 0.95mole/0.05L
Molarity of = 19 M
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Answer:
Explanation:
Hello!
In this case, when we want to balance chemical reactions such as in this case, the idea is to equal to number of atoms of each element at each side of the equation according to the lay of conservation of mass, just as shown below:
Because we have four phosphorous and ten oxygen atoms at each side.
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We have to get the relationship between metallic character and atomic radius.
Metallic character increases with increase in atomic radius and decrease with decrease of atomic radius.
If electrons from outermost shell of an element can be removed easily, that atom can be considered to have more metallic character.
With increase in atomic radius, nuclear force of attraction towards outermost shell electron decreases which facilitates the release of electron.
With decrease in atomic radius, nuclear force of attraction towards outermost shell electrons increases, so electrons are hold tightly to nucleus. Hence, removal of electron from outermost shell becomes difficult making the atom less metallic in nature.
