Aredox reaction involves the reduction of one substance (gaining electron(s)) and the oxidation (losing electron(s)) of another. examples of reduction and oxidation are shown below. combine the two reactions below like algebraic equations and write the net, or overall, redox equation.

the terms  ionic  and  covalent  describe the extremes of a continuum of bonding. there is some covalent character in even the most ionic compounds and vice versa.

it is useful to think about the compounds of the main group metals as if they contained positive and negative ions. the chemistry of magnesium oxide, for example, is easy to understand if we assume that mgo contains mg2+  and o2-  ions. but no compounds are 100% ionic. there is experimental evidence, for example, that the true charge on the magnesium and oxygen atoms in mgo is +1.5 and -1.5.

oxidation states provide a compromise between a powerful model of oxidation-reduction reactions based on the assumption that these compounds contain ions and our knowledge that the true charge on the ions in these compounds is not as large as this model predicts. by definition, the oxidation state of an atom is the charge that atom would carry if the compound were purely ionic.

for the active metals in groups ia and iia, the difference between the oxidation state of the metal atom and the charge on this atom is small enough to be ignored. the main group metals in groups iiia and iva, however, form compounds that have a significant amount of covalent character. it is misleading, for example, to assume that aluminum bromide contains al3+  and br-  ions. it actually exists as al2br6  molecules.

this problem becomes even more severe when we turn to the chemistry of the transition metals. mno, for example, is ionic enough to be considered a salt that contains mn2+  and o2-  ions. mn2o7, on the other hand, is a covalent compound that boils at room temperature. it is therefore more useful to think about this compound as if it contained manganese in a +7 oxidation state, not mn7+  ions.

electron microscopy and scanning tunneling microscopes gives us a greater understanding of atoms by being able to see the atom itself, and lets us see some atomic reactions.

explanation:

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm