How many deciliter a of a .88m lif solution be made using 25.5 cg lif? lif molar mass = 25.94 g/mol

0.112 deciliter

In this case we are given;

Molarity of LiF as 0.88 M

Mass of LiF as 25.5 cg

but, 1 g = 100 cg

Therefore, mass of LiF = 0.255 g

Molar mass of LiF = 25.94 g/mol

We are required to calculate the volume of LiF in deciliter

Moles = Mass ÷ Molar mass

          = 0.255 g ÷ 25.94 g/mol

          = 0.00983 mole

Molarity = Moles ÷ Volume

Rearranging the formula;

Volume = Moles ÷ Molarity

Volume of LiF = 0.00983 mole ÷ 0.88 moles/liter

                        = 0.0112 L

But, 1 L = 10 DL

Therefore, Volume = 0.112 DL

Thus, the volume of LiF is 0.112 deciliter

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.