5. When a ruthenium chloride solution was electrolyzed for 500 s with a 120mA current, 31 mg of ruthenium was deposited. What is the oxidation number of ruthenium in the ruthenium chloride?

Explanation:

Relation between equivalent weight, current, time is as follows.

         W =

where,    W = mass of the metal deposited = 31.0 mg =

               E = Equivalent weight of Ru = ?

               C = current = 120 mA =

               t = time taken = 500 sec

Therefore, calculate the value of equivalent weight as follows.

               E =

                  =

                  = 49.86 g eqv

Since, we know that the molar mass of ruthenium is 101.1 g/mol.

Therefore, the oxidation number of ruthenium is calculated as follows.

                   = 2.027

                   = 2 (approximately)

Thus, we can conclude that the oxidation number of ruthenium in the ruthenium chloride is 2.

1. 1 mole of anything contains 6.022x10^23 pieces  

d  

2. a  

3. c, number  

5. 1.24x10^24atoms / 6.022x10^23atoms/mole = 2moles h2 atoms  

6. fe2br3 mass = 2moles x mass fe + 3 moles x mass br  

molar mass = grams / number of moles = g/mole  

to find moles = g / molar mass  

to find mass = g = molar mass / moles

i think it is true since you can test the answer and can use observations to complete it.