Now, let's finish the calculation and the determination of the formula of the iron compound: Calculate the % water of hydration : Tries 0/3 Calculate the following for Fe3 : g in 100 g sample mol in 100 g sample mol/mol Fe (3 sig figs) mol/mol Fe (whole number) Tries 0/3 Calculate the following for K : g in 100 g sample mol in 100 g sample mol/mol Fe (3 sig figs) mol/mol Fe (whole number) Tries 0/3 Calculate the following for C2O42-: g in 100 g sample mol in 100 g sample mol/mol Fe (3 sig figs) mol/mol Fe (whole number) Tries 0/3 Calculate the following for H2O g in 100 g sample mol in 100 g sample mol/mol Fe (3 sig figs) mol/mol Fe (whole number)

Complete Question

The complete question is shown on the first and second uploaded image

A

The percentage of water of hydration is   %

Mass of in 100mg is 10.60mg

Moles of   in 100mg is

mol / mol Fe (3 sig figs) is

mol / mol Fe (whole number) is = 1

B

Mass of in 100mg is 27.70mg

Moles of   in 100mg is

mol  of K / mol of Fe (3 sig figs) is

mol  of K / mol of Fe (whole number) is

C

Mass of in 100mg is 55.69 mg

Moles of  in 100mg is

mol  of   / mol of Fe (3 sig figs) is

mol  of / mol of Fe (whole number) is

D

Mass of water in 100mg is 11.01 mg

Moles of water  in 100mg is

mol  of  water / mol of Fe (3 sig figs) is

mol  of water / mol of Fe (whole number) is

Explanation:

The percentage of water of hydration is mathematically represented as

Now substituting 10.60% for (percentage of iron ) , 22.70% for (Percentage of potassium) , 55.69% for (percentage of Oxlate)

             %

For IRON

Since the percentage of is 10.60% then in a 100 mg of the sample the amount of would be 10.60 mg

  Now the no of moles is mathematically denoted as

The molar mass of is  55.485 g/mol

     So the number of moles of in 100mg of he sample is

mol / mol Fe (3 sig figs) is

FOR POTASSIUM

Since the percentage of is 22.70% then in a 100mg of the sample the amount of would be 22.70mg

The molar mass of is  39.1 g/mol

   So the number of moles of in 100mg of he sample is

mol  of K / mol of Fe (3 sig figs) is

FOR OXILATE

Since the percentage of  is 55.69% then in a 100mg of the sample the amount of would be 55.69 mg

The molar mass of is  88.02 g/mol

  So the number of moles of   in 100mg of he sample is

mol  of / mol of Fe (3 sig figs) is  

FOR WATER OF HYDRATION

        Since the percentage of water  is 11.01% then in a 100mg of the sample the amount of water would be 11.0 mg

   The molar mass of water  is  18.0 g/mol

  So the number of moles of water   in 100mg of he sample is

mol  of  water / mol of Fe (3 sig figs) is  

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong

c. 4.54 x 10–12 j happens to be correct