Chemistry, 04.04.2020 11:32 rico126090

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)

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Answer from: Giabear23

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The percentage of water of hydration is P_h= 11.01 %

Mass of $Fe^{3+}$ in 100mg is 10.60mg

Moles of $Fe^{3+}$ in 100mg is n_i= 0.19

mol / mol Fe (3 sig figs) is = 1.00

mol / mol Fe (whole number) is = 1

Mass of $K^{+}$ in 100mg is 27.70mg

Moles of $K^{+}$ in 100mg is n_i= 0.581 moles

mol of K / mol of Fe (3 sig figs) is = 3.05

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

Mass of $C_2O_4^{-2}$ in 100mg is 55.69 mg

Moles of $C_2O_4^{-2}$ in 100mg is n_i= 0.633 moles

mol of $C_2O_4^{-2}$ / mol of Fe (3 sig figs) is = 3.33

mol of $C_2O_4^{-2}$ / mol of Fe (whole number) is

Mass of water in 100mg is 11.01 mg

Moles of water in 100mg is n_i= 0.611 moles

mol of water / mol of Fe (3 sig figs) is = 3.21

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

Explanation:

The percentage of water of hydration is mathematically represented as

P_h = 100 - (Pi + P_p + P_o)

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

P_h =100 - (10.60 + 22.70+55.69)

P_h= 11.01 %

For IRON

Since the percentage of $Fe^{3+}$ is 10.60% then in a 100 mg of the sample the amount of $Fe^{3+}$ would be 10.60 mg

Now the no of moles is mathematically denoted as

$n = \frac{mass}{molar \ mass }$

The molar mass of is 55.485 g/mol

So the number of moles of $Fe^{3+}$ in 100mg of he sample is

$n_i = \frac{10.60}{55.485}$

n_i= 0.19

mol / mol Fe (3 sig figs) is $= \frac{0.19}{0.19} = 1.00$

FOR POTASSIUM

Since the percentage of $K^{+}$ is 22.70% then in a 100mg of the sample the amount of $K^{+}$ would be 22.70mg

The molar mass of is 39.1 g/mol

So the number of moles of $K^{+}$ in 100mg of he sample is

$n_i = \frac{22.70}{39.1}$

=0.581 moles

mol of K / mol of Fe (3 sig figs) is $= \frac{0.581}{0.19} = 3.05$

FOR OXILATE $C_2O_4^{-2}$

Since the percentage of $C_2O_4^{-2}$ is 55.69% then in a 100mg of the sample the amount of $C_2O_4^{-2}$ would be 55.69 mg

The molar mass of $C_2O_4^{-2}$ is 88.02 g/mol

So the number of moles of $C_2O_4^{-2}$ in 100mg of he sample is

$n_i = \frac{55.69}{88.02}$

=0.633 moles

mol of $C_2O_4^{-2}$ / mol of Fe (3 sig figs) is $= \frac{0.633}{0.19} = 3.33$

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

$n_i = \frac{11.01}{18.0}$

=0.611 moles

mol of water / mol of Fe (3 sig figs) is $= \frac{0.611}{0.19} = 3.21$

Now, let's finish the calculation and the determination of the formula of the iron compound: Calcula

Answer from: Quest

it is a good conductor

explanation:

Answer from: Quest

smooth muscle is composed of involuntary muscles found within the walls of organs and structures such as the esophagus, stomach, intestines, and blood vessels. those muscles push materials like food or blood through organs. unlike skeletal muscle, smooth muscle can never be under your control.