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0.0200 M Fe3+ is initially mixed with 1.00 M oxalate ion, C2O42-, and they react according to the equation: Fe3+(aq) + 3 C2O42-(

Studentka2010 [4]

Answer : The concentration of $Fe^{3+}$ at equilibrium is 0 M.

Solution : Given,

Concentration of $Fe^{3+}$ = 0.0200 M

Concentration of $C_2O_4^{2-}$ = 1.00 M

The given equilibrium reaction is,

$Fe^{3+}(aq)+3C_2O_4^{2-}(aq)\rightleftharpoons [Fe(C_2O_4)_3]^{3-}(aq)$

Initially conc. 0.02 1.00 0

At eqm. (0.02-x) (1.00-3x) x

The expression of $K_c$ will be,

$K_c=\frac{[[Fe(C_2O_4)_3]^{3-}]}{[C_2O_4^{2-}]^3[Fe^{3+}]}$

$1.67\times 10^{20}=\frac{(x)^2}{(1.00-3x)^3\times (0.02-x)}$

By solving the term, we get:

$x=0.02M$

Concentration of $Fe^{3+}$ at equilibrium = 0.02 - x = 0.02 - 0.02 = 0 M

Therefore, the concentration of $Fe^{3+}$ at equilibrium is 0 M.

4 0

3 years ago

7. Wind, moving water, falling rocks, and roller-skating children are all examples of -

cluponka [151]

Answer:

kinetic energy

the is because all of them are moving at once

5 0

3 years ago

Calculate the molecular weight of a dibasic acid.0.56gm of which is required 250ml of N/20 sodium hydroxide solution for neutral

Alekssandra [29.7K]

<u>Answer: </u>The molecular weight of the dibasic acid is 89.6 g/mol

<u>Explanation:</u>

Normality is defined as the amount of solute expressed in the number of gram equivalents present per liter of solution. The units of normality are eq/L. The formula used to calculate normality:

$\text{Normality}=\frac{\text{Given mass of solute}\times 1000}{\text{Equivalent mass of solute}\times \text{Volume of solution (mL)}}$ ....(1)

We are given:

Normality of solution = $\frac{1}{20}=0.05N$

Given mass of solute = 0.56 g

Volume of solution = 250 mL

Putting values in equation 1, we get:

$0.05=\frac{0.56\times 1000}{\text{Equivalent mass of solute}\times 250}\\\\\text{Equivalent mass of solute}=\frac{0.56\times 1000}{0.05\times 250}=44.8g/eq$