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3 years ago

Kc is 1.67 x 10^20 at 25 °C for the formation of iron(III) oxalate complex ion:

Chemistry

1 answer:

Liula [17]3 years ago

7 0

Answer:

$n_{C_2O_4^{2-}}=1.59mol C_2O_4^{2-}$

Explanation:

Hello there!

In this case, according to the given information and chemical equation, it turns out possible for us to calculate the moles of C2O4^2- by firstly setting up the equilibrium expression:

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

However, according to the question, we just need to apply the given 1:3 mole ratio in the chemical reaction, of iron (III) ions to oxalate ions to obtain:

$n_{C_2O_4^{2-}}=0.53molFe^{3+}*\frac{3molC_2O_4^{2-}}{1molFe^{3+}}=1.59mol C_2O_4^{2-}$

Regards!

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Answer:

neutron, proton, and electrons

Explanation:

The protons and the neutrons make up the center of an atom which is called the nucleus

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Explanation:

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4 years ago

An aqueous solution 10 g of an optically pure substance diluted to 500ml with water and placed in a polarimeter tube 20 cm long.

Gennadij [26K]

Explanation:

Formula to calculate specific rotation is as follows.

Specific rotation ( $[\alpha]$ ) = $\frac{\alpha}{c} \times l$

where, $\alpha$ = observed rotation

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It is given that,

$\alpha = -6.16^{o}$

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l = 20 cm = 2 dm (as 1 dm = 10 cm)

Therefore, calculate the specific rotation as follows.

Specific rotation ( $[\alpha]$ ) = $\frac{\alpha}{c} \times l$

= $\frac{-6.16^{o}}{0.02 g/ml} \times 2 dm$

= $-616^{o}$

Thus, we can conclude that the specific rotation of this compound is $-616^{o}$ .

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3 years ago

In the laboratory a student combines 40.6 mL of a 0.113 M copper(II) sulfate solution with 26.4 mL of a 0.329 M copper(II) iodid

Arisa [49]

Answer : The final concentration of copper(II) ion is, 0.198 M

Explanation :

First we have to calculate the moles of $CuSO_4$ and $CuI_2$ .

$\text{Moles of }CuSO_4=\text{Concentration of }CuSO_4\times \text{Volume of solution}$

$\text{Moles of }CuSO_4=0.113mol/L\times 0.0406L=0.00459mol$

Moles of $CuSO_4$ = Moles of $Cu^{2+}$ = 0.00459 mol

and,

$\text{Moles of }CuI_2=\text{Concentration of }CuI_2\times \text{Volume of solution}$

$\text{Moles of }CuI_2=0.329mol/L\times 0.0264L=0.00869mol$

Moles of $CuI_2$ = Moles of $Cu^{2+}$ = 0.00869 mol

Now we have to calculate the total moles of copper(II) ion and total volume of solution.

Total moles copper(II) ion = 0.00459 mol + 0.00869 mol

Total moles copper(II) ion = 0.0133 mol

and,

Total volume of solution = 40.6 mL + 26.4 mL = 64.0 mL = 0.067 L (1 L = 1000 mL)

Now we have to calculate the final concentration of copper(II) ion.

$\text{Final concentration of copper(II) ion}=\frac{\text{Total moles}}{\text{Total volume}}$

$\text{Final concentration of copper(II) ion}=\frac{0.0133mol}{0.067L}=0.198mol/L=0.198M$

Thus, the final concentration of copper(II) ion is, 0.198 M

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3 years ago

Which type of hybridization leads to a trigonal planar electron domain geometry? s sp sp2 sp3

kolezko [41]

Electron domain geometry takes into account the bonds and the free electrons around the atom to predict the geometry.

A tigonal planar geometry means that all the electron domains are in the plane and that they are as far away as they can. That means that there are three electron domains in the plane separated at 120 ° each other.

sp2 means that there are three hybrid electron domains, and the geometrh that leads to the minimal repulsion force is they placed at 120°, which exactly the trigonal planar geometry described up.

Then the answer is sp2

You can check in your book, that sp corresponds to two domains, located linearly, at 180°; and sp3 is four domains with the form of tetrahedral with 109.5° angles.

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3 years ago