Astandard solution of fescn2+ is prepared by combining 9.0 ml of 0.20 m fe(no3)3 with 1.0 ml of 0.0020 m kscn . the standard solution had an absorbance of 0.480 . fe3+(aq)+scn−(aq)↽−−⇀fescn2+(aq) a trial solution was made in a similar manner, but with a more dilute fe(no3)3 reagent. the initial scn− concentration, immediately after mixing, was 0.00050 m . this trial solution had absorbance of 0.220 . what is the equilibrium concentration of scn− in the trial solution?

IN Δ MLN:

∠M = 18.3 , ∠L = 98.6 AND ∠N = 180 - (∠M + ∠L) = 180 - (18.3 + 98.6 ) = 63.1

IN Δ FGH:

∠F = 98.6 , ∠G = 61.1 AND ∠H = 180 - (∠F + ∠G ) = 180 - (98.6 + 61.1 ) = 20.3

∴ ONLY ∠N = ∠F = 98.6

There is no other congruent angles

So, The correct statement is :

Inaccurate. The triangles are not similar because angle M is not congruent to angle H, and angle N is not congruent to angle G.

The equilibrium concentration of SCN- in the trial solution is 0.0007059 M

Explanation:

Step 1: The equation

Fe(NO3)3 + KSCN ==> FeSCN^2+ + 3NO3^- +K+

Step 2: First we have to calculate the initial moles of Fe3+ and SCN-

Moles of Fe3+ = Concentration of Fe3+ * Volume of the solution

Moles of Fe3+ = 0.2M * 9*10-3 L = 0.0018 moles

Moles of SCN- = Concentration of SCN- * volume of the solution

Moles of SCN- = 0.0020 M* 1*10^-3 = 0.000002 moles

When 1 mole of Fe3+ is consumed, there is needed 1 mole of SCN- to produce 1 mole of FeSCN2+

SCN- os the limiting reagens, this means it will completely react.

Since there is 1 mole of SCN- needed to produce 1 mole of FeSCN2+

So if there is consumed 0.000002 moles of SCN-, there is also produced 0.000002 mole of FeSCN2+

Step 3: Calculate the concentration of FeSCN2+

Concentration of FeSCN2+ = 0.000002 mole of FeSCN2+/ (9+1 *10^-3L) = 0.0002 M

Step 4: using Lambert-beer's law:

A = c*C*l

with A= the absorbance of the solution

with C = concentration of the solution

with l= the path length

with c = molar absorptivity coefficient

c and l are same for stock solution and dilute solution.

c*l = A/C = 0.510/0.0002M = 2550M^-1

For the trial solution:

The equilibrium concentration of SCN^- is:

[SCN-]eq = [SCN-]initial - [FeSCN2+]

Step 5: Calculate concentration of FeSCN2+

C= A/cl

C = 0.240/2550 = 9.41*10 ^-5 M

Step 6: Calculate concentration of SCN-

[SCN-]eq = [SCN-]initial - [FeSCN2+]

[SCN-]eq = 0.00080 M - 9.41*10 ^-5 M

[SCN-]eq = 0.0007059 M

The equilibrium concentration of SCN- in the trial solution is 0.0007059 M

It is "D" just took the quiz.

For instance, sin(x) = 1/csc(x) is an identity. To "prove" an identity, you have to use logical steps to show that one side of the equation can be transformed into the other side of the equation. You do not plug values into the identity to "prove" anything.

Explanation:

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Answer : The equilibrium concentration of [tex]SCN^-[/tex] in the trial solution is [tex]4.58\times 10^{-8}M[/tex]

Explanation :

First we have to calculate the initial moles of [tex]Fe^{3+}[/tex] and [tex]SCN^-[/tex].

[tex]\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}[/tex]

[tex]\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol[/tex]

and,

[tex]\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}[/tex]

[tex]\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol[/tex]

The given balanced chemical reaction is,

[tex]Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)[/tex]

Since 1 mole of [tex]Fe^{3+}[/tex] reacts with 1 mole of [tex]SCN^-[/tex] to give 1 mole of [tex]FeSCN^{2+}[/tex]

The limiting reagent is, [tex]SCN^-[/tex]

So, the number of moles of [tex]FeSCN^{2+}[/tex] = 0.0020 mmole

Now we have to calculate the concentration of [tex]FeSCN^{2+}[/tex].

[tex]\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M[/tex]

Using Beer-Lambert's law :

[tex]A=\epsilon \times C\times l[/tex]

where,

A = absorbance of solution

C = concentration of solution

l = path length

[tex]\epsilon[/tex] = molar absorptivity coefficient

[tex]\epsilon[/tex] and l are same for stock solution and dilute solution. So,

[tex]\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}[/tex]

For trial solution:

The equilibrium concentration of [tex]SCN^-[/tex] is,

[tex][SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}][/tex]

[tex][SCN^-]_{initial}[/tex] = 0.00050 M

Now calculate the [tex][FeSCN^{2+}][/tex].

[tex]C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M[/tex]

Now calculate the concentration of [tex]SCN^-[/tex].

[tex][SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}][/tex]

[tex][SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)[/tex]

[tex][SCN^-]_{eqm}=4.58\times 10^{-8}M[/tex]

Therefore, the equilibrium concentration of [tex]SCN^-[/tex] in the trial solution is [tex]4.58\times 10^{-8}M[/tex]