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Molodets [167]
3 years ago
9

Fe3+(aq) (yellow) + SCN-(aq) (colorless) FeSCN2+(aq) (blood-red) Chloride ions are colorless. Potassium ions are also colorless.

The above equilibrium can be created by mixing an Iron (III) chloride solution with a potassium thiocyanate solution. Based on this information and the colors in the equilibrium above answer the first two questions. 1. What color would an FeCl3 solution be? 2. What color would a KSCN solution be? 3. What color do you get when you mix FeCl3 and KSCN as shown in the video for the control test tube?
Chemistry
1 answer:
harina [27]3 years ago
5 0

Explanation:

Fe^{3+}\text{(aq)(yellow)}+SCN^-\text{(aq)(colorless)}\rightleftharpoons FeSCN^{2+}\text{(aq)(blood-red)}

K^+\text{(aq)(colorless)}+Cl^-\text{(aq)(colorless)}\rightarrow KCl

The above two reactions xcan also be written in form of single chemcial equation:

FeCl_3+K(SCN)\rightleftharpoons KCl+[Fe(SCN)]Cl_2(blood-red)

1. Color of ferric chloride solution is yellow. This is due to presence of ferric ions which have yellow color in their aqueous solutions.

2. KSCN has the colorless solution. This due to potassium ion forms colorless aqueous solution.

3. On mixing, KSCN with FeCl_3 we will get blood red color solution of [FeSCN]Cl_2.

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<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

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rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt

integrating factor

=e^{\int\limits (1/40) \, dt}

integrating factor =e^{(1/40)t}

multiply on both sides by  =e^{(1/40)t}

dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t

integrate on both sides

\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})

b)

after long period of time means t - > ∞

{t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>
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