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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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kondaur [170]

Answer: The most likely partial pressures are 98.7MPa for NO₂ and 101.3MPa for N₂O₄

Explanation: To determine the partial pressures of each gas after the increase of pressure, it can be used the equilibrium constant Kp.

For the reaction 2NO₂ ⇄ N₂O₄, the equilibrium constant is:

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

where:

P(N₂O₄) and P(NO₂) are the partial pressure of each gas.

Calculating constant:

Kp = \frac{38.8}{61.2^{2} }

Kp = 0.0104

After the weights, the total pressure increase to 200 MPa. However, at equilibrium, the constant is the same.

P(N₂O₄) + P(NO₂) = 200

P(N₂O₄) = 200 - P(NO₂)

Kp = \frac{P(N_{2}O_{4} )}{P(NO_{2} ^{2}) }

0.0104 = \frac{200 - P(NO_{2})  }{[P(NO_{2} )]^{2}}

0.0104[P(NO_{2} )]^{2} + P(NO_{2} ) - 200 = 0

Resolving the second degree equation:

P(NO_{2} ) = \frac{-1+\sqrt{9.32} }{0.0208}

P(NO_{2} ) = 98.7

Find partial pressure of N₂O₄:

P(N₂O₄) = 200 - P(NO₂)

P(N₂O₄) = 200 - 98.7

P(N₂O₄) = 101.3

The partial pressures are P(NO_{2} ) = 98.7 MPa and P(N₂O₄) = 101.3 MPa

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3 years ago
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6 0
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gogolik [260]
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initial    0.2 M            0           0
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4.9x10^-10 = X^2 / (0.2-X) by solving this equation for X 
∴X ≈ 10^-6
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and PH = -㏒[H+]
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3 0
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