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lawyer [7]
2 years ago
13

When a mixture containing cations of Analytical Groups I–III is treated with H2S in acidic solution, which cations are expected

to precipitate? Analytical Groups I and II Analytical Group I only Analytical Group III only Analytical Groups II and III Analytical Group II only
Chemistry
2 answers:
Fantom [35]2 years ago
8 0

Answer:

Analytical Groups I and II

Explanation:

Precipitation reactions happen when anions and cations in aqueous solution mix together to form an insoluble ionic solid which is now referred to as a precipitate.

Whether or not a type of reaction like that takes place can be determined by utilizing the solubility principles for common ionic solids.

Tema [17]2 years ago
8 0

Answer:

Option D is the answer :

Analytical Groups II and III

Explanation:

Group II (Cu2+, Bi3+, Cd2+, Hg2+, As3+, Sb3+, Sn4+) cations produce very insoluble sulfides

(Ksp values less than 10-30) so they can be precipitated by low amounts of sulfide ion; this can be

achieved by adding an acidic solution of H2S.

Group III (Al3+, Cr3+, Fe3+, Zn2+, Ni2+, Co2+, Mn2+) cations produce slightly soluble sulfides

(Ksp values more than 10-20) so they can be precipitated by relatively high amounts of sulfide ion;

this can be achieved by adding a basic solution of H2S

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Ne4ueva [31]

Answer:

B. 111 J

Explanation:

The change in internal energy is the sum of the heat absorbed and the work done on the system:

ΔU = Q + W

At constant pressure, work is:

W = P ΔV

Given:

P = 0.5 atm = 50662.5 Pa

ΔV = 4 L − 2L = 2 L = 0.002 m³

Plugging in:

W = (50662.5 Pa) (0.002 m³)

W = 101.325 J

Therefore:

ΔU = 10 J + 101.325 J

ΔU = 111.325 J

Rounded to three significant figures, the change in internal energy is 111 J.

7 0
3 years ago
A container has a mixture of NO2 gas and N2O4 gas in equilibrium. The chemical reaction between the two gases is described by th
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}

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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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Answer:

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Answer:

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