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Nutka1998 [239]
3 years ago
9

A gas has a solubility of 2.45 g/L at a pressure of 0.750 atm. What pressure wold be required to produce an aqueous solution con

taining 6.25 g/L of this gas at constant temperature
Chemistry
1 answer:
Lelechka [254]3 years ago
3 0

Answer:

1.91 atm

Explanation:

Step 1: Calculate Henry's constant (k)

A gas has a solubility (C) of 2.45 g/L at a pressure (P) of 0.750 atm. These two variables are related to each other through Henry's law.

C = k × P

K = C/P

K = (2.45 g/L)/0.750 atm = 3.27 g/L.atm

Step 2: Calculate the pressure required to produce an aqueous solution containing 6.25 g/L of this gas at constant temperature.

We have C = 6.25 g/L and k = 3.27 g/L.atm. The required pressure is:

C = k × P

P = C/k

P = (6.25 g/L)/(3.27 g/L.atm) = 1.91 atm

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An copper mass is heated and placed in a foam cup calorimeter containing 45.0 mL of water at 24.0 degrees celcius. The water rea
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Answer:

= 9,593.1 Joules

Explanation:

Heat absorbed by water is equivalent to heat released by copper.

Heat absorbed is given by;

Q = mcΔT

where m is the mass, c is the specific capacity and ΔT is the change in temperature.

Therefore;

Since dnsity of water is 1 g/mL, and specific heat capacity is 4.18 J/g°C while the change in temperature is (75-24) = 51°C.

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4 years ago
The following anions can be separated by precipitation as silver salts: Cl- , Br- , I- , CrO4 2-. If Ag is added to a solution c
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Answer:

AgI, AgBr, AgCl and Ag₂CrO₄

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Ksp (product solubility constant) is defined as the equilibrium constant of the general reaction:

XₐYₙ(s) → aXⁿ⁺(aq) + nYᵃ⁻(aq)

<em>Where X is cation and Y is anion.</em>

Ksp = [aXⁿ⁺]ᵃ [nYᵃ⁻]ⁿ

The presence of XₐYₙ(s) produce ax moles of aXⁿ⁺ and nx moles of Yᵃ⁻. <em>Where X is the solubility of the compound.</em>

Replacing in Ksp:

Ksp = [ax]ᵃ [nx]ⁿ

Solving for x, Solubility (S) is defined as:

S = \sqrt[n+a]{\frac{Ksp}{a^{a} n^n} }

For AgCl, Ag₂CrO₄, AgBr and AgI solubilities are:

S = \sqrt[2]{\frac{1.8x10^{-10}}{1} } = 1.34x10⁻⁵M

S = \sqrt[3]{\frac{1.1x10^{-12}}{4} } = 6.50x10⁻⁵M

S = \sqrt[2]{\frac{5.4x10^{-13}}{1} } = 7.35x10⁻⁷M

S = \sqrt[2]{\frac{8.5^{-17}}{1} } = 9.22x10⁻⁹M

The lower solubility is the first compound in precipitate, thus, order of precipitation is:

<em>AgI, AgBr, AgCl and Ag₂CrO₄</em>

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