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LuckyWell [14K]
3 years ago
15

Calculate the pressure, in atmospheres, required to compress a sample of helium gas from 27.3 L (at 1.00 atm) to 3.36 L at const

ant temperature. Enter your answer in the box provided. atm
Chemistry
1 answer:
inn [45]3 years ago
4 0

Answer:

8.125 atm

Explanation:

Hello,

Boyle's law states that at constant temperature:

P_1V_1=P_2V_2

In this case:

P_1=1 atm, V_1=27.3L, V_2=3.36L

Solving for P_2:

P_2=\frac{P_1V_1}{V_2}=\frac{1atm*27.3L}{3.36L}\\P_2=8.125atm

Best regards!

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3 years ago
Calculate the molar solubility of ag2so4 in each solution below. the ksp of silver sulfate is 1.5 × 10−5. (a) 0.36 m agno3: (b)
Lana71 [14]

Answer:

a) 1.157 x 10⁻⁴ M.

b) 0.0032 M.

Explanation:

  • Ag₂SO₄ is sparingly soluble salt in water which is dissociate according to:

<em>Ag₂SO₄ ⇄ 2Ag⁺ + SO₄²⁻.</em>

Ksp = [Ag⁺][SO₄²⁻] = (2s)²(s) = 1.5 x 10⁻⁵.

4s³ = 1.5 x 10⁻⁵.

s³ = 1.5 x 10⁻⁵/4 = 3.75 x 10⁻⁶.

∴ s = ∛(3.75 x 10⁻⁶) = 0.0155 M.

<em>a) 0.36 M AgNO₃:</em>

  • The dissociation of Ag₂SO₄: Ag₂SO₄ ⇄ 2Ag⁺ + SO₄²⁻, will be accompanied by the ionization of AgNO₃:

<em>AgNO₃ → Ag⁺ + NO₃⁻.</em>

  • So, the concentration of Ag⁺ will be that of AgNO₃  and Ag₂SO₄:

[Ag⁺] = (2s + 0.36)², s is neglected with respect to 0.36 M the concentration resulted from AgNO₃.

So, [Ag⁺] = (0.36)².

∴ Ksp = [Ag⁺][SO₄²⁻] = (0.36)²(s) = 1.5 x 10⁻⁵.

<em>∴ s </em>= 1.5 x 10⁻⁵/(0.36)² =<em> 1.157 x 10⁻⁴ M.</em>

<em>b) 0.36 M Na₂SO₄:</em>

  • The dissociation of Ag₂SO₄: Ag₂SO₄ ⇄ 2Ag⁺ + SO₄²⁻, will be accompanied by the ionization of Na₂SO₄:

Na₂SO₄ → 2Na⁺ + SO₄²⁻.

  • So, the concentration of SO₄²⁻ will be that of Na₂SO₄  and Ag₂SO₄:

[SO₄²⁻] = (s + 0.36), s is neglected with respect to 0.36 M the concentration resulted from Na₂SO₄.

<em>So, [SO₄²⁻] = (0.36).</em>

∴ Ksp = [Ag⁺][SO₄²⁻] = (2s)²(0.36) = 1.5 x 10⁻⁵.

∴ 4s² = 1.5 x 10⁻⁵/(0.36) = 4.166 x 10⁻⁵.

∴ s² = 4.166 x 10⁻⁵/4 = 1.042 x 10⁻⁵.

<em>∴ s</em> = √(1.042 x 10⁻⁵) = <em>0.0032 M.</em>

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