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tia_tia [17]
2 years ago
11

The vapor pressure of liquid iodomethane, CH3I, is 100. mm Hg at 266 K. A 0.453 g sample of liquid CH3I is placed in a closed, e

vacuated 370. mL container at a temperature of 266 K.a.Assuming that the temperature remains constant, will all of the liquid evaporate? _____(yes/no)b.What will the pressure in the container be when equilibrium is reached? ________ mm Hg
Chemistry
1 answer:
Elanso [62]2 years ago
3 0

Answer:

a. Yes

b. 143.5 mmHg

Explanation:

The vapor pressure is the pressure of the vapor that is in equilibrium with the liquid. At a constant temperature, some molecules of the liquid will vaporize, and then will do pressure at the surface of the liquid.

If the pressure at the container is higher then the vapor pressure, the liquid will evaporate.

a. Let's calculate the pressure at the container by the ideal gas law:

PV = nRT

Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (62,364 mmHg.mL/mol.K), and T is the temperature.

The molar mass of CH₃I is 142 g/mol

n = mass/ molar mass

n = 0.453/142

n = 0.0032 mol

P*370 = 0.0032*62,364*266

370P = 53,084.24

P = 143.5 mmHg

So, all the liquid will evaporate.

b. Because all liquid evaporates, when the equilibrium is reached, the pressure is the gas pressure: 143.5 mmHg.

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Crank

Answer:

1.8 × 10⁻¹⁶ mol  

Explanation:

(a) Calculate the solubility of the Sr₃(PO₄)₂

Let s = the solubility of Sr₃(PO₄)₂.

The equation for the equilibrium is

Sr₃(PO₄)₂(s) ⇌ 3Sr²⁺(aq) + 2PO₄³⁻(aq); Ksp = 1.0 × 10⁻³¹

                         1.2 + 3s          2s

K_{sp} =\text{[Sr$^{2+}$]$^{3}$[PO$_{4}^{3-}$]$^{2}$} = (1.2 + 3s)^{3}\times (2s)^{2} =  1.0 \times 10^{-31}\\\text{Assume } 3s \ll 1.2\\1.2^{3} \times 4s^{2} = 1.0 \times 10^{-31}\\6.91s^{2} = 1.0 \times 10^{-31}\\s^{2} = \dfrac{1.0 \times 10^{-31}}{6.91} = 1.45 \times 10^{-32}\\\\s = \sqrt{ 1.45 \times 10^{-32}} = 1.20 \times 10^{-16} \text{ mol/L}\\

(b) Concentration of PO₄³⁻

[PO₄³⁻] = 2s = 2 × 1.20× 10⁻¹⁶ mol·L⁻¹ = 2.41× 10⁻¹⁶ mol·L⁻¹

(c) Moles of PO₄³⁻

Moles = 0.750 L × 2.41 × 10⁻¹⁶ mol·L⁻¹ = 1.8 × 10⁻¹⁶ mol

7 0
2 years ago
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Quantity of Medrol = 20 mg/mL 
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The graph shows the amount of a gaseous product formed over time during two trials of a reaction. A different concentration of a
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Answer:

Trial 2, because the amount of product formed per unit time is higher.

Explanation:

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8 0
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<h2>Answer:</h2>

The correct answer is option C which is, "Electrons in the orbit closest to the nucleus have the least amount of energy".

<h3>Explanation:</h3>
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