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3 years ago

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Consider an ideal gas enclosed in a 1.00 L container at an internal pressure of 18.0 atm. Calculate the work, w , if the gas exp

ands against a constant external pressure of 1.00 atm to a final volume of 18.0 L.

1 answer:

Dominik [7]3 years ago

6 0

Answer:

-32821.2 J (negative sign implies that work is done by the system)

Explanation:

The expression for the calculation of work done is shown below as:

$w=-P\times \Delta V$

Where, P is the pressure

$\Delta V$ is the change in volume

From the question,

$\Delta V$ = 18.0 - 1.00 L = 18.0 L

P = 18.0 atm

$w=-18.0\times18.0\ atmL$

Also, 1 atmL = 101.3 J

So,

$w=-18.0\times 18.0\times 101.3\ J=-32821.2\ J$ (negative sign implies that work is done by the system)

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Suppose 0.245 g of sodium chloride is dissolved in 50. mL of a 18.0 m M aqueous solution of silver nitrate.

Bezzdna [24]

Answer:

$\large \boxed{\text{ 0.066 mol/L}}$

Explanation:

We are given the amounts of two reactants, so this is a limiting reactant problem.

1. Assemble all the data in one place, with molar masses above the formulas and other information below them.

Mᵣ: 58.44

NaCl + AgNO₃ ⟶ NaNO₃ + AgCl

m/g: 0.245

V/mL: 50.

c/mmol·mL⁻¹: 0.0180

2. Calculate the moles of each reactant

$\text{Moles of NaCl} = \text{245 mg NaCl} \times \dfrac{\text{1 mmol NaCl}}{\text{58.44 mg NaCl}} = \text{4.192 mmol NaCl}\\\\\text{ Moles of AgNO}_{3}= \text{50. mL AgNO}_{3} \times \dfrac{\text{0.0180 mmol AgNO}_{3}}{\text{1 mL AgNO}_{3}} = \text{0.900 mmol AgNO}_{3}$

3. Identify the limiting reactant

Calculate the moles of AgCl we can obtain from each reactant.

From NaCl:

The molar ratio of NaCl to AgCl is 1:1.

$\text{Moles of AgCl} = \text{4.192 mmol NaCl} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol NaCl}} = \text{4.192 mmol AgCl}$

From AgNO₃:

The molar ratio of AgNO₃ to AgCl is 1:1.

$\text{Moles of AgCl} = \text{0.900 mmol AgNO}_{3} \times \dfrac{\text{1 mmol AgCl}}{\text{1 mmol AgNO}_{3}} = \text{0.900 mmol AgCl}$

AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.

4. Calculate the moles of excess reactant

Ag⁺(aq) + Cl⁻(aq) ⟶ AgCl(s)

I/mmol: 0.900 4.192 0

C/mmol: -0.900 -0.900 +0.900

E/mmol: 0 3.292 0.900

So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.

5. Calculate the concentration of Cl⁻

$\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}$

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3 years ago

The 1995 Nobel Prize in chemistry was shared by Paul Crutzen, F. Sherwood Rowland, and Mario Molina for their work concerning th

Nikolay [14]

Answer:

$\Delta H_{rxn3}=-162.5 kJ/mol$

Explanation:

The reaction we need to calculate:

$O_3 (g) + Cl (g) \longrightarrow ClO (g) + O_2 (g)$

1) $O_3 (g) + ClO (g) \longrightarrow Cl (g) +2 O_2 (g)$

$\Delta H_{rxn}=-122.8 kJ/mol$

We need the ClO in the products side, so we use the inverse of this reaction:

$Cl (g) +2 O_2 (g) \longrightarrow O_3 (g) + ClO (g)$

$\Delta H_{rxn1}=122.8 kJ/mol$

2) $2 O_3 (g) \longrightarrow 3 O_2 (g)$

$\Delta H_{rxn2}=-285.3 kJ/mol$

Now we need to combine this two:

$Cl (g) +2 O_2 (g) + 2 O_3 (g)\longrightarrow O_3 (g) + ClO (g) + 3 O_2 (g)$

$Cl (g) + O_3 (g)\longrightarrow ClO (g) + O_2 (g)$

The enthalpy of reaction:

$\Delta H_{rxn3}=\Delta H_{rxn1}+ \Delta H_{rxn2}=122.8 kJ/mol-285.3 kJ/mol$

$\Delta H_{rxn3}=-162.5 kJ/mol$

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