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3 years ago

Consider the reversible reaction.

Chemistry

1 answer:

Dmitry [639]3 years ago

3 0

Answer:

Option D is correct.

The concentrations of both PCl₅ and PCl₃ are changing at equilibrium

Explanation:

Chemical equilibrium during a reversible chemical reaction, is characterised by an equal rate of forward reaction and backward reaction. It is better described as dynamic equilibrium.

This is because, the concentration of the elements and compounds involved in the reversible chemical reaction at equilibrium changes, but the rate of change of the reactants is always equal to the rate of change of products.

Hence, the concentration of reactants and products, such as PCl₅ and PCl₃ are allowed to change at equilibrium, but alas, the rate of forward reaction must always match the rate of backward reaction for the process to remain in a state of Chemical equilibrium.

Hope this Helps!!!

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Explain why you can boil water in a pot without the pot also boiling

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Briefly describe how the apparatus work

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3 years ago

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If 7.4 moles of carbon dioxide is produced in this reaction, how many moles of oxygen gas would be needed?

Nimfa-mama [501]

Answer:

11.6 mol O₂

Explanation:

C₇H₁₆ + 11 O₂ → 7 CO₂ + 8 H₂O

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4 0

3 years ago

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

3 years ago

You drop an unknown substance that weighs 8.3g into a graduated cylinder with 6ml of water. The water rises to 8ml when you drop

valkas [14]

Answer:

<h3>The answer is 4.15 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

$density = \frac{mass}{volume} \\$

From the question

mass of object = 8.3 g

volume = final volume of water - initial volume of water

volume = 8 - 6 = 2 mL

So we have

$density = \frac{8.3}{2} \\$

We have the final answer as

Hope this helps you

7 0

3 years ago