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3 years ago

10

Consider the following chemical equilibrium: Now write an equation below that shows how to calculate from for this reaction at a

n absolute temperature . You can assume is comfortably above room temperature. If you include any common physical constants in your equation be sure you use their standard symbols, found in the ALEKS Calculator.

1 answer:

borishaifa [10]3 years ago

6 0

Answer:

Kp=Kc *(RT)+-3

Explanation:

The relation between Kp and Kc is given below:

Where,

Kp is the pressure equilibrium constant

Kc is the molar equilibrium constant

R is gas constant , R = 0.082057 L atm.mol⁻¹K⁻¹

T is the temperature in Kelvins

Δn = (No. of moles of gaseous products)-(No. of moles of gaseous reactants)

For the first equilibrium reaction:

Δn = (0)-(2+1) = -3

Thus, Kp is:

Kp=Kc *(RT)+-3

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Identify the information that can be included in a chemical equation.

tester [92]

Mole ratios
Reactants and products
Type of reaction eg equilibrium
Enthalpy
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3 years ago

Any help with these questions would help i'm lost

SSSSS [86.1K]

Answer:

1. 31.25 mL

2. 1.98 g/L

3. 0.45 g/mL

Explanation:

For each of the problems, you need to perform unit conversions. You need to use the information given to you to convert to a specific unit.

1. You need volume (mL). You have density (g/mL) and mass (g). Divide mass by density. You will cancel out mL and be left with g.

(50.0 g)/(1.60 g/mL) = 31.25 mL

2. You are given grams and liters. You need to find density with units g/L. This means that you have to divide grams by liters.

(0.891 g)/(0.450 L) = 1.98 g/L

3. You have to find density again but this time with units g/mL. Divide the given mass by the volume.

(10.0 g)/(22.0 mL) = 0.45 g/mL

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3 years ago

What the atmosphere mean

Alex_Xolod [135]

The envelope of gases surrounding the earth or another planet.

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4 years ago

What is the molar mass of NH4?

astraxan [27]

Answer:

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3 years ago

In their notebook you see that it takes 9 hours for a sixth of a 0.5M solution of BC2 to react. Unfortunately, you have somewher

trasher [3.6K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The concentration of $BC_2$ that should used originally is $C_Z_o = 0.4492M$

Explanation:

From the question we are told that

The necessary elementary step is

$2BC_2 ----->4C + B_2$

The time taken for sixth of 0.5 M of reactant to react $t = 9 hr$

The time available is $t_a = 3.5 hr$

The desired concentration to remain $C = 0.42M$

Let Z be the reactant , Y be the first product and X the second product

Generally the elementary rate law is mathematically as

$-r_Z = kC_Z^2 = - \frac{d C_Z}{dt}$

Where k is the rate constant , $C_Z$ is the concentration of Z

From the elementary rate law we see that the reaction is second order (This because the concentration of the reactant is raised to power 2 )

For second order reaction

$\frac{1}{C_Z} - \frac{1}{C_Z_o} = kt$

Where $C_Z_o$ is the initial concentration of Z which a value of $C_Z_o = 0.5M$

From the question we are told that it take 9 hours for the concentration of the reactant to become

$C_Z = C_Z_o - \frac{1}{6} C_Z_o$

$C_Z = 0.5 - \frac{0.5}{6}$

$= 0.4167 M$

So

$\frac{1}{0.4167} - \frac{1}{0.50} = 9 k$

$0.400 = 9 k$

=> $k = 0.044\ L/ mol \cdot hr^{-1}$

For $C_Z = 0.42M$

$\frac{1}{0.42} - \frac{1}{C_Z_o} = 3.5 * 0.044$

$2.38 - 0.154 = \frac{1}{C_Z_o}$

$2.226 = \frac{1}{C_Z_o}$

$C_Z_o = \frac{1}{2.226}$

$C_Z_o = 0.4492M$

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3 years ago