Please help me this is due by midnight!!

Determine the value of the equilibrium constant, Kgoal, for the reaction CO2(g)⇌C(s)+O2(g), Kgoal=? by making use of the followi

marta [7]

Answer : The value of $K_{goal}$ for the final reaction is, $1.238\times 10^{-7}$

Explanation :

The following equilibrium reactions are :

(1) $2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2$ $K_1=5.40\times 10^{-16}$

(2) $2H_2(g)+O_2(g)\rightleftharpoons 2H_2O(l)$ $K_2=1.06\times 10^{10}$

(3) $CH_3COOH(l)\rightleftharpoons 2C(s)+O_2(g)$ $K_3=2.68\times 10^{-9}$

The final equilibrium reaction is :

$CO_2(g)\rightleftharpoons C(s)+O_2(g)$ $K_{goal}=?$

Now we have to calculate the value of $K_{goal}$ for the final reaction.

First half the equation 1, 2 and 3 that means we are taking square root of equilibrium constant and then add all the equation 1, 2 and 3 that means we are multiplying all the equilibrium constant, we get the final equilibrium reaction and the expression of final equilibrium constant is:

$K_{goal}=\sqrt{K_1\times K_2\times K_3}$

Now put all the given values in this expression, we get :

$K_{goal}=\sqrt{(5.40\times 10^{-16})\times (1.06\times 10^{10})\times (2.68\times 10^{-9})}$

$K_{goal}=1.238\times 10^{-7}$

Therefore, the value of $K_{goal}$ for the final reaction is, $1.238\times 10^{-7}$

3 0

3 years ago

20 N force moving a box to 30 meters distance. How much work id done? if it takes 10 sec to do this work, how much power is used

dalvyx [7]

Answer:

Work done = 600 J

Power used = 60 W

Explanation:

Given:

Force acting on the box is, $F=20\ N$

Displacement of the box is, $S=30\ m$

Time taken for the work, $t=10\ s$

Now, we know that, work is said to be done by a force only when there is displacement caused by the force in its direction.

Here, the force acting on the box causes a displacement of 30 m in its direction. So, work done is equal to the product of force and displacement caused.

Therefore, work done on the box is given as:

$Work=Force\times Displacement\\Work=F\times S\\Work=(20\ N)\times (30\ m)\\Work=600\ J$

Therefore, the work done is 600 J.

Now, we know that, power is given as work done per unit time.

So, power used is given as:

$Power=\frac{Work}{Time}\\\\Power=\frac{600\ J}{10\ s}\\\\Power=60\ W$

Therefore, the power used is 60 W.

5 0

3 years ago

In IR spectroscopy, we normally talk about "frequencies" when in reality we are referring to wavenumbers. What is the mathematic

Svetach [21]

Answer:

Here's what I get.

Explanation:

(b) Wavenumber and wavelength

The wavenumber is the distance over which a cycle repeats, that is, it is the number of waves in a unit distance.

$\bar \nu = \dfrac{1}{\lambda}$

Thus, if λ = 3 µm,

$\bar \nu = \dfrac{1}{3 \times 10^{-6} \text{ m}}= 3.3 \times 10^{5}\text{ m}^{-1} = \textbf{3300 cm}^{-1}$

(a) Wavenumber and frequency

Since

λ = c/f and 1/λ = f/c

the relation between wavenumber and frequency is

$\bar \nu = \mathbf{\dfrac{f}{c}}$

Thus, if f = 90 THz

$\bar \nu = \dfrac{90 \times 10^{12} \text{ s}^{-1}}{3 \times 10^{8} \text{ m$\cdot$ s}^{-1}}= 3 \times 10^{5} \text{ m}^{-1} = \textbf{3000 cm}^{-1}$

(c) Units

(i) Frequency

The units are s⁻¹ or Hz.

(ii) Wavelength

The SI base unit is metres, but infrared wavelengths are usually measured in micrometres (roughly 2.5 µm to 20 µm).

(iii) Wavenumber

The SI base unit is m⁻¹, but infrared wavenumbers are usually measured in cm⁻¹ (roughly 4000 cm⁻¹ to 500 cm⁻¹).

8 0

3 years ago

How many moles are in 7.6 g of Ca

klasskru [66]

Your answer would be 0.024951344877489 but rounding it would be 0.025 moles

7 0

3 years ago

Electrons involved in bonding between atoms are __________________

Vitek1552 [10]

The name given to these electrons are that they are valence electrons or binding electrons as these are directly involved in chemical Bonding and allow for different compounds to be made.

5 0

3 years ago