Answer: Equilibrium constant is 0.70.
Explanation:
Initial moles of 
= 0.35 mole
Volume of container = 1 L
Initial concentration of 
Initial moles of 
= 0.40 mole
Volume of container = 1 L
Initial concentration of 
equilibrium concentration of 
[/tex]
The given balanced equilibrium reaction is,
Initial conc. 0.35 M 0.40M 0 0
At eqm. conc. (0.35-x) M (0.40-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CO_2]\times [H_2O]}{[CO]\times [H_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5Ctimes%20%5BH_2O%5D%7D%7B%5BCO%5D%5Ctimes%20%5BH_2O%5D%7D)
we are given : (0.35-x)= 0.18
x = 0.17
Now put all the given values in this expression, we get :
Thus the value of the equilibrium constant is 0.70.
If you divide 8.1 by 1.8 that leads to 4.5
Answer: <u><em>Option B; It traps light energy and converts it into chemical energy.</em></u>
<h2>
Explanation: This substance is chlorophyll. It is a pigment present in leaves of all plants. It absorbs light energy and provides it to carry out the process of photosynthesis. Light energy is converted into chemical energy, in form of NADPH and ATP, which can be used by plants for photosynthesis.</h2><h2>
</h2><h2>
This pigment is present only in plants, so option A is incorrect.</h2><h2>
</h2><h2>
This pigment only absorbs and transfers energy to other molecules, and is not associated with carbon dioxide directly, so option C and D are also incorrect.</h2>
Answer: The correct answer is -297 kJ.
Explanation:
To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:
2SO3 —> O2 + 2SO2 (196 kJ)
Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.
Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:
SO3 —>1/2O2 + SO2 (98 kJ)
S + 3/2O2 —> SO3 (-395 kJ)
Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.
S + O2 —> SO2
Now, we must add the enthalpies together to get our final answer.
-395 kJ + 98 kJ = -297 kJ
Hope this helps!
