Answer: The concentrations of 
at equilibrium is 0.023 M
Explanation:
Moles of = 
Volume of solution = 1 L
Initial concentration of = 
The given balanced equilibrium reaction is,
Initial conc. 0.14 M 0 M 0M
At eqm. conc. (0.14-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO%5D%5Ctimes%20%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Now put all the given values in this expression, we get :
By solving the term 'x', we get :
x = 0.023 M
Thus, the concentrations of at equilibrium is 0.023 M
Answer:
Explanation:
turn over number = R max / [E]t = K2
From given , R max = 249 * 10 ^ -6 mol. L^-1
T [E]t = 2.23 n mol. L^-1
= 2.23 * 10^-9 mol. L^-1
Putting values in above equation,
= 111.65 * 10^3 S^-1
Turn over number is maximum no of substrate molecule that can be converted into product molecules for unit time by enzyme molecule.
Answer:
-3.28 × 10⁴ J
Explanation:
Step 1: Given data
- Pressure exerted (P): 27.0 atm
- Initial volume (Vi): 88.0 L
- Final volume (Vf): 100.0 L
Step 2: Calculate the work (w) done by the gaseous mixture
We will use the following expression.
w = -P × ΔV = -P × (Vf - Vi)
w = -27.0 atm × (100.0 L - 88.0 L)
w = -324 atm.L
Step 3: Convert w to Joule (SI unit)
We will use the conversion factor 1 atm.L = 101.325 J.
-324 atm.L × 101.325 J/1 atm.L = -3.28 × 10⁴ J
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