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
The SI unit for temperature is Kelvin.
Robert A. Millikan and Harvey Fletcher performed the oil drop experiment.
Answer:
[CaCl₂] = 1.32 M
Explanation:
We know the volume of solution → 0.30 L
We know the mass of solute → 44 g of CaCl₂
Let's convert the mass of solute to moles.
44 g . 1 mol / 110.98 g = 0.396 moles
Molarity (mol/L) → 0.396 mol / 0.3 L = 1.32 M
2 H₂O₂ --> 2 H₂ + 2 O₂
2 moles H₂O₂ ------> 2 moles O₂
8 moles H₂O₂ ------> ?
moles O₂ = 8 x 2 / 2
moles O₂ = 16 / 2
= 8 moles
Answer C
hope this helps!
