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 correct answer is c. Temperature is the average kinetic energy of a sample so if two samples have the same temperature they will also have the same average kinetic energy. I hope this helps. Let me know if anything is unclear.
Answer:
Pray to ur God to give you some peace
Explanation:
and then listen to some peace full music
I only know that
Answer:
Mass = 785.9 g
Explanation:
Given data:
Atoms of gold = 2.4 × 10²⁴ atoms
Mass of gold = ?
Solution:
First of all we will convert the number of atoms into moles.
2.4 × 10²⁴ atoms × 1 mol/ 6.02 × 10²³ atoms
number of moles = 3.99 mol
Now we will determine the mass of gold.
Mass = number of moles × molar mass
Mass = 3.99 mol × 196.97 g/mol
Mass = 785.9 g
Answer:
there are 3 significant figures
Explanation:
do not count the first 2 zeros. only the nimbers after the zero
