Answer:
d. 99.63% and 0.364%, respectively
Explanation:
The average mass of a nitrogen atom is 14.006 amu
14N-14.00307 amu with 0.9963 fractional abundance
1SN- 15.000108 amu with 0.00364 fractional abundance.
The relative abundance definition in chemistry is the percentage of a particular isotope that occurs in nature. The atomic mass listed for an element on the periodic table is an average mass of all known isotopes of that element.
In this problem, the fractional abundance has areaady been given. All we need to do now is convert them to percentage abundance.
14N-14.00307 = 0.9963* 100% = 99.63%
1SN- 15.000108 = 0.00364 * 100% = 0.364%
Answer:
A. It is the ratio of the concentrations of products to the concentrations of reactants.
Explanation:
The equilibrium constant of a chemical reaction is the ratio of the concentration of products to the concentration of reactants.
This equilibrium constant can be expressed in many different formats.
- For any system, the molar concentration of all the species on the right side are related to the molar concentrations of those on the left side by the equilibrium constant.
- The equilibrium constant is a constant at a given temperature and it is temperature dependent.
- The derivation of the equilibrium constant is based on the law of mass action.
- It states that "the rate of a chemical reaction is proportional to the product of the concentration of the reacting substances. "
Answer:
Chemical Properties
Explanation:
Chemical properties are properties that are observed during chemical reactions. Some examples of chemical properties are reactivity, flammability and chemical stability.
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