Answer: The equilibrium constant for the given reaction is 0.0421.
Explanation:

Concentration of
= 0.0095 M
Concentration of
= 0.020 M
Concentration of
= 0.020 M
The expression of the equilibrium constant is given as:
![K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}=\frac{0.020 M\times 0.020 M}{0.0095 M}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D%3D%5Cfrac%7B0.020%20M%5Ctimes%200.020%20M%7D%7B0.0095%20M%7D)
(An equilibrium constant is an unit less constant)
The equilibrium constant for the given reaction is 0.0421.
Answer:
Potential energy
Explanation:
The thrown baseball is converting from kinetic energy into potential energy. When it finally stops at a particular height, it attains its maximum potential energy at the position or point.
- Potential energy is the energy at rest of body.
- Kinetic energy is the energy due to the motion of body.
The more a body speeds, the higher its kinetic energy attained.
As a body comes to rest, at a height, it attains potential energy.
The body during flight decreases in kinetic energy but increases its potential energy due to gravity pulling it to rest.
M=11.20 g
m(H₂)=0.6854 g
M(H₂)=2.016 g/mol
M(Mg)=24.305 g/mol
M(Zn)=65.39 g/mol
w-?
m(Mg)=wm
m(Zn)=(1-w)m
Zn + 2HCl = ZnCl₂ + H₂
m₁(H₂)=M(H₂)m(Zn)/M(Zn)=M(H₂)(1-w)m/M(Zn)
Mg + 2HCl = MgCl₂ + H₂
m₂(H₂)=M(H₂)m(Mg)/M(Mg)=M(H₂)wm/M(Mg)
m(H₂)=m₁(H₂)+m₂(H₂)
m(H₂)=M(H₂)(1-w)m/M(Zn)+M(H₂)wm/M(Mg)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}
m(H₂)=M(H₂)m{(1-w)/M(Zn)+w/M(Mg)}
(1-w)/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}
1/M(Zn)-w/M(Zn)+w/M(Mg)=m(H₂)/{M(H₂)m}
w(1/M(Mg)-1/M(Zn))=m(H₂)/{M(H₂)m}-1/M(Zn)
w=[m(H₂)/{M(H₂)m}-1/M(Zn)]/(1/M(Mg)-1/M(Zn))
w=0.583 (58.3%)
Answer:
anthropogenic sources like landfills, agricultural activities,coal mining and combustion and some sources are natural ones
The correct answer is A. A test tube of zinc oxide. Zinc oxide is a compound which is a pure substance composed of two or more elements. They can only be separated or broken down into its individual elements or smaller compounds by chemical means. Therefore, the answer is a test tube of zinc oxide.