Answer:
D) The equilibrium lies far to the left
Explanation:
According to the law of mass action, the equilibrium constant K for the reaction at 373K can be calculated as follows:
K = ![\frac{[CO][Cl_{2}]}{[COCl_{2}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO%5D%5BCl_%7B2%7D%5D%7D%7B%5BCOCl_%7B2%7D%5D%7D)
= 2.19×10^{-10}
([X] means = concentration of X)
This means that in the equilibrium the concentration of the reactant (that is in the denominator) will be much higher (around 10^{10} fold) than the concentrations of the products (that are in the numerator), and this means that the equilibrium lies far to the left (to the reactants side) as very small amount of product is being formed.
Hey there!
When work is done on an object, the amount of energy it has changes.
Hope this helps!
~Autumly
Answer:
Work = 90.65 j
Explanation:
Given data:
Mass = 0.500 Kg
Distance = 18.5 m
Work done = ?
Solution:
Work = force . distance
Force = mg
Work = mg.distance
Work = mgh
Work = 0.500 Kg × 9.8 m/s²× 18.5 m
Work = 90.65 Kg .m²/s²
Kg .m²/s² = j
Work = 90.65 j
Reactions happen instantly even if some are slower than others they all happen instantly so the answer would be A.
Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation.
To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign.
Pb(s) --> Pb2+ +2e- E0 = +0.13 V
Ag+ + e- ---> Ag E0 = +0.80 V
Adding up the E0's would yield an overall electric cell potential of +0.93 V.
