Answer:
- 278.85 J
Explanation:
Given that:
Pressure = 1.1 atm
The initial volume V₁ = 0.0 L
The final volume V₂ = 2.5 L
The work that takes place in a reaction at constant pressure can be expressed by using the equation:
W = P(V₂ - V₁ )
Since the volume of the gas is expanded from 0 to 2.5 L when 1.1 atm pressure is applied. Then, the work can be given by the expression:
W = - P(V₂ - V₁ )
W = -1.1 atm ( 2.5 - 0.0) L
W = -1.1 atm (2.5 L)
W = -2.75 atm L
Recall that:
1 atm L = 101.4 J
Therefore;
-2.75 atm L = ( -2.75 × 101.4 )J
= -278.85 J
Thus, the work required at the chemical reaction when the pressure applied is 1.1 atm = - 278.85 J
Answer: The equilibrium concentration of 
will be much smaller than the equilibrium concentration of 
, because Keq<<1
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as 
K is the constant of a certain reaction when it is in equilibrium, while Q is the quotient of activities of products and reactants at any stage other than equilibrium of a reaction.
For the given chemical reaction:
The expression for is written as:
![K=\frac{[H_3O^+]\times [BrO^-]}{[HBrO]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5Ctimes%20%5BBrO%5E-%5D%7D%7B%5BHBrO%5D%7D)
Concentration of pure solids and liquids is taken as 1.
Thus as 
, That means the concentration of products is less as the reaction does not proceed much towards the forward direction.
Correct answer is 20 kJ; 50 kJ
potential energy is resting energy, that peak that's 75 kJ would be the kinetic energy
D. CH and C2H2
Both compounds have the empirical formula CH. The trick is to see if you can multiply one compound by a number to turn it into the other. In this case, you can multiply CH by 2 to make C2H2.
