The answer i got was 2.55 M KOH
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
To see the mole to mole ratio. If you have this then you see how many moles are used for other reactants/products in the equation
<span>As can be seen from the 1st MO diagram on the website the overlap of the H1s AO on one H atom with the H1s AO on another H atom results in a Ď MO at lower energy to the constituent AOs, and a Ď* MO at higher energy than the starting AOs. Each MO can hold two eâ»s so the H2^- has the configuration Ď(2eâ») Ď*(1eâ») or Ď(↑↓) Ď*(↑) Bond Order = ½[ÎŁ (bonding eâ») - ÎŁ (antibonding eâ»)]
bo = ½[ÎŁ (2eâ») - ÎŁ (1 eâ»)] = 0.5 the H2^- is predicted to be bound.
The lowest energy electronic transition is Ď â†’ Ď*: Ď(↑↓) Ď*(↑) → hν → Ď((↑)) Ď*(↑↓)
This guy has a bond order of -0.5 and hence is unbound.</span>
