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
The answer is caso4 and ca(C2H3O2)2
KOH + HBr ---> KBr + H2O
0,3 moles of HBr ---in-------1000ml
x moles of HBr-------in------75ml
x = 0,0225 moles of HBr
according to the reaction: 1 mole of KOH = 1 mole of HBr
so
0,0225 moles of HBr = 0,0225 moles of KOH
0,0225 mole of KOH------in-----45ml
x moles of KOH -----------in------1000ml
x = 0,5 moles of KOH
answer: 0,5 mol/dm³ KOH (molarity)
It's being dissolved so it's the solute
