The answer to the question is b
Answer:
dG will be the same -20 kcal/mol
Explanation:
The dG can be expressed in terms of the G(products) - G(reactants). If the amount of enzyme is doubled the Gibbs energy of the reactants and products will be the same, so the substraction dG has the same value
In the reaction,
Cr3+(aq) + 6H2O -------> [Cr(H2O)6]3+(aq)
Cr3+ IS THE LEWIS ACID H2O is the Lewis base.
A Lewis acid is a compound or chemical that accepts a lone pair of electrons. In the above equation, Cr3+ is accepting the lone pair of electrons.
112.5 g. The production of 50.00 g O2 requires 112.5 g H2O.
a) Write the partially balanced equation for the decomposition of water.
MM = 18.02 32.00
2H2O → O2 + …
Mass/g = 50.00
b) Calculate the <em>moles of O2
</em>
Moles of O2 = 50.00 g O2 × (1 mol O2/16.00 g O2) = 3.1250 mol O2
c) Calculate the <em>moles of water</em>
Moles of H2O = 3.1250 mol O2 × (2 mol H2O/1 mol O2)
= 6.2500 mol H2O
d) Calculate the mass of water
Mass of H2O = 6.2500 mol H2O × (18.02 g H2O/1 mol H2O)
= 112.5 g H2O
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
