Answer:
(a) The equilibrium partial pressure of BrCl (g) will be greater than 2.00 atm.
Explanation:
Q is the coefficient of the reaction and is calculated the same of the way of the equilibrium constant, but using the concentrations or partial pressures in any moment of the reaction, so, for the reaction given:
Q = (pBrCl)²/(pBr₂*pCl₂)
Q = 2²/(1x1)
Q = 4
As Q < Kp, the reaction didn't reach the equilibrium, and the value must increase. As we can notice by the equation, Q is directly proportional to the partial pressure of BrCl, so it must increase, and be greater than 2.00 atm in the equilibrium.
The partial pressures of Br₂ and Cl₂ must decrease, so they will be smaller than 1.00 atm. And the total pressure must not change because of the stoichiometry of the reaction: there are 2 moles of the gas reactants for 2 moles of the gas products.
Because is a reversible reaction, it will not go to completion, it will reach an equilibrium, and as discussed above, the partial pressures will change.
Answer:
5 L
Explanation:
Given data
- Initial pressure (P₁): 1 atm
- Initial volume (V₁): 2.5 L
- Final pressure (P₂): 0.50 atm
For a gas, there is an inverse relationship between the pressure and the volume. Mathematically, for an ideal gas that undergoes an isothermic change, this is expressed through Boyle's law.
Oxidizing agent is that which is reduced and the reducing agent is that which is oxidized. Reduced is when the charged is decreased and oxidized when the charge is increased.
(1) 2Na + 2H2O(l) --> 2NaOH(aq) + H2(g)
The charge of Na in the reactant is 0 and the charge of Na in the NaOH is +1. Na is oxidized. Hence, it is the reducing agent.
The charge of H in H2O is +1 while that in H2 is 0. H is reduced. Hence, it is the oxidizing agent.
(2) C(s) + O2(g) --> CO2(g)
The charge of C in the reactant side is 0 and that its charge in CO2 is +4. C is oxidized. Hence, it is the reducing agent.
The charge of O in O2 is 0 while in CO2, its charge is -2. O is reduced. Hence, it is the oxidizing agent.
(3) 2MnO⁻⁴ + SO2 + 2H2O --> 2Mn²⁺ + 5SO2⁻⁴ 4H⁺
The charge of Mn in MnO⁻⁴ is 4+ while its charge in Mn²⁺ is 2+. Mn is reduced. Hence, it is the oxidizing agent.
The charged of S in SO2 is -4 while its charge in SO₂⁻⁴ is 0. S is oxidized. Hence, it is the reducing agent.
1) - it can become a theory if the hypothesis is <span>tested extensively and competing hypotheses are eliminated - other options are not enough
2) This is a hypothesis. A theory is basically almost an established truth, which can still be changed with new data, but which as far as we know is true.</span>
