Methylene blue indicates the presence of oxidizing agents because it is oxidized itself by these compounds. When electrons are stripped from methylene blue, the resulting molecule imparts a blue color to the solution--giving a clear sign of a chemical change.
Answer:
The answer in this question is show you made Sodium Hydroxide and Hydrogen Gas.In order to do the products of the reaction relate to the phenol red test and the splint test you need to show that you made Sodium Hydroxide and Hydrogen Gas. Show that you made Sodium Hydroxide and Hydrogen Gas so that the products of the reaction relate to the phenol red test and the splint test.
Gas, as the particles have the most energy, and thus move the most.
Electrons absorb energy, as they absorb energy they go from ground state to excited state and to return to ground state electrons release energy in the form of photons producing that color.
Our reaction balanced equation at equilibrium N2(g) + 3 H2(g) ↔ 2 NH3(g)
and we have the Kp value at equilibrium = 4.51 X 10^-5
A) 98 atm NH3, 45 atm N2, 55 atm H2
when Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
= 98^2 / (45 * 55^3) = 1.28 x 10^-3
by comparing the Kp by the Kp at equilibrium(the given value) So,
Kp > Kp equ So the mixture is not equilibrium,
it will shift leftward (to decrease its value) towards the reactants to achieve equilibrium.
B) 57 atm NH3, 143 atm N2, no H2
∴ Kp = [P(NH3)]^2 / [P(N2)]
= 57^2 / 143 = 22.7
∴Kp> Kp equ (the given value)
∴it will shift leftward (to decrease its value) towards reactants to achieve equilibrium.
c) 13 atm NH3, 27 atm N2, 82 atm H2
∴Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
= 13^2 / (27* 82^3) = 1.14 X 10^-5
∴ Kp< Kp equ (the given value)
∴it will shift rightward (to increase its value) towards porducts to achieve equilibrium.
