Solve the equation 5m+4=7m+6
1 answer:
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<u>Answer:</u> The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.
<u>Explanation:</u>
Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.
The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.
A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.
The given redox reaction follows:
To balance the given redox reaction in basic medium, there are few steps to be followed:
- Writing the given oxidation and reduction half-reactions for the given equation with the correct number of electrons
Oxidation half-reaction:
Reduction half-reaction:
- Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions
Oxidation half-reaction: ( × 3)
Reduction half-reaction: ( × 2)
The half-reactions now become:
Oxidation half-reaction:
Reduction half-reaction:
- Add the equations and simplify to get a balanced equation
Overall redox reaction:
As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution
Answer:
0.4 M
Explanation:
Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.
Because there's no O₂ in the beginning, the NO will decompose:
N₂(g) + O₂(g) ⇄ 2NO(g)
0.30 0 0.70 Initial
+x +x -2x Reacts (the stoichiometry is 1:1:2)
0.30+x x 0.70-2x Equilibrium
The equilibrium concentrations are the number of moles divided by the volume (0.250 L):
[N₂] = (0.30 + x)/0.250
[O₂] = x/0.25
[NO] = (0.70 - 2x)/0.250
K = [NO]²/([N₂]*[O₂])
K =
7.70 = (0.70-2x)²/[(0.30+x)*x]
7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)
4x² - 2.80x + 0.49 = 2.31x + 7.70x²
3.7x² + 5.11x - 0.49 = 0
Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70
x = 0.09 mol
Thus,
[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M