<em>An is formed when an atom loses or gains one or more electrons. Because the number of electrons in an ion is different from the number of protons, an ion does have an overall electric charge. Consider how a positive ion can form from an atom. The left side of the illustration below represents a sodium (Na) atom</em>
The equilibrium conditions allow to find the results for the balance forces are:
When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.
∑ F = 0
∑ τ = 0
Where F are the forces and τ the torques.
The torque is the product of the force and the perpendicular distance to the point of support,
The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.
We write the translational equilibrium condition.
F₁ - W₁ - W₂ + F₂ = 0
We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.
F₂ 2 - W₁ 1 - W₂ 1.5 = 0
Let's calculate F₂
F₂ =
F₂ = (m g + M g 1.5)/ 2
F₂ =
F₂ = 558.6 N
We substitute in the translational equilibrium equation.
F₁ = W₁ + W₂ - F₂
F₁ = (m + M) g - F₂
F₁ = (12 +68) 9.8 - 558.6
F₁ = 225.4 N
In conclusion using the equilibrium conditions we can find the forces of the balance are:
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Complete question:
Consider the hypothetical reaction 4A + 2B → C + 3D
Over an interval of 4.0 s the average rate of change of the concentration of B was measured to be -0.0760 M/s. What is the final concentration of A at the end of this same interval if its concentration was initially 1.600 M?
Answer:
the final concentration of A is 0.992 M.
Explanation:
Given;
time of reaction, t = 4.0 s
rate of change of the concentration of B = -0.0760 M/s
initial concentration of A = 1.600 M
⇒Determine the rate of change of the concentration of A.
From the given reaction: 4A + 2B → C + 3D
2 moles of B ---------------> 4 moles of A
-0.0760 M/s of B -----------> x

⇒Determine the change in concentration of A after 4s;
ΔA = -0.152 M/s x 4s
ΔA = -0.608 M
⇒ Determine the final concentration of A after 4s
A = A₀ + ΔA
A = 1.6 M + (-0.608 M)
A = 1.6 M - 0.608 M
A = 0.992 M
Therefore, the final concentration of A is 0.992 M.