If you create an equilibrium mixture from f e 3 and s c n − ions, adding c l − ions will choose. Highlight_off because choose. H
2 answers:
The chloride ions are not part of the ionic equilibrium, adding them to the solution will merely raise the concentration of anions.
<h3>What is equilibrium?</h3>When a chemical reaction does not convert all reactants to products, it reaches equilibrium. Many processes attain a state of balance, or dynamic equilibrium, in which both reactants and products are present.
The following is the ionic equilibrium between iron (III) ion+ and thiocyanate ions:
The addition of either ion in the ionic equilibrium will shift the equilibrium position to the left or right, according to the principle of equilibrium.
The inclusion of chloride ions, Cl-, which are not components of the ionic equilibrium, increases the levels of anions in the solution.
Thus, the chloride ions are not part of the ionic equilibrium, adding them to the solution will merely raise the concentration of anions.
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Since chloride ions are not part of the ionic equilibrium, addition of chloride ions will only increase the concentration of the anions in the solution.
<h3>What is ionic equilibrium?</h3>An ionic equilibrium is an equilibrium that exists between two or more ions in an aqueous solution.
The ionic equilibrium between iron (iii) ion and thiocyanate ions is given below:
From the principle of equilibrium, addition of either ions in the ionic equilibrium will alter equilibrium position to the left or right.
Addition of chloride ions, Cl- which is not a part of the ionic equilibrium will only increase the concentration of the anions in the solution.
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Refer to the figure shown below.
Let m₁ and m₂ e the two masses.
Let a = the acceleration.
Let T = tension over the frictionless pulley.
Write the equations of motion.
m₂g - T = m₂a (1)
T - m₁g = m₁a (2)
Add equations (1) and (2).
m₂g - T + T - m₁g = (m₁ + m₂)a
(m₂ - m₁)g = (m₁ + m₂)a
Divide through by m₁.
(m₂/m₁ - 1)g = (1 + m₂/m₁)a
Define r = m₂/m₁ as the ratio of the two masses. Then
(r - 1)g = (1 +r)a
r(g-a) = a + g
r = (g - a)/(g + a)
With = 2 ft/s from rest, the acceleration is
a = 2/32.2 = 0.062 ft/s²
Therefore
r = (32.2 - 0.062)/(32.2 + 0.062) = 0.9962
Answer:
The ratio of masses is 0.9962 (heavier mass divided by the lighter mass).
Let m₁ and m₂ e the two masses.
Let a = the acceleration.
Let T = tension over the frictionless pulley.
Write the equations of motion.
m₂g - T = m₂a (1)
T - m₁g = m₁a (2)
Add equations (1) and (2).
m₂g - T + T - m₁g = (m₁ + m₂)a
(m₂ - m₁)g = (m₁ + m₂)a
Divide through by m₁.
(m₂/m₁ - 1)g = (1 + m₂/m₁)a
Define r = m₂/m₁ as the ratio of the two masses. Then
(r - 1)g = (1 +r)a
r(g-a) = a + g
r = (g - a)/(g + a)
With = 2 ft/s from rest, the acceleration is
a = 2/32.2 = 0.062 ft/s²
Therefore
r = (32.2 - 0.062)/(32.2 + 0.062) = 0.9962
Answer:
The ratio of masses is 0.9962 (heavier mass divided by the lighter mass).
A) The position at t = 2.0 sec is 43.0 m east
B) The position is 55 m east
Explanation:
A)
In order to solve the problem, we take the east direction as positive direction.
We know that:
- at t = 0, the motorcyclist is at a position of
- at t = 0, the initial velocity of the motorcyclist is east
- The acceleration of the motorcyclist is constant and it is
Since the motion is a uniformly accelerated motion, the position of the motorcylist is given by the expression
where t is the time.
Substituting t = 2.0 s, we find the position:
B)
The velocity of the motoryclist can be found by calculating the derivative of the position. Therefore, it is:
where:
is the initial velocity
is the acceleration
We want to find the time t at which the velocity is
v = 25 m/s
Solving the equation for t,
And therefore, the position at t = 2.5 s is:
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Answer:
1 angstrom = 0.1nm
5000 angstrom = 5000/1 × 0.1nm
<h3>= 500nm</h3>5000 angstrom = 5000 × 1 × 10^-10
<h3>= 5 × 10^-7 m</h3>Hope this helps you