Two pith balls hang side by side close to each other without touching as shown in the figure below. They are both neutral to beg
1 answer:
a)
When a plastic rod is rubbed with wool, charging by friction occurs.
This is one of the three mechanisms of charging an object: in this particular case, as the rod is rubbed with the wool, electrons jump from the wool to the rod.
As a result, the plastic rod will remain with an excess of electrons (negative charge), while the wool will remain with a lack of electrons (positive charge).
However, the number of electrons gained by the rod is equal to the number of electrons given off by the wool: therefore, the total net charge of the system rod + cloth has not changed.
b)
When the rod touches sphere A, some electrons are transferred from the rod to the sphere (this method of charging is called charging by conduction).
Then, the sphere A will also remain negatively charged, as well as the plastic rod.
Later, the rod is brought close to sphere A without touching it: here, the electrons in the rod repel the electrons in the sphere (like charges repel each other), so the rod and the sphere will repel each other.
c)
Before part b), the negative charge on the plastic rod was equal to the positive charge on the wool.
After part b), the plastic rod has given off part of its electrons to sphere A. As a result, the negative charge on the plastic rod will be now less than the positive charge on the wool.
Therefore, the net charge of the rod + wool system will now be positive.
d)
The electrostatic force between two charged object is:
- Attractive if the charges on the objects have opposite sign
- Repulsive if the charges on the objects have same sign
Since sphere A is charged negatively, when it is brought close to sphere B, the electrons in sphere B will be attracted on the opposite side relative to sphere A, while the positive charges will migrate close to sphere A. Therefore, the two spheres will attract each other, since opposite charges attract each other.
e)
When the plastic rod (which is negatively charged) touches sphere B, some electrons are transferred to sphere B (charging by induction). As a result, charge B will also be negatively charged now, as it has an excess of electrons.
At the same time, sphere A is also negatively charged. Therefore, as like charges repel each other, this means that the two spheres will now repel each other.
f)
The net charge of the rod and the two spheres is negative.
In fact, the electrons given off by the rod to the two spheres were initially on the rod itself: this means that the total net charge of the rod + 2 spheres is equal to the net charge on the rod before the process.
Since the net charge on the rod before the process was negative, then the net charge of the system consisting of the rod + 2 spheres must be negative as well.
g)
When a glass rod is rubbed with silk, charging by friction occurs, similar to part a). However in this case, the glass rod gives off electrons to the silk: therefore, the glass rod remains positively charged.
Later, the glass rod touches sphere B: as a result, positive charge is transferred from the rod to sphere B, so sphere B will remain positively charged.
In the meantime, sphere A is negatively charged: since opposite charges attract each other, this means that now sphere A and sphere B will attract each other.
h)
When the two spheres are moved a bit further apart, the attractive force between them decreases a bit. In fact, the electrostatic force between two charged objects is given by
where
k is the Coulomb's constant
q1, q2 are the charges on the two objects
r is the separation between the objects
As we can see, the force is inversely proportional to the square of the distance: therefore, if the separation between the spheres increases, the force will decrease.
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Answer:
The volume of the final solution, V = 0.0305L
Explanation:
Number of moles = Concentration * volume
Concentration of HA = 1.00 * 10⁻⁴M
Volume of HA = 1000mL = 1 L
Number of moles of HA = 1.00 * 10⁻⁴ * 1
Number of moles of HA = 1.00 * 10⁻⁴ mols
Equation of reaction:
HA → H⁺ + A⁻
If 1 mol of HA produces 1 mol of H⁺ and A⁻, 1.00 * 10⁻⁴ mol of HA will produce 1.00 * 10⁻⁴ mol of H⁺ and A⁻.
Since only 16% dissociation occurs = 0.16
Number of moles of H⁺ produced = 0.16 * 1.00 * 10⁻⁴
Number of moles of H⁺ produced = 1.6 * 10⁻⁵mols
Number of moles of A⁻ produced = 0.16 * 1.00 * 10⁻⁴
Number of moles of A⁻ produced = 1.6 * 10⁻⁵mols
Since 16% of HA dissociated into H⁺ and A⁻, 84% of HA is left
Number of mols of HA left = 0.84 * 1.00 * 10⁻⁴
Number of mols of HA left = 8.4 * 10⁻⁵mols
Concentration = num of moles/volume
Let the volume of the final solution be V
Conc of HA = 8.4 * 10⁻⁵/V
Conc of H⁺ = 1.6 * 10⁻⁵/V
Conc of A⁻ = 1.6 * 10⁻⁵/V
To calculate the dissociation constant