What’s the question? Is it true or false?
Answer:
The electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts
Explanation:
Electric potential is given as;
V = E*r
where;
E is the electric field strength, = kq/r²
V = ( kq/r²)*r
V = kq/r
k is coulomb's constant = 8.99 X 10⁹ Nm²/C²
q is the charge of the particles = 1.6 X 10⁻¹⁹ C
r is the distance between the particles = 859 nm
At midpoint, the distance = r/2 = 859nm/2 = 429.5 nm
V = (8.99 X 10⁹ * 1.6 X 10⁻¹⁹)/ (429.5 X 10⁻⁹)
V = 3.349 X 10⁻³ Volts
Therefore, the electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts
Ionic Compound is the answer
From the problem statement, we are given a solution thus the solute in the solution would have an effect on some of the properties of the whole system. These properties are called the colligative properties. To calculate the freezing point of the solution, we use the freezing point depression equation which is expressed as follows:
ΔTf = kf(m)i
where ΔTf represents the freezing point depression, kf is a constant which 4.90 C/m for benzene, i is the vant hoff factor which is 1 for the given solute since it does not dissociate into ions and m is the molality of the solution. We calculate as follows:
ΔTf = kf(m)i
ΔTf = 4.90 (40.00 / .800 (122.13)) (1)
ΔTf = 2.01 C
ΔTf = Tf - Tfs
Tfs = 5.5 - 2.01
Tfs = 3.49 C
The correct answer would be the first option.
