If I leave 750 mL of 0.75 M sodium chloride solution uncovered on a windowsill and 150 mL of the solvent evaporates, what will t
1 answer:
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The electric potential due to ammonia at a point away along the axis of a dipole is 1.44 10^-5 V.
<u>Explanation:</u>
Given that 1 D = 1 debye unit = 3.34 × 10-30 C-m.
Given p = 1.47 D = 1.47 3.34
10^-30 = 4.90
10^-30.
V = 1 / (4π∈о) (p cos(θ)) / (r^2)
where p is a permanent electric dipole,
∈ο is permittivity,
r is the radius from the axis of a dipole,
V is the electric potential.
V = 1 / (4 3.14
8.85
10^-12)
(4.90
10^-30
1) / (55.3
10^-9)^2
V = 1.44 10^-5 V.