1.Calculate the radius of a spherical drop of liquid mercury of mass 0.649 g. The density of mercury is 1.36 × 104 kg m-3.
Correct answer:
a) 2.25 mm
Feedback:
The volume of a sphere is
The density, from eqn 0.7, is
so that, rearranging,
and therefore
Thus, rearranging
The density, from eqn 0.7, is
so that, rearranging,
and therefore
Thus, rearranging
2.Calculate the Coulombic potential energy of an electron at a distance of 52.9 pm from the nucleus of a helium atom.
Correct answer:
c) -8.72 × 10-18 J
Feedback:
The Coulombic potential energy between two charges Q1 and Q2 separated by a distance r is given by eqn 0.3
where ε0 = 8.854 × 10-12 J-1 C2 m-1 is the vacuum permittivity. The charges must be expressed in the SI unit of coulombs. Thus, for the interaction between an electron and a helium nucleus,
and, because the helium nucleus has a double positive charge,
so that
where ε0 = 8.854 × 10-12 J-1 C2 m-1 is the vacuum permittivity. The charges must be expressed in the SI unit of coulombs. Thus, for the interaction between an electron and a helium nucleus,
and, because the helium nucleus has a double positive charge,
so that
3.A photoelectron has a kinetic energy of 3.588 × 10-19 J. What is its speed?
Correct answer:
b) 883.9 × 103 m s-1
Feedback:
The kinetic energy of a body of mass m is given by eqn 0.1
The expression may be rearranged to find the speed v
so that
The expression may be rearranged to find the speed v
so that
4.A lump of gold of volume 127 mm3 has a mass of 2.45 mg. Calculate the density of gold.
Correct answer:
c) 19.3 × 103 kg m-3
Feedback:
Density is the ratio of the mass of a sample to the volume that it occupies. Thus the density, from eqn 0.7, is
A mass of 1 mg is equivalent to 10-6kg, whilst a volume of 1 mm3 is equivalent to 1 × (10-3m)3 = 10-9 m3, so that
A mass of 1 mg is equivalent to 10-6kg, whilst a volume of 1 mm3 is equivalent to 1 × (10-3m)3 = 10-9 m3, so that
5.In a barometer, pressure is determined by measuring the height of a column of liquid. Calculate the pressure, in units of Pa, at the bottom of a 347 mm high column of paraffin oil, of density 0.798 g cm-3.
Correct answer:
b) 2720 Pa
Feedback:
The hydrostatic pressure exerted by a column of liquid depends upon the height of the column, h, the density of the liquid, ρ, and the acceleration due to gravity g, and is given by eqn 0.6.
For a pressure in Pa, the height of the column must be in units of m, the density in units of kg-3 and the acceleration due to gravity in m s-2. A density of 0.798 g cm-3 is equivalent to 798 kg m-3 because 1 g = 10-3kg but 1 cm-3 = (10-2m)-3 = 10-6 m3
Thus
For a pressure in Pa, the height of the column must be in units of m, the density in units of kg-3 and the acceleration due to gravity in m s-2. A density of 0.798 g cm-3 is equivalent to 798 kg m-3 because 1 g = 10-3kg but 1 cm-3 = (10-2m)-3 = 10-6 m3
Thus
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