(a) Atoms are very small compared to objects on the...

(a) Atoms are very small compared to objects on the macroscopic scale. The radius of a gold atom is 144 pm. What is this value in meters and in centimeters? m cm (b) The mass of a single gold atom is 3.27!10^-22 g. Suppose enough Au atoms were lined up like beads on a string to span a distance of 30.2 cm ( 12 inches). How many would be required? atoms What mass in grams of Au would be used? g Could you weigh out this amount of gold using a typical laboratory balance? (c) Taking the density of gold metal to be 19.3 g/cm^3, calculate the mass of metal needed to form a piece of Au wire with the same length as the distance in b, but with a diameter of 1.00 mm. Hint: The volume of a cylinder is ? times its radius squared times its height. (V = ? r^2 h) g How many gold atoms does this represent? atoms

Transcript

00:01 In this question in the first part, the radius is given to us in pm pipometer and we have to convert it into centimeter.

00:08 To the radius in piper meter is 144 pm.

00:14 And if we convert it into meter, it will be 144 multiplied by 10 raised to the power minus 12 meter.

00:24 And further, if we convert it into centimeter, it will be 144 multiplied by 10 raised to the power minus 12 meter.

00:30 10 raised to the power minus 10 centimeter.

00:36 So this is the value of the radius in centimeter.

00:42 Now for the second part, we have to calculate the number of atoms first, the number of atom required.

00:52 So we will let the number of atoms, the number of atoms required to be n and we know that the total distance is equal to n multiplied by 2 n here 2n is the diameter of single gold atom this 2n is the diameter of single gold atom so now the total distance is 38 .4 centimeter this will be equal to to n multiplied by 2 multiplied by 1 44 multiplied by 10 0 .10 cm.

02:00 So the value of n from here comes out to be 38 .4 divided by 288 multiplied by 10 raised to the power minus 10 centimeter.

02:16 And again to solve this we we will get 38384 divided by 288 multiplied by 10 raised to the power 9 atoms.

02:30 And on further solving this we will get 1 .33 multiplied by 10 raised to the power 9 atoms.

02:44 This will be the number of atoms required.

02:49 And now further, the mass of the mass of the power of the atoms required.

02:52 Of a .u used, mass of au used will be number of atoms multiplied by mass of one atom.

03:11 Mass of one atom.

03:15 So it will be 1 .33 multiplied by 10 raised to the power 9 multiplied by 3 .27 multiplied by 10 raise to the power, minus 22 gram and finally the mass of auu's will be 4 .35 multiplied by 10 range to the power minus 13 gram this will be the mass of au huge and this amount of gold this amount of gold cannot be weighed cannot be weighed using a typical laboratory balance using a typical laboratory balance.

04:26 And now for the c part of the question, what all has been given to us are we are given length of the wire.

04:36 L is given as 38 .4 cm and diameter of wire...

Question

Please give Ace some feedback