All categories

3 years ago

How does the density of aluminum compare with the density of gold

1 answer:

3 0

Gold has a very high density of about 19.32g/cm^3 while Aluminum has a low density of 2.7 gm/cm^3 which means gold can pack more amount of matter in a comparatively small space as compared to Aluminum.

You might be interested in

At constant pressure, which of these systems do work on the surroundings? A ( s ) + B ( s ) ⟶ C ( g ) A(s)+B(s)⟶C(g) 2 A ( g ) +

Tju [1.3M]

Correct question:

At constant pressure, which of these systems do work on the surroundings?

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

(d) 2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

Answer:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

Explanation:

Work done by a system on the surroundings at a constant pressure is given as;

W = -PΔV

Where;

ΔV is gas expansion, that is final volume of the gas minus initial volume of the gas must be greater than zero.

Part (a)

A ( s ) + B ( s ) ⟶ C ( g )

ΔV = 1 - (0) = 1 (expansion)

Part (b)

2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

ΔV = 5 - ( 2+ 2) = 1 (expansion)

Part (c)

A ( g ) + B ( g ) ⟶ C ( g )

ΔV = 1 - ( 1 + 1) = -1 (compression)

Part (d)

2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

ΔV = 3 - ( 4) = -1 (compression)

Thus, systems where there is gas expansion are in part (a) and part (b). The correct answers are:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

4 0

3 years ago

Please show all work for full credit.

koban [17]

Answer:

No, it is not sufficient

Please find the workings below

Explanation:

Using E = hf

Where;

E = energy of a photon (J)

h = Planck's constant (6.626 × 10^-34 J/s)

f = frequency

However, λ = v/f

f = v/λ

Where; λ = wavelength of light = 325nm = 325 × 10^-9m

v = speed of light (3 × 10^8 m/s)

Hence, E = hv/λ

E = 6.626 × 10^-34 × 3 × 10^8 ÷ 325 × 10^-9

E = 19.878 × 10^-26 ÷ 325 × 10^-9

E = 19.878/325 × 10^ (-26+9)

E = 0.061 × 10^-17

E = 6.1 × 10^-19J

Next, we work out the energy required to dissociate 1 mole of N=N. Since the bond energy is 418 kJ/mol.

E = 418 × 10³ ÷ 6.022 × 10^23

E = 69.412 × 10^(3-23)

E = 69.412 × 10^-20

E = 6.9412 × 10^-19J

6.9412 × 10^-19J is required to break one mole of N=N bond.

Based on the workings above, the photon, which has an energy of 6.1 × 10^-19J is not sufficient to break a N=N bond that has an energy of 6.9412 × 10^-19J

8 0

3 years ago

Answer (example of a bar graph)

Vilka [71]

Here's an example of a bar graph.