Guess and check, test, trial and error, completion.
These events are actually sorted right, according to the time they occurred.
1. Democritus proposes the existence of atoms - this happened in the 5th century BC
2. Dalton's atomic theory - it was first presented in 1803
3. J.J. Thomson discovers the electron - happened in 1897
4. Rutherford's gold foil experiment - somewhere between 1908 and 1913
5. Bohr model - it was introduced in 1913
6. Schrodinger's wave - the equation was published in 1925
To compute the energy, Q, needed to melt a certain amount of ice, n, in moles, we have

where the latent heat of fusion for ice is equal to 6.009 kJ/mol.
Now, since we have a 20.0 lb ice, we must first convert its mass to grams. Thus mass = (20.0)(1000)(0.4536) = 9072 g.
To find the number of moles present, we must recall that the molar mass of water (ice) is equal to <span>1.00794(2) + </span><span>15.9994 </span>≈ 18.01 g/mol. Hence, we have

Now, to compute for the molar heat of fusion, Q,

Therefore, the amount of heat needed to melt the 20-lb bag of ice is equal to 3026.9 kJ.
Answer: 3026.9 kJ
Answer:
The time required to melt the frost is 3.25 hours.
Explanation:
The time required to melt the frost dependes on the latent heat of the frost and the amount of heat it is transfered by convection to the air .
The heat transferred per unit area can be expressed as:

being hc the convective heat transfer coefficient (2 Wm^-2K^-1) and ΔT the difference of temperature (20-0=20 °C or K).

If we take 1 m^2 of ice, with 2 mm of thickness, we have this volume

The mass of the frost can be estimated as

Then, the amount of heat needed to melt this surface (1 m²) of frost is

The time needed to melt the frost can be calculated as

When The rate of effusion is inversely proportional to the √molar mass of the substance.
and we have R(He) = 1L / 4.5 min so,
R(He)/R(Cl2) = (molar mass of Cl2/ molar mass of He)^0.5
and when we have the molar mass of Cl2 = 70.9 & the molar mass of He = 4
so by substitution:
(1L/4.5 min)/ R(Cl2) = (70.9 / 4)^0.5
(1L/4.5 min) / R(Cl2) = 4.21
∴R(Cl2) = (1L/4.5 min) / 4.21 = 1L/ (4.5*4.21)min = 1 L / 18.945 min
∴Cl2 will take 18.945 min for 1 L to effuse under identical conditions