This problem is easily solvable because radioactivity equations are common and well-established. The pseudo-first reaction is written below:
A = A₀(1/2)^(t/h)
where
A is the final amount
A₀ is the original amount
t is the time
h is the half life
5,000 = A₀(1/2)^(24,000/6,000)
Solving for A₀,
<em>A₀ = 80,000 atoms</em>
The SI unit for specific heat capacity is the joule per kilogram Kelvin, J?kg-1?K-1 or J/(kg?K), which is the amount of energy required to raise the temperature of one kilogram of the substance by one Kelvin
The nulear charge is the number of protons.
As the number of protons increases, the nuclear charge grows ant thhe pulling electrostatic force between them and electrons also grows, given that the electrostatic force is proportional to the magnitude of the charges.
As the number of electrons grows, they occupy outer shelss (farther from the nucleus). And the outer electrons will feel not only the atraction of the protons from the nucleus, but the repulsion of the inner electrons.
Then, we see that the increase of nuclear charge is opposed by the increase of core electrons, and the outer (valence) electrons are not so tied to the nucleus as the core electrons are.
This is called shielding effect. A way to quantify the shielding effect is through the effective nuclear charge which is the number of protons (Z) less the number of core electrons.
The more the number of core shells the greater the shielding effect experience by electros in the outermost shells.
The shielding effect, explains why the valence eletrons are more easily removed from the atom than core electrons, and also explains some trends of the periodic table: variationof the size of the atoms in a row, the greater the shielding efect, the less the atraction force felt by the outermos electron, the farther they are and the larger the atom.
<span>294400 cal
The heating of the water will have 3 phases
1. Melting of the ice, the temperature will remain constant at 0 degrees C
2. Heating of water to boiling, the temperature will rise
3. Boiling of water, temperature will remain constant at 100 degrees C
So, let's see how many cal are needed for each phase.
We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion.
80 cal/g * 320 g = 25600 cal
Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C
420 * 100 = 42000 cal
Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away.
420 g * 540 cal/g = 226800 cal
So the total number of cal used is
25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
