There will be 1.11 grams of the sample of radon-222 left. This answer can be obtained using the formula of half life to get the rate constant which will be used in another equation later on.
Half-life (t) = ln2/k = 3.8 days
k = 0.182407/day
Using the general equation of a first order reaction:
Ca/Cao = 1/e^(kt)
Ca/Cao = 0.01506 --> fraction of radon-222 left
This means that 1.51% of the original amount remains, so 1.51% of 73.9 is 1.11 grams.
ΔH = Enthalpy of formation of products - <span>Enthalpy of formation of rectants.</span>
ΔH = [ΔHf (CO2)(s) + ΔHf (H2)(g) ] − [ ΔHf (CO)(g) + ΔHf (H2O)(g) ]
ΔHf (CO2) = <span>−393.509 kJ/mol
</span>ΔHf (H2) = 0 kJ/mol
ΔHf (CO) = −110.525 kJ/mol
ΔHf (H2O) = −241.818 kJ/mol
ΔH = [−393.509 + 0 ] − [ −110.525 + (−241.818) ] =
−41.166 kJ/mol , answer.
The air particles in the bubble are forced to expand when we pull up on the plunger.
<h3><u>Explanation:</u></h3>
Pulling creates a large amount of volume , when the volume of the air bubble is increased, air particles inside the bubble tries to accumulate all of the volume by expanding the size of the bubble. Since according to Ideal gas law,
P V = n R T
where P = Pressure of the gas
V = Volume of the gas
n = No. of moles
R = Boltzmann's constant
T = Temperature
We can observe that pressure is inversely proportional to the volume of the gas. Therefore, when we pull up the plunger, the volume of the air bubble is increased and the pressure inside it is decreased.