A total of 1.505×10^23 lead atoms
In the lungs, the volume of lead equals the total lung volume, which is 5.60L
1 mole corresponds to 22.4L
Thus, 5.6L of lead converts to 5.6/22.4 = 0.25 moles
According to Avogadro's law
1 mole of lead contains 6.02×10^23 lead atoms
Thus, 0.25 moles of lead equates to 0.25×6.02×10^23 = 1.505×10^23 lead atoms
Answer:
The adjustable legs along with the sand table.
Note: The question is incomplete. The full question is presented below.
Using Models to Address Questions Regarding Systems
Armando’s class was examining images of rivers shaped by flowing water. Most rivers appeared wide and shallow, except for one, which was narrow and deep. The students theorized that this river's narrowness and depth are due to:
- the steepness of the hill from which the water descends, or
- the diminutive size of the sand grains the water flows through.
To explore the answer to the question of why this river is so narrow and deep, Armando created the model outlined below.
Explanation:
The model constructed by Armando will facilitate addressing the question due to specific features:
1. Adjustable leg - as one theory proposed by the class suggests that the steep hill affecting the water's path could be the reason for the river's dimensions, the adjustable legs are designed to be raised or lowered to alter the slope, allowing testing of this theory.
2. Sand table - this acts as the streambed. By modifying the size of the sand grains, students can examine the second hypothesis that smaller sand grains contribute to the river's narrowness and depth.
The outcomes of their experimentation will lead them to a conclusion.
Answer:
180.56 kilojoules of heat energy is extracted when 1.00 kg of freon-11 evaporates.
Explanation:
The molar mass of freon-11 is 137.35 g/mol
The enthalpy of vaporization for freon-11 is 24.8 kJ/mol at its normal boiling point of 24°C. Given that,
Mass of freon-11 evaporated = 1.00 kg = 1000 g
Moles of freon-11 evaporated can be calculated as
The energy removed in the form of heat when 1.00 kg of freon-11 vaporizes is:
N₀ signifies the quantity of C-14 atoms per kg of carbon in the original sample at time = 0 seconds, when the carbon composition matched that in today’s atmosphere. As time progresses to ts, the number of C-14 atoms per kg declines to N, due to radioactive decay. λ indicates the decay constant.
Hence, we have N = N₀e - λt, which is the equation for radioactive decay. Rearranging gives us N₀/N = e λt, or In(N₀/N) = - λt, which becomes equation 1.
The sample contains mc kg of carbon, leading to an activity measured as A/mc decay per kg. The variable r represents the initial mass of C-14 in the sample at t=0 relative to the total mass of carbon which is calculated as [(total number of C-14 atoms at t = 0) × ma] / total mass of carbon. Thus, N₀ equates to r/ma, which becomes equation 2.
The activity of the radioactive element is directly related to the atom count at the moment. The activity equation A = dN/dt = λ(N) indicates that: A = λ₁(N × mc). Rearranging provides N = A / (λmc), represented in equation 3.
By integrating equations 2 and 3, we can solve for t yielding
t = (1/λ) In(rλmc/m₀A).
1. Stars originate within clouds of gas and dust referred to as nebulas. 2. These clouds are drawn together by gravitational forces. 3. Once sufficient heat and pressure accumulate, nuclear fusion commences, representing the birth of a star.
