Response: 1000
Rationale: because 5 cubic meters equals 5000 liters
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).
Response: The rate constant at 525 K is, 
Rationale:
Based on the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant when 
= 
= rate constant when 
=?
= activation energy for the process = 
R = gas constant = 8.314 J/mole.K
= initial temperature = 701 K
= final temperature = 525 K
Substituting the provided values into this formula yields:
![\log (\frac{K_2}{2.57M^{-1}s^{-1}})=\frac{1.5\times 10^5J/mol}{2.303\times 8.314J/mole.K}[\frac{1}{701K}-\frac{1}{525K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7B2.57M%5E%7B-1%7Ds%5E%7B-1%7D%7D%29%3D%5Cfrac%7B1.5%5Ctimes%2010%5E5J%2Fmol%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B701K%7D-%5Cfrac%7B1%7D%7B525K%7D%5D)
Thus, the rate constant at 525 K is,
The ________ orbital shares degeneracy with 5py in a multi-electron atom.
5px is the right choice
Answer:
C₄F₈
Explanation:
To find the mass, we will use the mole ratio.
The molar mass of carbon is 12.0107 g/mol.
For fluorine gas, the molar mass is 37.99681 g/mol.
Let x represent the mass of carbon.
The provided mass of fluorine is 1.70 g.
Setting up the proportion: x / 12.01067 = 1.70 / 37.99687.
Now, we will cross-multiply.
This results in: x = (1.70 × 12) / 37.99687 = 20.4 / 37.99687 = 0.53688 g.
To find the mass of one mole of CF₂, we add: 0.53688 + 1.70 = 2.23688 g.
The moles of CF₂ are calculated as: 8.93 g / 2.23688 = 3.992, which rounds to approximately 4.
The molecular formula for CF₂ is then 4(CF₂) = C₄F₈
