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).
Janice's teacher recommended using temperature to differentiate between the alcohol and water mixture. The relevant property illustrated through this experiment is D. boiling. The boiling point is the temperature at which a liquid transitions into vapor. Water boils at 100°C under atmospheric pressure, while most alcohols have lower boiling points. Recognizing the significant boiling temperature disparity between the two substances was essential for their separation.
The proportion of component A by mass for substance AB is given by 75%.
Further explanation
According to Proust's Comparative Law, compounds are made from elements that maintain the same Mass Comparison, ensuring that compounds have a consistent ratio of elements.
The empirical formula presents the mole ratio of elements forming a compound.
In the case of substance AB₂, 60.0% of its mass is attributed to A.
For instance, if the mass of AB₂ is 100 grams, then the mass of A would be 60 grams, and the mass of B would be 40 grams, divided according to the coefficient in compound AB₂ which is 2, leading to 20 grams.
Thus, for compound AB, the total mass is the sum of mass A and mass B, which equals 60 grams plus 20 grams, resulting in a total mass of 80 grams.
Consequently, the percentage of compound A calculates to (60: 80) = 75%.
D.) Gas X possesses a lower density and effuses more quickly than Gas Y.
Explanation:
Gas X has a lower molar mass compared to Gas Y, and the density of a gas is directly proportional to its molar mass. Therefore, it has a lower density than Gas Y. Thomas Graham discovered that, under constant temperature and pressure, the effusion rates of various gases are inversely related to the square root of their molar masses (M). Consequently, a gas with a lower molar mass will effuse faster than one with a higher molar mass. Thus, Gas X effuses faster than Gas Y.
So, the correct selection is: D.) Gas X has a lower density and effuses faster than Gas Y.
Answer: The enthalpy change for the reaction is, 201.9 kJ
Explanation:
Based on Hess’s law of constant heat summation, the energy released or absorbed in a chemical reaction stays constant, regardless of whether the process unfolds in one step or multiple steps.
This principle implies, that chemical equations can be treated analogously to algebraic expressions, allowing addition or subtraction to create the needed equation. Thus, the overall enthalpy change corresponds to the summation of the individual enthalpy changes of the reactions occurring in between.
The balanced equation for 
appears as follows,
The intermediate balanced reactions are outlined as follows,
(1) 
(2) 
(3) 
(4) 
Next, we will multiply the first reaction by 2, reverse the second, and reverse and halve the third and fourth reactions before combining them. This gives us:
(1) 
(2) 
(3) 
(4) 
Therefore, the expression for the enthalpy of the reaction is,
Hence, the enthalpy change for this reaction is, 201.9 kJ
