Respuesta:
D
Explicación:
Utilizamos la relación de moles para calcular la presión parcial. El número total de moles es 0.2 + 0.2 + 0.1 = 0.5 moles
Ahora, sabemos que la fracción molar del gas argón es 0.2/0.5
La presión parcial se determina así. Para calcular esto, simplemente multiplicamos el número de moles por la presión total.
0.2/0.5 * 5 = 1.0/0.5 = 2.00atm
D
Answer:
Chemists observe phenomena on a macroscopic level which informs their understanding of microscopic aspects.
Explanation:
Many critical chemical insights arise from macroscopic observations because most scientific instruments currently cannot directly evidence microscopic events. Data gathered from these larger-scale observations can yield valuable insights into the nature of specific microscopic interactions.
This is particularly true in atomic structure studies. The majority of evidence that contributed to our understanding of atomic structure was obtained from macroscopic observations and subsequently provided crucial information regarding the atom's microscopic configuration.
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).
62.0g divided by 125g equals 0.496, then multiplied by 100 gives 49.6%.
A indicates the power lines are what truly conducts (or transmits) the power.
Options B or E are incorrect, as wood is not a good conductor. That's why it's commonly used in homes to retain heat (which comes from electricity), preventing it from escaping.
Option C is also not correct since rubber, similarly, is a poor conductor. Like wood, it acts as an insulator, not transmitting heat (or electricity). This is why rubber gloves are utilized during electrical work.
Option D is valid—most metals are excellent electricity conductors. For instance, copper pans are efficient in cooking because copper effectively conducts heat. Being a metal, this is also why wire cutters have rubber grips; it isolates the user from potential electric shock from the conductive metal. The rubber serves as a barrier to protect against electrocution when handling wires or electricity.
