Beta decay occurs in two forms: β⁻ decay and β⁺ decay. In β⁻ decay, a neutron is transformed into a proton through the emission of an electron. If β⁻ decay occurs, the mass number of the daughter nucleus remains unchanged, but the number of protons increases by 1 and the number of neutrons decreases by 1, compared to the parent nucleus. On the other hand, β⁺ decay involves the conversion of a proton into a neutron via the emission of a positron. In this case, the mass number of the daughter nucleus stays the same, while the number of protons decreases by 1 and the number of neutrons increases by 1 compared to the parent nucleus.
To find the temperature at which the volume of the gas would be 0.550 L, given that it is 0.432 L at -20.0 °C, apply Charles’s Law.
The formula is v1/T1 = v2/T2
Known values:
V1 = 0.550 L
T1 = ?
T2 = -20°C + 273 = 253 K
V2 = 0.432 L
Rearranging for T1:
T1 = (V1 × T2) / V2
Calculating:
T1 = (0.55 L × 253) / 0.432 L = 322.11 K or 49.11°C
Explanation:
The rate at which gases effuse is inversely related to the square root of their molar masses.
In this case, half of the helium (1.5 L) passed through the membrane in 24 hours. Therefore, we can calculate the effusion rate of He gas as follows.
= 0.0625 L/hr
Given that the molar mass of He is 4 g/mol and for 
it is 32 g/mol.
Now,
= 2.83
Thus, the effusion rate of 
= 
Rate of = 0.022 L/hr.
This implies that 0.022 L of gas will effuse in one hour.
Consequently, to find the duration needed for 1.5 L of gas to effuse, we calculate as follows.
= 68.18 hours
Thus, we can conclude that it will require 68.18 hours for half of the oxygen to effuse through the membrane.
For the first-order decomposition, the equation is: ln(x0 / x) = kt. At t = 200, x = 0.0300 M, we have ln(x0 / 0.03) = 200k. At t = 400, when x = 0.0200 M, we utilize ln(x0 / 0.02) = 400k. By multiplying the first equation by 2, we get 2ln(x0 / 0.03) = 400k, which aligns with the second equation, leading us to conclude that 2ln(x0 / 0.03) = ln(x0 / 0.02). This suggests (x0 / 0.03)^2 = x0 / 0.02, allowing us to find x0 = 0.045 M as the initial concentration. Plugging this back into the first equation yields: ln(0.045 / 0.03) = 200k, from which it follows that k = 0.0020273 (rate constant). The half-life can be calculated with x = 0.5x0: ln(x0 / 0.5x0) = 0.0020273t, resulting in ln(2) = 0.0020273t, which simplifies to t = 341.90 minutes (half-life).
