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).
<span>To determine the balanced chemical equation, it is necessary to identify the symbols and charges for each element and then balance the equation. This results in:
Al(OH)3(aq) + 3HBr(aq) ---> AlBr3(aq) + 3H2O(l)
The net ionic equation involves summing the charges and breaking them down into fundamental parts.
Al3+ + 3OH- + 3H+ + 3Br- --> Al3+ + 3Br + H3O+
After eliminating aluminum and bromine, we have:
3OH-(aq) + 3H+(aq) --> 3H2O(l)</span>
Vapor pressure refers to the force exerted by vapor or gas molecules above the surface of a liquid. It is inversely related to the concentration of solute particles; an increase in solute concentration results in a decrease in vapor pressure, and vice versa. For (a), it dissociates into two particles. In (b), the total count of particles from dissociation becomes 1 + 2, totaling three. For (c), dissociation yields 1 + 3 for a total of four particles. (d) Since sucrose is a covalent compound, it does not break apart into ions, so it remains as one particle. For (e), dissociation results in 1 + 1, equating to two particles.
Answer: The right choice is (c) application of both a mobile phase and a stationary phase.
Explanation:
Chromatography: This refers to a technique for separating a mixture where the mixture is distributed between two phases at varying rates, one being stationary and the other moving.
Mobile phase: The component in which the mixture is dissolved is referred to as the mobile phase.
Stationary phase: This is an adsorbent medium that remains in place while a liquid or gas passes over its surface, thus remaining stationary.
Consequently, a key characteristic of any chromatography technique involves utilizing both a mobile and a stationary phase.
