(a)
Write the balanced half-reactions for the overall process:
Oxidation: Se^2- (aq) → Se
(s) + 2e-
Reduction: 2So3^2- (aq) + 3H2O (l) + 4e- →
S2O3^2- + 6OH- (aq)
(b)
Assuming E sulfite is 0.57 V, compute E selenium:
E anode = E cathode – E cell
= -0.57 – 0.35
=
-.092
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.
As the ball descends down the hill, its potential energy diminishes while its kinetic energy rises.
The ball's potential energy will decrease as it moves down the slope, and its kinetic energy will experience an increase.
Kinetic energy refers to the energy possessed by an object in motion.
K. E = 
m v²
where m is the mass of the ball
and v represents the ball's velocity.
Potential energy is the energy associated with an object's position as it traverses down a slope, expressed as:
P.E = mgh
with m as the mass of the ball,
g as gravitational acceleration, and h as the height.
It is clear that as the object descends, its height decreases, while its velocity increases.
learn more:
Potential energy
Solution:
Molality measures the concentration of a solute in a solution, defined by the amount of solute per specific mass of solvent.
Thus,
Molality = moles of solute / kg of solvent.
Therefore, kg of solvent = moles of solute / molality.
moles of solute = mass / molar mass
= 25.31 g / 101.1 g/mole
= 0.2503 mole.
kg of solvent = 0.2503 mole / 0.1982 m
= 1.263 kg
= 1263 g.
This is the final answer.
The balloon's volume is 128 ml when the gas temperature rises to 320.0 K. Explanation: Given the following: T1 (initial temperature) = 300K, V1 (initial volume) = 120ml, T2 (final temperature) = 320 K, V2 (final volume) =?. Pressure is kept constant during this process. From the equation: Given that the pressure stays constant, we have: V2 = Putting the values into this formula yields: V2 = 128 ml, which indicates the volume of the gas when the temperature increases from 300 K to 320 K.
