(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
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.
To solve for density, you can use the formula--> Density= PM/ RT, where P stands for pressure, M for molar mass, R represents the gas constant, and T is temperature.
P= 1.75 atm
M= 16.01 g/ mol
R= 0.0821 atm·L/ mol·K
T=337 k
Thus, the density calculation becomes: density= (1.75 x 16.01)/ (0.0821 x 337)= 1.01 g/L
To calculate the moles of MgSO4.7H2O, we find the molar mass equals 246, thus moles = 32 / 246 = 0.13 moles. Upon heating, all 7 H2O from one molecule will evaporate. The total moles of H2O present amount to 7 x 0.13 = 0.91, and the mass of that H2O is 0.91 x 18 = 16.38g. Therefore, the mass of the anhydrous MgSO4 that remains is 32 - 16.38 = 15.62 g.
Answer:
The temperature of the gas rises.
Explanation:
This is classified as an ISOCHORIC process where the volume remains unchanged. There is no work done by the system.
The gas only receives internal energy from the heat transferred to it from the surroundings.
In this situation, the pressure also increases.