Answer:
W=-37.6kJ, therefore, work is done on the system.
Explanation:
Hello,
In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:
Next, the total moles:
After that, since the process is isobaric, we can compute the work as:
Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)
Thereby, the magnitude and direction of work turn out:
Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).
Regards.
Answer:
The molarity of the solution increases.
Explanation:
Molarity is the measure of the concentration of the solute in the solution. In this case, the solvent is the sugar solution and the solute is the sugar.
If sugar is ADDED to the already sugary solution, then there would be more sugar. Therefore, the sugar (solute) would increase in number.
This means that the answer is the third choice: The molarity of the solution increases.
The answer would not be the first or second choice because there isn't anything in the question that implies water. It just says sugar solution.
The answer is not the last choice because the sugar concentration does not decrease after you have added more sugar to it. It increases.
The SAME number of molecules are in ANY “mole” of a compound or element. So, you only need to ... 24 g116 g/mol=0.207 moles of FeF3.
Answer: from the Zn anode to the Cu cathode
Justification:
1) The reaction given is: Zn(s) + Cu₂⁺ (aq) -> Zn²⁺ (aq) +Cu(s)
2) From that, you can see the Zn(s) is losing electrons, since it is being oxidized (from 0 to 2⁺), while Cu²⁺, is gaining electrons, since it is being reduced (from 2⁺ to 0).
3) Then, you can already tell that electrons go from Zn to Cu.
4) The plate where oxidation occurs is called anode, and the plate where reduction occus is called cathode.
So you get that the electrons flow from the anode (Zn) to the cathode (Cu).
Always oxidation occurs at the anode, and reduction occurs at the cathode.
