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:
Internally reversible is the answer.
Explanation:
"Silver chloride is essentially insoluble in water" this statement is true for the equilibrium constant for the dissolution of silver chloride.
Option: b
<u>Explanation</u>:
As silver chloride is essentially insoluble in water but also show sparing solubility, its reason is explained through Fajan's rule. Therefore when AgCl added in water, equilibrium take place between undissolved and dissolved ions. While solubility product constant 
for silver chloride is determined by equilibrium concentrations of dissolved ions. But solubility may vary also at different temperatures. Complete solubility is possible in ammonia solution as it form stable complex as water is not good ligand for Ag+.
To calculate firstly molarity of ions are needed to be found with formula: 
Then at equilibrium cations and anions concentration is considered same hence:
![\left[\mathbf{A} \mathbf{g}^{+}\right]=[\mathbf{C} \mathbf{I}]=\text { molarity of ions }](https://tex.z-dn.net/?f=%5Cleft%5B%5Cmathbf%7BA%7D%20%5Cmathbf%7Bg%7D%5E%7B%2B%7D%5Cright%5D%3D%5B%5Cmathbf%7BC%7D%20%5Cmathbf%7BI%7D%5D%3D%5Ctext%20%7B%20molarity%20of%20ions%20%7D)
Hence from above data can be calculated by: = ![\left[\mathbf{A} \mathbf{g}^{+}\right] \cdot[\mathbf{C} \mathbf{I}]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cmathbf%7BA%7D%20%5Cmathbf%7Bg%7D%5E%7B%2B%7D%5Cright%5D%20%5Ccdot%5B%5Cmathbf%7BC%7D%20%5Cmathbf%7BI%7D%5D)
The correct answer from the choices given is the third option. Covalent compounds have low boiling points. Also, their melting points are low. Covalent bonds have relatively low attractions which results to these properties. The bonds are easily broken by taking energy or adding energy.
