Following reaction arise between Br2 and Cl2
Br2 + Cl2 → 2BrCl
(1mole) (1mole) (2moles)
From above balanced reaction, it can be seen that 1 mole of Br2 reacts with 1 mole of Cl2 to form 2 mole of BrCl
Thus, when <span>2.74 mol Cl2 reacts with excess Br2, 2.74 X 2 = 5.48 moles of BrCl will be formed. </span>
Answer:
The volume of the gas is determined, which will allow you to calculate the temperature.
Explanation:
According to Charles law; the volume of a given mass of an ideal gas is directly proportional to its temperature at constant pressure.
This implies that, when the volume of an ideal gas is measured at constant pressure, the temperature of the ideal gas can be calculated from it according to Charles law.
Hence in the Ideal Gas Law lab, the temperature of an ideal gas is measured by determining the volume of the ideal gas.
Answers:
1) <span>Breaking Solvent-Solvent Attractions is an Endothermic Process.
2) </span><span>Breaking Solute-Solute Attractions is an Endothermic Process.
3) </span><span>Forming Solute-Solvent Attractions is an Exothermic Process.
Explanation:
When a solute is dissolved in solvent it either releases heat or absorbs heat depending upon the the interactions broken and interactions formed. At first, the solvent solvent interactions are broken , this process requires heat which is provided either from external source or is provided by the forming of solute solvent bond forming process which is exothermic.
When the solvent molecules get apart the solute particles enter to form interactions with elimination of heat. So, if the heat required to break solvent solvent interactions is greater than the heat provided by solute solvent interactions formation then the solute will not dissolve at room temperature and vice versa.</span>
Answer:
a) 
b) entropy of the sistem equal to a), entropy of the universe grater than a).
Explanation:
a) The change of entropy for a reversible process:
The energy balance:
![\delta U=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=%5Cdelta%20U%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)
If the process is isothermical the U doesn't change:
![0=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=0%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)
The work:
If it is an ideal gas:
Solving:
Replacing:
Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:
b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.
