Answer: ΔH for the reaction is -277.4 kJ
Explanation:
The balanced chemical reaction is,

The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%28products%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%28reactant%29%5D)
![\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCCl_4%7D%5Ctimes%20%5CDelta%20H_%7BCCl_4%7D%29%2B%28n_%7BHCl%7D%5Ctimes%20B.E_%7BHCl%7D%29%20%5D-%5B%28n_%7BCH_4%7D%5Ctimes%20%5CDelta%20H_%7BCH_4%7D%29%2Bn_%7BCl_2%7D%5Ctimes%20%5CDelta%20H_%7BCl_2%7D%5D)
where,
n = number of moles
Now put all the given values in this expression, we get
![\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%281%5Ctimes%20-139%29%2B%281%5Ctimes%20-92.31%29%20%5D-%5B%281%5Ctimes%20-74.87%29%2B%281%5Ctimes%20121.0%5D)

Therefore, the enthalpy change for this reaction is, -277.4 kJ
Explanation:
Upon dissolution of KCl heat is generated and temperature of the solution raises.
Therefore, heat generated by dissolving 0.25 moles of KCl will be as follows.

= 4.31 kJ
or, = 4310 J (as 1 kJ = 1000 J)
Mass of solution will be the sum of mass of water and mass of KCl.
Mass of Solution = mass of water + (no. of moles of KCl × molar mass)
= 200 g + 
= 200 g + 13.625 g
= 213.625 g
Relation between heat, mass and change in temperature is as follows.
Q = 
where, C = specific heat of water = 
Therefore, putting the given values into the above formula as follows.
Q = 
4310 J =
Thus, we can conclude that rise in temperature will be
.
Answer:
Motile bacteria have flagella, while nonmotile bacteria do not.
Explanation:
Answer:
Y is a 3-chloro-3-methylpentane.
The structure is shown in the figure attached.
Explanation:
The radical chlorination of 3-methylpentane can lead to a tertiary substituted carbon (Y) and to a secondary one (X).
The E2 elimination mechanism, as shown in the figure, will happen with a simulyaneous attack from the base and elimination of the chlorine. This means that primary and secondary substracts undergo the E2 mechanism faster than tertiary substracts.