Helium can be liquefy through a very
low temperature because of the weakness of attractions between the helium
atoms. In addition, helium is a noble gas that has a very weak interatomic London
dispersion forces. Thus, this element would remain liquid at atmospheric pressure
all the way to its liquefaction point going to absolute zero.
<span> </span>
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
Answer:
B
Explanation:
If the student needs one gram but so far only has 0.37 grams, then the amount they need is the difference between what they need and how much they already have. 1-0.37=0.63 grams.
...which isn't actually an option because none of them have decimal points but I would say it is B anyway because it is the equivalent ratio and maybe there was a typo.
Hope this helped!
<span>Celsius scale: 100 degrees.
Fahrenheit scale: 180 degrees.</span>
