Answer : The energy removed must be, 29.4 kJ
Explanation :
The process involved in this problem are :

The expression used will be:
![Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=Q%3D%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2B%5Bm%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%5D%2B%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= heat released for the reaction = ?
m = mass of benzene = 94.4 g
= specific heat of solid benzene = 
= specific heat of liquid benzene = 
= enthalpy change for fusion = 
Now put all the given values in the above expression, we get:
![Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]](https://tex.z-dn.net/?f=Q%3D%5B94.4g%5Ctimes%201.73J%2Fg.K%5Ctimes%20%28279-322%29K%5D%2B%5B94.4g%5Ctimes%20-125.6J%2Fg%5D%2B%5B94.4g%5Ctimes%201.51J%2Fg.K%5Ctimes%20%28205-279%29K%5D)

Negative sign indicates that the heat is removed from the system.
Therefore, the energy removed must be, 29.4 kJ
Answer:
It really depends on whether you are trying to eat healthy but if you excersize then you should be able to burn off the calories from the soda.
Answer:
For chemical reactions involving gases, gas volume measurements provide a convenient means of determining stoichiometric relationships.
Explanation:
Hope this helps! ∩ω∩
Answer:They are not permanently altered by the reaction they catalyze.
Explanation: Enzymes are usually in lower concentration than substrate molecules they catalyze. Hence an enzyme catalyzes as many substrate molecules as it can. So when an enzyme binds a substrate to it's active site, it does this so as to increase the reaction rate which otherwise would not have been possible without the enzyme. It doesn't mean that the enzyme itself takes part in the chemical reaction. Hence, once an ES(Enzyme-substrate) moves to P(product), the product leaves the active site and the enzyme returns to it's original confirmation ready for binding another molecule of the substrate. Therefore, the enzyme is altered transiently in order to allow the substrate fit into it's active site. Its never altered permanently