Answer:
4.Current and static electricity
Explanation:
Answer : The energy removed must be, -67.7 kJ
Solution :
The process involved in this problem are :

The expression used will be:
![\Delta H=[m\times c_{p,g}\times (T_{final}-T_{initial})]+m\times \Delta H_{vap}+[m\times c_{p,l}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bp%2Cg%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bm%5Ctimes%20%5CDelta%20H_%7Bvap%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= heat released by the reaction = ?
m = mass of benzene = 125 g
= specific heat of gaseous benzene = 
= specific heat of liquid benzene = 
= enthalpy change for vaporization = 
Molar mass of benzene = 78.11 g/mole
Now put all the given values in the above expression, we get:
![\Delta H=[125g\times 1.06J/g.K\times (353.0-(425.0))K]+125g\times -434.0J/g+[125g\times 1.73J/g.K\times (335.0-353.0)K]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B125g%5Ctimes%201.06J%2Fg.K%5Ctimes%20%28353.0-%28425.0%29%29K%5D%2B125g%5Ctimes%20-434.0J%2Fg%2B%5B125g%5Ctimes%201.73J%2Fg.K%5Ctimes%20%28335.0-353.0%29K%5D)

Therefore, the energy removed must be, -67.7 kJ
Explanation:
An electron current, the flow of electrons, contributes to an electric current since the electron 'carries' negative electric charge. ... The flow of ions (either positively or negatively charged) also contributes to an electric current in, for example, the electrolyte of an electrochemical cell.
hope it will work well?!
Answer:
C5H5N is the base and C5H5NH+ is the conjugate acid
H2O is the acid and OH− is the conjugate base
Explanation:
<u>Hydrogen + is also called a proton</u>
C5H5N is the base because it receives the proton (H+) and C5H5NH+ is its conjugate acid
H2O is the acid because it gives up the proton and OH− is the conjugate base because it is capable of receiving the proton
Answer:
HNO3 is the acid and NO3- is the conjugate base
H2O is the base and H3O+ is the conjugate acid
Explanation
HNO3 is the acid and NO3− is its conjugate base, capable of receiving a proton
H2O is the base because it receives the proton and H3O+ is a conjugate acid capable of giving up the proton.