g 8. A hydrogen atom is initially a rest and in the ground state. A proton with kinetic energy 1000 eV scatters off the atom. In the process the hydrogen atom is ionized. The kinetic energy of the ionized electron is measured to be 15.2 eV and the kinetic energy of the proton produced from the ionization of hydrogen is 4.3 eV. What would be the final kinetic energy of the projectile proton after the collision

Explanation:

Ionization energy of hydrogen atom is 13.6 eV . This energy will be provided by energetic proton , the kinetic energy of which is 1000 eV.  The kinetic energy of ionized electron is 15.2 eV . Kinetic energy of proton produced from from the ionization of hydrogen or the nucleus of the hydrogen atom is 4.3 eV . All these energy must have come from kinetic energy of initial proton.

So kinetic energy of projectile proton after collision

= 1000 - ( 13.6 + 15.2 + 4.3 ) eV.

= 966.9 eV .

1.)    2H2  +  O2  --->  2H2O

2.)   CH4  +  4Cl2 --->  CCl4  +  4HCl

Explanation:

In order to balance the equation, you have to make sure there are the same amount of each element on both sides.

Ex (#1).  By adding the coefficients infront of H2 and H20, You now have 4 hydrogen's and 2 oxygens on the left side as well as 4 hydrogens and 2 oxygens on the right side.

Sorry if that's confusing, but hope this helps! :)