As batteries discharge, their voltage decreases. the manufacturer of a cell phone battery would like it to maintain 90.0% of its initial voltage. if a battery has an initial voltage of 6.45 v and loses an average of 0.040 v per minute, how long will it remain at 90.0% or more of the initial voltage?

t = 16,1 minutes

Explanation:

For a battery of 6,45V the 90,0% of its voltage is:

6,45V×0,900 = 5,805V

The voltage decreasing of the batteries could be written as:

Where t is discharge time .

As initial voltage is 6,45V and Vt is 5,805V:

t = 16,1 minutes

I hope it helps!

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.