A piece of purple plastic is charged with 1.13×1061.13×106 extra electrons compared to its neutral state. What is its net electric charge (including its sign) in coulombs?
net electric charge:
×10
CC
A glittering glass globe is given a net electric charge of 4.61×10−64.61×10−6 C. Does the globe now have more or fewer electrons than it does in its neutral state?
fewer
more
How many more or fewer?
amount:
electronselectrons
1) Each electron carries a charge of

The plastic is charged with
Hence, net electric charge is

2) In the neutral state, number of protons and number of neutrons are equal. As protons and electrons are charged oppositely, they cancel out making the object electrically neutral. Adding positive charge equals to adding more protons or removing electrons.
Hence, when the globe is given a net positive electric charge it will have fewer electrons than it's neutral state.
Net electric charge given is

Dividing this with the charge of one electron we will have the number of fewer electrons.
Hence, number of electrons removed is

A piece of purple plastic is charged with 1.13×1061.13×106 extra electrons compared to its neutral state....
Need help calculating the how many more fewer part, please show
work.
A glittering glass globe is given a net electric charge of 7.19 × 10-6 C. Does the globe now have more or fewer electrons than it does in its neutral state? more O fewer How many more or fewer? amount: x10 electrons
A piece of plastic has a net charge of +2.00 μC. How many fewer electrons than protons does this piece of plastic have? (e = 1.60 × 10-19 C)
5. What was the purpose of the NaNO3 solution in this experiment? 6. Could a solution of NaCl be used instead of NaNO3? 7. What was the purpose of FeSO4 solution in this experiment? 8. Could a solution of FeCl, be used instead of FeSO4? 9. Could a solution of NaSO4 be used instead of FeSO4? 10. Calculate the standard cell potential for the spontaneous redox reaction between a Pb(s)/Pb(NO3)2(aq) half-cell and a Ag(s)/AgNO3(aq) half-cell. Which metal would be oxidized?...