How many electrons must be removed from a piece of metal to give it a positive charge of 1/10 7?

Answer: removing each electron will give u 1.603*10^-19 C positive charge. So if u want +1*10^-7 C u need to remove +1*10^-7 / 1.603*10^-19 = 6.2383*10^11 electrons.

How many electrons would have to be added to or removed from a coin to leave it with a charge of 1.0 x10 6 C?

Answer Expert Verified 625 billion electrons should be removed from [neutral] coin.

How many mega coulombs are positive and negative charges?

of electrons in 1 mole is = 2*6.022*1023 . As charge on 1 e is 1.6*10-19 coulombs. Thus total charge = 2*6.022*1023*1.6*10-19 coulombs=19.2704*104 coulombs=19.2704*10-2 mega coulombs.

How many mega coulombs are present in 2 moles of neutral hydrogen gas?

Since 1 molecule of Hydrogen has 2 electrons , then 2 mole hydrogen gas will have 2× 2× 6.023 ×10^23. Therefore, the no. of megacoulomb charge present in 2.0 mol of neutral hydrogen gas is 385,472/10^6 = 0.385 MC.

What is the charge on 75kg of electrons?

The total charge q on n number of electrons is the product of the number of electrons and the charge on a single electron. Substitute 8.33×1031 for n and 1.6×10−19C for qe in the above equation. Therefore, the total charge on 75 kg of electrons is −1.33×1013C.

How do you find the number of electrons in 100g of water?

Weight of water = 100g.
molecular weight of water = 18g/mol.
∴ number of mole of water = given weight/molecule weight.
= 100/18 = 5.55 moles.
∵ 1 mole of water contains = 10 mole of electrons.
[ H₂O , 2 electrons of H and 8 electrons of O ]
∴ 5.55 moles of water contain 10 × 5.55 = 55.5 mole of electrons.

What is the total positive charge in coulombs of all the protons in 1 mole of hydrogen atoms?

Both protons and electrons have charge that is quantized. That is, the magnitude of their respective charges, which are equal each other, is 1. This standard value is equal to approximately 1.6×10-19 Coulombs.