Suppose a 1.30 g nugget of pure gold has zero net charge. what would be its net charge after it has 1.68% of its electrons removed?

The net charge of 1.3 g nugget of pure gold after 1.68% of its electrons are removed is 559 C

Explanation:

When an atom gains electrons it becomes negatively charged. Conversely, when it looses electrons the atoms becomes positively charged thus

To solve this question, we rely on the relationship between the nmber of particles present in a given mass of an atom, Avogadro's number and number of moles, n

The given variables are

mass of pure gold nugget = 1.30 g

Quantity of electrons removed = 1.68% of electrons present in the gold sample

Molar mass of gold = 197 g/mol

Avogadro's number = 6.02 × 10²³ atoms/mole  

qc = one electron charge = -1.06 × 10⁻¹⁹ C/electron

Electrical charge of gold nugget = 0 C

Number of electrons in one gold atom = 79 electrons

Solving for the number of prticles or gold atoms in 1.3 grams of gold we get

n mass/(molar mass) = 1.3/197 moles of gold =  0.0066 moles

number of particles in 0.0066 moles of gold N = n× = 0.0066 × 6.02 × 10²³  = 3.97 × 10²¹ atoms

since 79 electrons are present per particle we have

3.97 × 10²¹ × 79 = 3.14 × 10²³ electrons

quantity of elecrtrons removed = 1.68% of  3.14 × 10²³ electrons =1.68/100 × 3.14 × 10²³ electrons = 0.0168 × 3.14 × 10²³ electrons = 5.3 × 10²¹ electrons

The net charge of 5.3 × 10²¹ electrons = 5.3 × 10²¹ electrons × -1.06 × 10⁻¹⁹ C/electron =

5.59 × 10² C = 559 C

beta decay is the answer

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of c is as shown here in the image.