Question

find the no.of electron present in an 8 g gold pin given molar mass of gold =197g\mol electron per atom in gold =79 and avogadro no.=6.023*1023​

Answer 1

Hello Dear,

◆ Answer -

No of electrons = 2.445×10²²

◆ Explanation -

No of moles of gold is calculated by -

n = W/M

n = 8/197

No of electrons in 8 g gold is calculated by -

Ne = n × NA

Ne = 8/197 × 6.022×10^23

Ne = 2.445×10^22 electrons

Therefore, 2.445×10²² electrons are present in 8 g gold.

Hope this was helpful..

Answer 2

Answer:

Explanation:

193.55x10^22e

1mol=79x6.023x10^23e

Mol=mass given÷molar mass

Mol=8÷197=0.041

Total electorn in 8g gold= 0.041x79x6.023x10^23

=193.55x10^22e

Answer.