Question

how many atoms are present in 50g of gold ring​

Answer 1

Answer:

50= (aco3)

$100g = 3 \times 6.021 \times {10}^{23}$

54= ?

$3 \times 6.023 \times {10}^{23}$

$18.069 \times {10}^{23}$

$- 9.0345 \times \\ {10}^{23}$

Answer 2

Answer:

How many atoms are present in 50g of gold ring​ is 1.52 × 10²³.

Explanation:

The Atomic Mass of Gold is 197.

What is Avagadro's number?

One mole of a substance is equal to 6.022 × 10²³ units of that substance (such as atoms, molecules, or ions). The number 6.022 × 10²³ is known as Avogadro's number or Avogadro's constant. The concept of the mole can be used to convert between mass and number of particles.

Calculations-

197 gram of gold contains 6.022×10²³ number of atoms.

Thus,

50gram of Gold Contains.

[$6.022$×$10$²³$/197$]×$50$ number of atoms.

3.18×10²³×53.

1.52×10²³.

So, The total number of atoms present in 50g of gold ring is 1.52 × 10²³.

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https://brainly.in/question/19625838#:~:text=Avogadro's%20number%20is%20defined%20as,a%20dozen%20or%20a%20gross.