Question

How many grams of CaO2 are in 6.90x10^23 molecules of CaO2

Answer 1

Answer: 82.8 grams of
`CaO_2` are there in
`6.90* 10^(23)` molecules of

Step-by-step explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number
`6.023* 10^(23)` of particles.

To calculate the moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given molecules}}{\text {avogadro's number}}=(6.90* 10^(23))/(6.023* 10^(23))=1.15moles`

1 mole of
`CaO_2` weighs = 72 g

1.15 moles of
`CaO_2` weighs =
`(72)/(1)* 1.15=82.8g`

Thus 82.8 grams of
`CaO_2` are there in
`6.90* 10^(23)` molecules of