Question

Please help me solve this!

Answer 1

Answer : The image is attached below.

Explanation :

For
`O_3`:

Molar mass, M = 48 g/mol

Mass, m = 24 g

Moles, n =
`(m)/(M)=(24g)/(48g/mol)=0.5mol`

Number of particles, N =
`n* 6.022* 10^(23)=0.5* 6.022* 10^(23)=3.0* 10^(23)`

For
`NH_3`:

Molar mass, M = 17 g/mol

Mass, m = 170 g

Moles, n =
`(m)/(M)=(170g)/(17g/mol)=10mol`

Number of particles, N =
`n* 6.022* 10^(23)=10* 6.022* 10^(23)=6.0* 10^(24)`

For
`F_2`:

Molar mass, M = 38 g/mol

Mass, m = 38 g

Moles, n =
`(m)/(M)=(38g)/(38g/mol)=1mol`

Number of particles, N =
`n* 6.022* 10^(23)=1* 6.022* 10^(23)=6.0* 10^(23)`

For
`CO_2`:

Molar mass, M = 44 g/mol

Moles, n = 0.10 mol

Mass, m =
`n* M=0.10mol* 44g/mol=4.4g`

Number of particles, N =
`n* 6.022* 10^(23)=0.10* 6.022* 10^(23)=6.0* 10^(22)`

For
`NO_2`:

Molar mass, M = 46 g/mol

Moles, n = 0.20 mol

Mass, m =
`n* M=0.20mol* 46g/mol=9.2g`

Number of particles, N =
`n* 6.022* 10^(23)=0.20* 6.022* 10^(23)=1.2* 10^(23)`

For
`Ne`:

Molar mass, M = 20 g/mol

Number of particles =
`1.5* 10^(23)`

Moles, n =
`(N)/(6.022* 10^(23))=(1.5* 10^(23))/(6.022* 10^(23))=0.25mol`

Mass, m =
`n* M=0.25mol* 20g/mol=5g`

For
`N_2O`:

Molar mass, M = 44 g/mol

Number of particles =
`1.2* 10^(24)`

Moles, n =
`(N)/(6.022* 10^(23))=(1.2* 10^(24))/(6.022* 10^(23))=1.9mol`

Mass, m =
`n* M=1.9mol* 44g/mol=83.6g`

For unknown substance:

Number of particles =
`3.0* 10^(23)`

Mass, m = 8.5 g

Moles, n =
`(N)/(6.022* 10^(23))=(3.0* 10^(23))/(6.022* 10^(23))=0.50mol`

Molar mass, M =
`(m)/(n)=(8.5g)/(0.50mol)=17g/mol`

The substance is
`NH_3`.