If 4.644.64 mol of NO2NO2 occupy 25.925.9 L, how many liters will 2.732.73 mol of NO2NO2 occupy at the same temperature and pressure

I will assume that the moles are all mistyped and will change them to:

4.64, 25.9, and 2.73.

15.2 liters

Explanation:

Both situations can be described using the ideal gas law:  

PV=nRT,

where P, V, and T are the pressure, volume, and temperature (in degrees K).  n is the number of moles.  R is the gas constant.

First NO2 :      P1(25.9L) = (4.64moles)RT

Second NO2: P2V2 = (2.73moles)RT2

Divide one equation by the other:

(P1(25.9L)  =  (4.64)RT1)      

  P2V2 = (2.73)RT2          

The temperatures, pressures, and gas constants are unchanged, so we can cancel those to obtain:

(25.9L)/V2 = (4.64/2.73)

V2 = (25.9L)(2.73/4.64)

V2 = 15.2 Liters

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b