Aballoon filled with air has a volume of 0.500 l at 25.0 ºc and an internal pressure of 1.20 atm. what will the volume of the balloon be after it is placed in a freezer at –10.0 ºc if the internal pressure does not change?

0.441 L

Explanation:

At constant pressure, Using Charle's law  :-

Given ,  

V₁ = 0.500 L

V₂ = ?

T₁ = 25.0 °C

T₂ = -10.0 °C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (25 + 273.15) K = 298.15 K  

T₂ = (-10 + 273.15) K = 263.15 K  

Using above equation as:

New volume = 0.441 L

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i