A weather balloon is inflated to a volume of 25.4 L at a pressure of 743 mmHg and a temperature of 25.8 ∘C. The balloon rises in the atmosphere to an altitude where the pressure is 400. mmHg and the temperature is -13.0 ∘C. Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude

41.45 l.

Explanation:

Using gas equation,

(P1V1)/T1 = (P2V2)/T2

P1 = initial pressure

= 743 mmhg

= 99.73 kPa

V1 = initial volume

= 25.4 l

T1 = initial absolute temperature

= 25.8 + 273.15

= 298.95 K

P2 = final pressure

= 400 mmhg

= 53.33 kPa

T2 = final absolute temperature

= 273.15 + (-13)

= 260.15 K

V2 = (99.73 * 25.5 * 260.15)/(298.95 * 53.33)

= 41.45 l.

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

explanation:

58750.08068 is the answer i think