When 0.80 grams of Helium are added to a weather balloon, its volume is 6.0 L. If the pressure and temperature do not change, how many more grams are needed to make the balloon have a volume of 8.0 L? Enter 20 Acellus Corporation. All Rights Reserved. A Using Alternative Server g

More grams needed = 0.28 g

Explanation:

Given data:

Initial mass of He = 0.80 g

Initial volume = 6.0 L

Final volume = 8.0 L

More grams needed = ?

Solution:

Number of moles of He:

Number of moles = mass/molar mass

Number of moles = 0.8 g/ 4 g/mol

Number of moles = 0.2 mol

Formula:

V₁/n₁   = V₂/n₂

6.0 L/ 0.2 mol   =  8.0 L/ n₂

n₂ =  8.0 L× 0.2 mol /6.0 L

n₂ = 1.6 L.mol /6.0 L

n₂ = 0.27 mol

Final Grams:

Mass = number of moles × molar mass

Mass = 0.27 mol × 4 g/mol

Mass = 1.08 g

More grams needed = Total - initial

More grams needed = 1.08 g - 0.80

More grams needed = 0.28 g

it's c (engineer's do their work with mechanical stuff not on any living being)