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How many grams of helium (He) are needed to fill a blimp with a volume of 2.00 x 10^6L?

A) 4 grams
B) 40 grams
C) 400 grams
D) 4,000 grams

User Nadina
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1 Answer

4 votes

Final answer:

To fill a blimp with a volume of 2.00 x 10^6L with helium, 4,000 grams of helium are needed.

Step-by-step explanation:

To calculate the grams of helium needed to fill a blimp with a given volume, we need to know the molar mass of helium.

The molar mass of helium is approximately 4 grams per mole.

Therefore, to find the number of grams needed, we can use the given volume and the molar mass of helium in a mole to gram conversion.

Given volume of the blimp: 2.00 x 10^6 L

Molar mass of helium: 4 g/mol

Using the conversion factor 1 mole = 22.4 L at standard temperature and pressure (STP), we can set up the following proportion:

(2.00 x 10^6 L)(1 mol/22.4 L)(4 g/mol) = 3.57 x 10^5 g

Therefore, the correct answer is option D) 4,000 grams.

User JoeR
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