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What is the volume occupied by 3.00 × 1023 molecules of bromine gas at STP? a 404.5 L B. 0.80 L C. 2.2 × 10-2 D. 11.2 L
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What is the volume occupied by 3.00 × 1023 molecules of bromine gas at STP?

a 404.5 L

B. 0.80 L

C. 2.2 × 10-2

D. 11.2 L

User Sergey Podobry
by
5.7k points

2 Answers

4 votes

Answer:

The answer is 11.2 L.

Step-by-step explanation:

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User Gijs Brandsma
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5.9k points
6 votes
D. 11.2 L , since we are at STP we will use the ideal gas law. 3.0 x 10²³ 6.02x10²³ / 1 mole and if you continue with this process it will give you 11 L.
User Ellin
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