What volume of oxygen is produced at STP when #6.58 times 10^24# molecules of water is decomposed?
Every two molecules of water make for one molecule of oxygen.
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To find the volume of oxygen produced at STP when 6.58 × 10^24 molecules of water are decomposed, you can use the stoichiometry of the reaction for the decomposition of water:
2 H₂O → 2 H₂ + O₂
Given that one mole of water produces one mole of oxygen, and STP conditions are 22.4 L/mol, the volume of oxygen produced is:
Volume of O₂ = (6.58 × 10^24 molecules) / (6.022 × 10^23 molecules/mol) * 22.4 L/mol
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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