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How would you correct the volume of 2.90 L of gas at -10°C to the volume it would occupy at standard temperature?

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Final answer:

To correct the volume of 2.90 L of gas from -10°C to 0°C (STP), Charles's Law is used, which gives a new volume of approximately 3.01 L after temperature conversion to kelvins and applying the formula V2 = (V1 × T2) / T1.

Step-by-step explanation:

To correct the volume of 2.90 L of gas at -10°C to the volume it would occupy at standard temperature (STP), you can use Charles's Law which states that the volume of a gas is directly proportional to its temperature in kelvins, assuming pressure and the amount of gas remain constant. The formula for Charles's Law is V1/T1 = V2/T2, where V1 is the initial volume, T1 is the initial temperature, T2 is the final temperature (standard temperature of 0°C or 273.15 K), and V2 is the final volume we want to find.

First, convert -10°C to kelvins by adding 273.15: -10 + 273.15 = 263.15 K. Next, set up the equation using Charles's Law:

V1/T1 = V2/T2

V2 = (V1 × T2) / T1

V2 = (2.90 L × 273.15 K) / 263.15 K

V2 = 2.90 L × (273.15 / 263.15)

V2 = 2.90 L × 1.0380

V2 ≈ 3.01 L

The corrected volume at standard temperature (0°C) would be approximately 3.01 L.

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