Final answer:The pressure of the gas at standard temperature, having started at .370 atm and 50 degrees Celsius, can be found using Gay-Lussac's Law. When the initial temperature and pressure are substituted into the law's formula and solved, the final pressure at the standard temperature is found to be approximately 0.315 atm.Explanation:The problem can be solved using the Gas Laws, specifically theGay-Lussac's Lawwhich states that the pressure of a gas is directly proportional to its temperature in kelvins. In your case, the initial pressure (P1) is 0.370 atm and the initial temperature (T1) is 50.0 Celsius, converted to Kelvin that's 273.15 + 50.0 = 323.15 K. The final temperature (T2) is standard temperature, which is 273.15 K.Applying Gay-Lussac's Law which is P1/T1 = P2/T2, where P2 is the final pressure which we are trying to find out. Substitute the known values into thisequationand solve for P2: P2 = P1*(T2/T1) = 0.370 atm*(273.15 K/323.15 K) = 0.315 atm. So, the final pressure of the gas at standard temperature is roughly0.315 atm.Learn more about Gas Laws here:brainly.com/question/36978528#SPJ3...

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