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Perform calculations using Gay-Lussacs Law Question A gas in a sealed container has an initial pressure of 125 kPa at 25.0°

I guess in a sealed container has an initial pressure of 1 to 5KPA at 25゚C the pressures increased to 150゚ KPA what will the new temperature be a pottery answer with 3 significant figures use negative 27 3.15゚C for absolute zero

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Initial pressure (P)=125 kPa Initial Temperature (Ti) = 25.0°C = 25+273.15 = 298.15K final pressure (P2) = 150kpa final Tempe

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