A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an elevation where the pressure is 0.71 atm, then the new volume is 1.8 L. What is the temperature (in °C) of the air at this elevation?

44.37 degrees C

Explanation:

Use combined gas law: (P1)(V1)/T1=(P2)(V2)/T2

For most gas laws, you  must convert to Kelvin:

K=deg C+273

K=25+273=298 K

Plug and chug:

(1.0 atm)(1.2 L)/(298 K)=(0.71 atm)(1.8 L)/(x)

Solve for x and get 317.37 K

Subtract 273 from this to convert to degrees Celsius. You will get 44.37 degrees Celsius.

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the answer is option a.

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.