(0.956 atm)(12.4 L) = (1.20 atm)(V 2 ) The new volume is 9.88 L.
The new volume will become 0.849 L. Remember to convert the temperatures to Kelvin. (1.30 L) / (24.7+273) = V 2 / (-78.5 + 273)
The new temperature is 696°C. (1.05 atm)(121 mL) / (27+273) = (1.40 atm)(293 mL) / (T 2 + 273)
The number of moles of air is 2.24 mol. (2.18 atm)(25.0 L) = n(0.0821)(23+273)
The pressure is 114 atm. Remember to convert kg of helium into moles of helium before using the ideal gas law. (P)(304.0) = ([5.670×1000]/4.003)(0.0821)(25+273)
The pressure is 2.25 atm. There are two ways to solve the problem: 1) Find the new pressure of the gas on the left side after the valve is opened (P 1 V 1 =P 2 V 2 ). The pressure becomes 1.50 atm. Find the new pressure of the gas on the right side after the valve is opened (P 1 V 1 =P 2 V 2 ). The pressure becomes 0.75 atm. The total pressure is therefore 1.50 + 0.75 = 2.25 atm. 2) Find the number of moles in each chamber separately (using PV = nRT), add the number of moles and volumes on each side, then use PV = nRT to solve for the new pressure.
The vapor pressure of water at 100 o C is 1 atm. You know this because atmospheric pressure is 1 atm and this is the temperature at which we observe water to boil.
Intramolecular bonds are stronger because it would take a lot more energy to overcome covalent bonds and break apart the molecule than to overcome intermolecular forces in between the atoms (to make it become a liquid or gas).
H 2 O has the stronger intermolecular forces because it exhibits hydrogen bonding, whereas N 2 only exhibits London dispersion forces.
One Lewis structure could be ethanol and one Lewis structure could be dimethyl ether. Ethanol will have a higher boiling point than dimethyl ether because ethanol exhibits hydrogen bonding and dimethyl ether exhibits dipole-dipole interactions. Hydrogen bonding is an especially strong type of dipole-dipole interaction and will thus raise the boiling point of ethanol.
N 2 would behave more ideally because it is nonpolar and only exhibits London dispersion forces, therefore the intermolecular forces between N 2 molecules are weak (and thus the collisions will be more “elastic”). CO also exhibits dipole-dipole interactions.