Question
Chemistry
(a) Avogadro’s law states that equal volumes of gas at the same temperature and pressure contain equal numbers of molecules.
(b) At STP, one mole of uranium hexafluoride (UF6 MW 352 amu), the gas used in uranium enrichment programs, occupies a volume of 352 L.
© If two gas samples have the same temperature, volume, and pressure, then both contain the same number of molecules.
(d) The value of Avogadro’s number is $6.02×{10}^{23}\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$6.02 xx10^(23)g//mol.
(e) Avogadro’s number is valid only for gases at STP.
(f) The ideal gas law is $PV=nRT$PV=nRT.
(g) When using the ideal gas law for calculations, temperature must be in degrees Celsius.
(h) If one mole of ethane $\left({\mathrm{C}\mathrm{H}}_{3}{\mathrm{C}\mathrm{H}}_{3}\right)$(CH_(3)CH_(3)) gas occupies $20.0\mathrm{L}$20.0L at 1.00 atm, the temperature of the gas is $244\mathrm{K}$244K(i) One mole of helium (MW 4.0 amu) gas at STP occupies twice the volume of one mole of hydrogen (MW 2.0 amu).
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(a) True. Avogadro’s law states that equal volumes of gas at the same temperature and pressure contain equal numbers of molecules.
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(b) False. At STP, one mole of any gas occupies a volume of 22.4 L, regardless of its molecular weight.
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© True. If two gas samples have the same temperature, volume, and pressure, then both contain the same number of molecules according to Avogadro’s law.
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(d) False. The value of Avogadro’s number is $6.02×{10}^{23}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}/\mathrm{m}\mathrm{o}\mathrm{l}$6.02 xx10^(23)molecules//mol, not grams per mole.
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(e) False. Avogadro’s number is valid for any substance, not just gases at STP.
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(f) True. The ideal gas law is $PV=nRT$PV=nRT.
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(g) False. When using the ideal gas law for calculations, temperature must be in Kelvin.
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(h) True. Using the ideal gas law, $PV=nRT$PV=nRT, we can calculate the temperature: $T=\frac{PV}{nR}=\frac{\left(1.00\mathrm{a}\mathrm{t}\mathrm{m}\right)\left(20.0\mathrm{L}\right)}{\left(1\mathrm{m}\mathrm{o}\mathrm{l}\right)\left(0.0821\mathrm{L}\cdot \mathrm{a}\mathrm{t}\mathrm{m}/\mathrm{m}\mathrm{o}\mathrm{l}\cdot \mathrm{K}\right)}=244\mathrm{K}$T=(PV)/(nR)=((1.00atm)(20.0L))/((1mol)(0.0821L*atm//mol*K))=244K.