How much energy is required to vaporize 14.27 grams of water at its boiling point? *
1 answer:
Energy required to vaporize : 32.3 kJ
<h3>Further explanation </h3>The heat to change the phase can be formulated :
Q = mLf (melting/freezing)
Q = mLv (vaporization/condensation)
Lf=latent heat of fusion
Lv=latent heat of vaporization
Mass of water = 14.27 g
Latent heat of vaporization ( boiling point of 100 ºC) : 2260 J/g
Energy required :
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Answer:
4.03dm³
Explanation:
The reaction expression is given as:
3H₂ + N₂ → 2NH₃
Volume of hydrogen = 12dm³
AT rtp:
1 mole of gas occupies volume of 22.4dm³
x mole of hydrogen will occupy a volume of 12dm³
Number of moles of hydrogen = = 0.54mole
From the balanced reaction equation:
3 mole of hydrogen gas combines with 1 mole of Nitrogen gas
0.54 mole of hydrogen as will therefore combine with = 0.18moles of nitrogen gas
Since ;
1 mole of gas occupies a volume of 22.4dm³
0.18moles of Nitrogen gas will occupy 0.18 x 22.4 = 4.03dm³