Answer: The energy required to vaporize 12.5 g of liquid water is 28.2 kJ
Explanation:
Latent heat of vaporization is the amount of heat required to convert 1 gram of liquid into its vapor state without change in its temperature.
Given : The enthalpy of vaporization of water is 40.65 kJ/mol.
n = number of moles =
Thus 1 mole of water requires heat = 40.65 kJ
0.694 moles of water requires heat =
Thus the energy required to vaporize 12.5 g of liquid water is 28.2 kJ
Metals:
<span>Distinguishing luster (shine)
</span><span>Malleable and ductile (flexible) as solids
</span><span>Conduct heat and electricity
</span><span>Metallic oxides are basic, ionic
</span><span>Cations in aqueous solution
</span>
Nonmetals:
<span>Non-lustrous, various colors
</span><span>Brittle, hard or soft
</span><span>Poor conductors
</span><span>Nonmetallic oxides are acidic, compounds
</span><span>Anions, oxyanions in aqueous solution
</span>
XeF6 hope this helps dawg
We need to first calculate the empirical formula. Empirical formula is the simplest ratio of whole numbers of components in a compound,
Mass percentages have been given. We need to then calculate for 100 g of the compound
C H O
mass 77.87 g 11.76 g <span>10.37 g
number of moles 77.87/12 11.76/1 10.37/16
moles = 6.48 = 11.76 =0.648
divide by least number of moles
6.48/0.648 11.76/0.648 0.648/0.648
= 10 =18.1 = 1
rounded off
C - 10 , H - 18 and O - 1
empirical formula - C</span>₁₀H₁₈O
mass of empirical unit = 12 x 10 + 1x 18 + 16 = 120 + 18 + 16 = 154
number of empirical units = molecular mass / mass of one empirical unit
= 154.25 / 154 = 1.00
Therefore molecular formula = C₁₀H₁₈O