Water: hydrogen bonding
carbon tetrafluoride: dispersion or Van del Waals
dichloromethane: dipole-dipole
The mass of atoms of carbon and 3 molecules of hydrogen : 18 g/mol
<h3>Further explanation
</h3>
An atomic mass unit ( amu or "u") is a relative atomic mass of 1/12 the mass of an atom of carbon-12.
The molar mass(molecular mass-formula mass-molecular weight(MW)) of a compound is the sum of the relative atomic mass (Ar) of the constituent elements of the compound
Can be formulated :
M AxBy = (x.Ar A + y. Ar B)
The mass of atom of Carbon(C)⇒Ar = 12 g/mol
The mass of 1 molecule of Hydrogen - H₂(MW) : 2 g/mol
The mass of 3 molecules of Hydrogen : 3 x 2 = 6 g/mol
So the mass of atoms of carbon and 3 molecules of hydrogen :
Average atomic mass of an element is a sum of the product of the isotope mass and its relative abundance.
For example: Chlorine has 2 isotopes with the following abundances
Cl(35): Atomic mass = 34.9688 amu; Abundance = 75.78%
Cl(37): Atomic mass = 36.9659 amu; Abundance = 24.22 %
Average atomic mass of Cl = 34.9688(0.7578) + 36.9659(0.2422) =
= 26.4993 + 8.9531 = 35.4524 amu
Thus, the term “ average atomic mass “ is a <u>weighted</u> average so it is calculated differently from a normal average
Answer: ΔH for the reaction is -277.4 kJ
Explanation:
The balanced chemical reaction is,
The expression for enthalpy change is,
![\Delta H=\sum [n\times \Delta H(products)]-\sum [n\times \Delta H(reactant)]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%28products%29%5D-%5Csum%20%5Bn%5Ctimes%20%5CDelta%20H%28reactant%29%5D)
![\Delta H=[(n_{CCl_4}\times \Delta H_{CCl_4})+(n_{HCl}\times B.E_{HCl}) ]-[(n_{CH_4}\times \Delta H_{CH_4})+n_{Cl_2}\times \Delta H_{Cl_2}]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%28n_%7BCCl_4%7D%5Ctimes%20%5CDelta%20H_%7BCCl_4%7D%29%2B%28n_%7BHCl%7D%5Ctimes%20B.E_%7BHCl%7D%29%20%5D-%5B%28n_%7BCH_4%7D%5Ctimes%20%5CDelta%20H_%7BCH_4%7D%29%2Bn_%7BCl_2%7D%5Ctimes%20%5CDelta%20H_%7BCl_2%7D%5D)
where,
n = number of moles
Now put all the given values in this expression, we get
![\Delta H=[(1\times -139)+(1\times -92.31) ]-[(1\times -74.87)+(1\times 121.0]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B%281%5Ctimes%20-139%29%2B%281%5Ctimes%20-92.31%29%20%5D-%5B%281%5Ctimes%20-74.87%29%2B%281%5Ctimes%20121.0%5D)
Therefore, the enthalpy change for this reaction is, -277.4 kJ
Answer:
-0.1767°C (Option A)
Explanation:
Let's apply the colligative property of freezing point depression.
ΔT = Kf . m. i
i = Van't Hoff factot (number of ions dissolved). Glucose is non electrolytic so i = 1
m = molality (mol of solute / 1kg of solvent)
We have this data → 0.095 m
Kf is the freezing-point-depression constantm 1.86 °C/m, for water
ΔT = T° frezzing pure solvent - T° freezing solution
(0° - T° freezing solution) = 1.86 °C/m . 0.095 m . 1
T° freezing solution = - 1.86 °C/m . 0.095 m . 1 → -0.1767°C
