In all chemical equations, regardless of the type of reaction that is occurring, the reagents that are on the left side of the arrow are the reactants and on the right your products.
B. Is the solution.
Calculate the Ag+ ion concentration in a saturated solution of Ag2CO3 Ksp=8.1X10^-12
Calculate the pH of a buffer solution that is 1.00 M CH3COOH (acetic acid) and 2.00 M CH3COONa (sodium acetate, a salt) Ka=1.8X-5
The mineral fluorite is composed of CAF2. The molar solubility of calcium fluoride in water is 2.1X106-4
moles/L. what is the Ksp of CaF2?
what ratio of benzoate ion to benzoic acid would be required to prepare a buffer with a pH of 7.20?
Ka(C6H5COOH)=6.5X10-5
<h3>Answer:</h3>
The lowest boiling point is of n-Butane because it only experiences London Dispersion Forces between molecules.
<h3>Explanation:</h3>
Lets take start with the melting point of both compounds.
n-Butane = - 140 °C
Trimethylamine = - 117 °C
Intermolecular Forces in n-Butane:
As we know n-Butane is made up of Carbon and Hydrogen atoms only bonded via single covalent bonds. The electronegativity difference between C and C atoms is zero while, that between C and H atoms is 0.35 which is less than 0.4. Hence, the bonds in n-Butane are purely non polar in nature. Therefore, only London Dispersion Forces are found in n-Butane which are considered as the weakest intermolecular interactions.
Intermolecular Forces in Trimethylamine:
Trimethylamine (a tertiary amine) is made up of Nitrogen, Carbon and Hydrogen atoms bonded via single covalent bonds. The electronegativity difference between N and C atoms is 0.49 which is greater than 0.4. Hence, the C-N bond is polar in nature. Therefore, Dipole-Dipole interactions will be formed along with London Dispersion Forces which are stronger than Dispersion Forces. Therefore, due to Dipole-Dipole interactions Trimethylamine will have greater melting point than n-Butane.
Kr 4d10 5s2 5p2 would be your answer.
