<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.
Answer:
2.50 g of AlCl3
Explanation:
Goodness, stoichiometry...
So, what we need to find first is the amount of grams of AlCl3. To do this we look at the formula of molarity.
M = mols/L of solvent
So we know two parts of this formula. We have the Molarity (0.150) and the mL (125).
Now, we can't forget that we must convert 125 mL into liters so we have 0.125 L ( I forgot and had to do the entire problem again...)
So if we do the backwards equation we get:
0.150 = x/0.125
If we do math (fun ikr) we get 18.75 mols of the solution.
Now, we have to plug this wonderful number into stoichiometry
<u>0.01875 mols | 133.5 g</u>
<u>| 1 mol AgCl3</u>
If you are unfamiliar with what I'm doing, I'm basically going to multiply 0.01875*133.5 then divide that whole thing by 1.
So, I got 2.503125 g AlCl3
If your teacher is a stickler for significant figures there are 3 sig figs for this problem so your final answer should be
2.50 g of AlCl3
Hope you have a great day and fun with chemistry!!!!
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