Answer:
7
Explanation:
It has 7 typing because it needs to be longer to submit
Answer:
Higher than 59 °C because dipole-dipole interactions in iodine monochloride are stronger than dispersion forces in bromine.
Explanation:
I just took the test and i got it right
<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:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
Answer:
92.26% of C
Explanation:
To solve this problem we must assume we have 1 mole of benzene. The mole contains 6 moles of C and 6 moles of H. We have to convert these moles to grams in order to find the total mass and mass percent will be:
Mass atom / Total mass * 100
<em>Mass C: </em>6mol C * (12.0107g / mol) = 72.0642g
<em>Mass H: </em>6mol H * (1.00794g / mol) = 6.04764g
<em>total mass: </em>72.0642g + 6.04764g = 78.11184g
Mass percent of C will be:
72.0642g C / 78.11184g* 100
<h3>92.26% of C</h3>
