<u>Answer:</u> The freezing point of solution is 2.6°C
<u>Explanation:</u>
To calculate the depression in freezing point, we use the equation:
Or,
where,
=
Freezing point of pure solution = 5.5°C
i = Vant hoff factor = 1 (For non-electrolytes)
= molal freezing point depression constant = 5.12 K/m = 5.12 °C/m
= Given mass of solute (anthracene) = 7.99 g
= Molar mass of solute (anthracene) = 178.23 g/mol
= Mass of solvent (benzene) = 79 g
Putting values in above equation, we get:
Hence, the freezing point of solution is 2.6°C
One of the causes of polarity is the unequal forces of the dipole forces due to differences in electronegativity. The more electronegative molecule tend to attract electrons, hence the polarity arrows point towards it. The electronegativity of Phosphorus and Fluorine is 2.19 and 3.98, respectively. Thus, it points outward towards the fluorine atoms.
Answer:
Explanation:
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1)Identify the atoms that are participating in a covalent bond.
2)Draw each atom by using its element symbol. The number of valence electrons is shown by placing up to two dots on each side of the element symbol, with each dot representing a single valence electron.
3)Predict the number of covalent bonds each atom will make using the octet rule.
4)Draw the bonding atoms next to each other, showing a single covalent bond as either a pair of dots or a line representing a shared valence electron pair. If the molecule forms a double or triple bond, use two or three lines to represent the shared electron pairs, respectively.
Answer:
It takes 1.32x10⁻⁷s for the concentration of A to fall by a factor of 8
Explanation:
The equation that represents a first-order kinetics is:
Ln ([A] / [A]₀] = -kt
<em>Where [A] is actual concentration, [A]₀ is initial concentration, K is rate constant (For the given problem, 1.57x10⁷s⁻¹ and t is time.</em>
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As you want the time when you have [A] in a factor of 8 = [A] / [A]₀ = 1/8
Replacing:
Ln ([A] / [A]₀] = -kt
Ln (1/8) = -1.57x10⁷s⁻¹*t
t = 1.32x10⁻⁷s
<h3>It takes 1.32x10⁻⁷s for the concentration of A to fall by a factor of 8</h3>