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AnnZ [28]
3 years ago
9

Please help !!

Chemistry
1 answer:
Strike441 [17]3 years ago
7 0
1) boiling points increase as molecular weight increase and vice versa. This is due to the increase in van der waals forces between molecules.
2) branching decreases the melting and boiling i.e increase in branching decrease boiling point and melting point. This is due to the fact that there are less point of contact between neighbouring molecules, so molecules are farther apart from each other, which means weaker van der waals(London forces) less energy is required to overcome these force of attraction.
3) In homolytic fission each of the fragment retain one of the bonded electron and radicals are made if the molecule is neutral. In heterolytic fission one fragment gets both bonding electron.
The energy for the heterolytic fission is higher because energy is not only needed to break the covalent bond but also to overcome the force of attraction between oppositely charged ions formed.
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Vitamin K is involved in normal blood clotting. When 0.802 g of vitamin K is dissolved in 25.0 g of camphor, the freezing point
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Answer:

Molar mass of vitamin K = 450.56\frac{g}{mol}[/tex]

Explanation:

The freezing point of camphor = 178.4 ⁰C

the Kf of camphor =  37.7°C/m

where : m = molality

the relation between freezing point depression and molality is

Depression in freezing point = Kf X molality

Where

Kf = cryoscopic constant of camphor

molality = moles of solute dissolved per kg of solvent.

putting values

2.69°C = 37.7°C/m X molality

molality = 0.0714 mol /kg

molality=\frac{molesofvitaminK}{massofcamphor(kg)}=\frac{moles}{0.025}

moles of vitamin K = 0.0714X0.025 = 0.00178 mol

we know that moles are related to mass and molar mass of a substance as:

moles=\frac{mass}{molarmass}

For vitamin K the mass is given = 0.802 grams

therefore molar mass = \frac{mass}{moles}=\frac{0.802}{0.00178}=450.56\frac{g}{mol}

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---------------------------------------------------

Steps To Solve:

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0

---------------------------------------------------

Answer:

\Large\boxed{\mathsf{0}}

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