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VikaD [51]
3 years ago
15

What property of hydrocarbon molecules correlates to the viscosity of a particular fraction when it is cooled to room temperatur

e
Chemistry
2 answers:
EastWind [94]3 years ago
8 0

Answer:

Catenation

Explanation:

What property of hydrocarbon molecules correlates to the viscosity of a particular fraction when it is cooled to room temperature

The property will be catenation

Catenation is the ability of carbon to form a long continuous chain from carbon to carbon connection. with this ,. they can form complex compound from simple ones. E.g hexene can form cyclohexene

bonding in alkenes and alkynes prevent saturation. the ability of carbon molecules to bond / form a long protruding chain correlates with the viscosity of a particular fraction when it is cooled to room temperature.

lora16 [44]3 years ago
6 0
Because of differences in molecular structure, the empirical formula remains different between hydrocarbons; in linear, or "straight-run" alkanes, alkenes and alkynes, the amount of bonded hydrogen lessens in alkenes and alkynes due to the "self-bonding" or catenation of carbon preventing entire saturation of the hydrocarbon by the formation of double or triple bonds. 

<span>This inherent ability of hydrocarbons to bond to themselves is referred to as catenation, and allows hydrocarbon to form more complex molecules, such as cyclohexane, and in rarer cases, arenes such as benzene. This ability comes from the fact that bond character between carbon atoms is entirely non-polar, in that the distribution of electrons between the two elements is somewhat even due to the same electronegativity values of the elements (~0.30), and does not result in the formation of an electrophile. 

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Acetic acid has a Kb of 2.93 °C/m and a normal boiling point of 118.1 °C. What would be the boiling point of a solution made by
alexdok [17]

Boiling point elevation is given as:

ΔTb=iKbm

Where,

ΔTb=elevation in the boiling point

that is given by expression:

ΔTb=Tb (solution) - Tb (pure solvent)

Here Tb (pure solvent)=118.1 °C

i for CaCO3= 2

Kb=2.93 °C/m

m=Molality of CaCO₃:

Molality of CaCO₃=Number of moles of CaCO₃/ Mass of solvent (Kg)

=(Given Mass of CaCO3/Molar mass of CaCO₃)/ Mass of solvent (Kg)

=(100.0÷100 g/mol)/0.4

= 2.5 m

So now putting value of m, i and Kb in the boiling point elevation equation we get:

ΔTb=iKbm

=2×2.93×2.5

=14.65 °C

boiling point of a solution can be calculated:

ΔTb=Tb (solution) - Tb (pure solvent)

14.65=Tb (solution)-118.1

Tb (solution)=118.1+14.65

=132.75

3 0
3 years ago
Titanium has an HCP crystal structure, a c/a ratio of 1.669, an atomic weight of 47.87 g/mol, and a density of 4.51 g/cm3. Compu
IrinaK [193]

Answer : The atomic radius for Ti is, 1.45\times 10^{-8}cm

Explanation :

Atomic weight = 47.87 g/mole

Avogadro's number (N_{A})=6.022\times 10^{23} mol^{-1}

First we have to calculate the volume of HCP crystal structure.

Formula used :  

\rho=\frac{Z\times M}{N_{A}\times V} .............(1)

where,

\rho = density  = 4.51g/cm^3

Z = number of atom in unit cell (for HCP = 6)

M = atomic mass  = 47.87 g/mole

(N_{A}) = Avogadro's number  

V = volume of HCP crystal structure = ?

Now put all the values in above formula (1), we get

4.51g/cm^3=\frac{6\times (47.87g/mol)}{(6.022\times 10^{23}mol^{-1}) \times V}

V=1.06\times 10^{-22}cm^3

Now we have to calculate the atomic radius for Ti.

Formula used :

V=6R^2c\sqrt{3}

Given:

c/a ratio = 1.669 that means,  c = 1.669 a

Now put (c = 1.669 a) and (a = 2R) in this formula, we get:

V=6R^2\times (1.669a)\sqrt{3}

V=6R^2\times (1.669\times 2R)\sqrt{3}

V=(1.669)\times (12\sqrt{3})R^3

Now put all the given values in this formula, we get:

1.06\times 10^{-22}cm^3=(1.669)\times (12\sqrt{3})R^3

R=1.45\times 10^{-8}cm

Therefore, the atomic radius for Ti is, 1.45\times 10^{-8}cm

3 0
3 years ago
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Sonbull [250]

The answer is: the pressure inside a can of deodorant is 1.28 atm.

Gay-Lussac's Law: the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature.

p₁/T₁ = p₂/T₂.  

p₁ = 1.0 atm.; initial pressure

T₁ = 15°C = 288.15 K; initial temperature.

T₂ = 95°C = 368.15 K, final temperature

p₂ = ?; final presure.

1.0 atm/288.15 K = p₂/368.15 K.  

1.0 atm · 368.15 K = 288.15 K · p₂.  

p₂ = 368.15 atm·K ÷ 288.15 K.  

p₂ = 1.28 atm.  

As the temperature goes up, the pressure also goes up and vice-versa.  

6 0
3 years ago
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