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Zolol [24]
2 years ago
12

The Society of Automotive Engineers has established an accepted numerical scale to measure the viscosity of motor oil. For examp

le, SAE 40 motor oil has a higher viscosity than an SAE 10 oil. Rank the following hydrocarbons by their expected viscosity.
CH3CH2CH2CH2CH3
CH3(CH2)8CH3
CH3(CH2)11CH3
Chemistry
2 answers:
Sphinxa [80]2 years ago
6 0

Answer:

CH3(CH2)11CH3 > CH3(CH2)8CH3 > CH3CH2CH2CH2CH3

Explanation:

KengaRu [80]2 years ago
6 0

Answer:

CH3CH2CH2CH2CH3.

CH3(CH2)8CH3.

CH3(CH2)11CH3.

Explanation:

Viscosity is a material property which describes the resistance of a fluid to shearing flows. It corresponds roughly to the intuitive notion of a fluid's thickness. For instance, honey has a much higher viscosity than

water .

Viscosity is measured using a

viscometer . Measured values span several orders of magnitude. Of all fluids, gases have the lowest viscosities, and thick liquids have the highest.

The aforementioned hydrocarbon are has different viscosity and so are arrange in order of their viscosity.

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CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is
BaLLatris [955]

Answer : The moles of O_2 left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of CH_4.

Using ideal gas equation:

PV=nRT

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = 27^oC=273+27=300K

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)

n=0.406mole

Now we have to calculate the moles of O_2.

The balanced chemical reaction will be:

CH_4+2O_2\rightarrow CO_2+2H_2O

From the balanced reaction we conclude that,

As, 1 mole of CH_4 react with 2 moles of O_2

So, 0.406 mole of CH_4 react with 2\times 0.406=0.812 moles of O_2

Now we have to calculate the excess moles of O_2.

O_2 is 20 % excess. That means,

Excess moles of O_2 = \frac{(100 + 20)}{100} × Required moles of O_2

Excess moles of O_2 = 1.2 × Required moles of O_2

Excess moles of O_2 = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of O_2 left in the products.

Moles of O_2 left in the products = Excess moles of O_2 - Required moles of O_2

Moles of O_2 left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of O_2 left in the products are 0.16 moles.

7 0
3 years ago
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