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Zolol [24]
3 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]3 years ago
6 0

Answer:

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

Explanation:

KengaRu [80]3 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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Camphor, a white solid with a pleasant odor, is extracted from the roots, branches, and trunk of the camphor tree. Assume you di
katrin [286]

<u>Answer:</u> The molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

<u>Explanation:</u>

  • <u>Calculating the molarity of solution:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}

Given mass of camphor = 70.0 g

Molar mass of camphor = 152.2 g/mol

Volume of solution = 575 mL

Putting values in above equation, we get:

\text{Molarity of camphor}=\frac{70\times 1000}{152.2\times 575}\\\\\text{Molarity of camphor}=0.799M

  • <u>Calculating the molarity of solution:</u>

To calculate the mass of ethanol, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

Density of ethanol = 0.785 g/mL

Volume of ethanol = 575 mL

Putting values in above equation, we get:

0.785g/mL=\frac{\text{Mass of ethanol}}{575mL}\\\\\text{Mass of ethanol}=(0.785g/mL\times 575mL)=451.38g

To calculate the molality of solution, we use the equation:

\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

m_{solute} = Given mass of solute (camphor) = 70 g

M_{solute} = Molar mass of solute (camphor) = 152.2  g/mol

W_{solvent} = Mass of solvent (ethanol) = 451.38 g

Putting values in above equation, we get:

\text{Molality of camphor}=\frac{70\times 1000}{152.2\times 451.38}\\\\\text{Molality of camphor}=1.02m

  • <u>Calculating the mole fraction of camphor:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

<u>For camphor:</u>

Given mass of camphor = 70 g

Molar mass of camphor = 152.2 g/mol

Putting values in equation 1, we get:

\text{Moles of camphor}=\frac{70g}{152.2g/mol}=0.459mol

<u>For ethanol:</u>

Given mass of ethanol = 451.38 g

Molar mass of ethanol = 46 g/mol

Putting values in equation 1, we get:

\text{Moles of ethanol}=\frac{451.38g}{46g/mol}=9.813mol

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

Moles of camphor = 0.459 moles

Total moles = [0.459 + 9.813] = 10.272 moles

Putting values in above equation, we get:

\chi_{(camphor)}=\frac{0.459}{10.272}=0.045\

  • <u>Calculating the mass percent of camphor:</u>

To calculate the mass percentage of camphor in solution, we use the equation:

\text{Mass percent of camphor}=\frac{\text{Mass of camphor}}{\text{Mass of solution}}\times 100

Mass of camphor = 70 g

Mass of solution = [70 + 451.38] = 521.38 g

Putting values in above equation, we get:

\text{Mass percent of camphor}=\frac{70g}{521.38g}\times 100=13.43\%

Hence, the molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

3 0
3 years ago
To convert from liters/second to cubic gallons/minute, multiply the number of liters/second by 15.850 0 0.0353 00.2642 0 60
ivanzaharov [21]

Answer: 15.850

Explanation:

The conversion used from liters to gallons is:

1 L = 0.264172 gallon

The conversion used from sec to min is:

60 sec = 1 min

1 sec =\frac{1}{60}\times 1=0.017min

We are asked: liters/sec = gallons/min

liters/sec=\frac{0.264172}{0.017}=15.850gallons/min

Therefore, to convert from liters/second to gallons/minute, multiply the number of liters/second by 15.850.

8 0
3 years ago
Inertia increases as an object's_______increases.
lilavasa [31]

Answer:

<h2><em><u>MASS</u></em></h2>

Explanation:

Inertia increases as an object's <u>Mass</u> increases.

3 0
3 years ago
Mr Singh is cooking using wo pans of oiling water. Both pans are made of steel, but one pan has a much thicker base than the oth
Sholpan [36]

Answer:

See explanation

Explanation:

When either pan is heated, energy is transferred via conduction. Conduction is the process by which heat is transferred through a material, the average position of the particles remaining the same.

When the pans are heated, the particles in each pan vibrate faster and transfer this energy rapidly to neighboring particles.

The pan with a thicker base has more particles in it than the pan with lighter weight base. Note that, The rate of heat transfer is inversely proportional to the thickness of the material in question. Hence, the thicker the base, the more the number of particles present and the longer the time it takes for the food to cook.

6 0
2 years ago
The pOH of a solution is 6.0. Which statement is correct? Use p O H equals negative logarithm StartBracket upper O upper H super
castortr0y [4]

Answer:

The pH of the solution is 8.0.

Explanation:

taking the test rn

6 0
3 years ago
Read 2 more answers
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