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2 years ago

Calculate each of the following quantities: (c) Mass (g) of 5.85 mol of glycerol (C₃H₈O₃)

Chemistry

1 answer:

Sergio039 [100]2 years ago

5 0

<u>538.73 g</u> is the mass of 5.85 mol of glycerol (C₃H₈O₃)

<h3>What is glycerol?</h3>

Glycerol is a simple polyol chemical, usually known as glycerine in British English and glycerin in American English. It is a sweet-tasting, colourless, odourless, viscous liquid that is non-toxic. Glycerides, a class of lipids, contain the glycerol backbone.

It is frequently utilised in FDA-approved wound and burn therapies due to its antibacterial and antiviral characteristics. On the other hand, it can also serve as a medium for bacterial cultivation. It can be used as a reliable indicator of liver illness.

Additionally, it is employed temporarily in the food sector as a sweetener and in drug formulations as a humectant. Glycerol is hygroscopic in nature and miscible with water due to its three primary hydroxyl groups.

Learn more about glycerol

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A car takes a sharp turn on a road. Which of the following provides the centripetal force on the car?

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Answer:

The centripetal force causing the car to turn in a circular path is due to friction between the tires and the road. A minimum coefficient of friction is needed, or the car will move in a larger-radius curve and leave the roadway.

Explanation:

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3 years ago

At the normal melting point of NaCl, 801 degrees C, its enthalpy of fusion is 28.8 kJ / mol. The density of the solid is 2.165 g

melisa1 [442]

<u>Answer:</u> The increase in pressure is 0.003 atm

<u>Explanation:</u>

To calculate the final pressure, we use the Clausius-Clayperon equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = initial pressure which is the pressure at normal boiling point = 1 atm

$P_2$ = final pressure = ?

$\Delta H$ = Enthalpy change of the reaction = 28.8 kJ/mol = 28800 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = $801^oC=[801+273]K=1074K$

$T_2$ = final temperature = $(801+1.00)^oC=802.00=[802+273]K=1075K$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1})=\frac{28800J/mol}{8.314J/mol.K}[\frac{1}{1074}-\frac{1}{1075}]\\\\\ln P_2=3\times 10^{-3}atm\\\\P_2=e^{3\times 10^{-3}}=1.003atm$

Change in pressure = $P_2-P_1=1.003-1.00=0.003atm$

Hence, the increase in pressure is 0.003 atm

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