CHEMISTRY COLLEGE

The heat capacity of chloroform (trichloromethane,CHCl3)in the range 240K to 330K is given
byCpm/(JK-1mol-1) = 91.47
+7.5x10-2(T/K). In a particular experiment,
1.0molCHCl3 is heated from 273K to 300K. Calculate the
changein molar entropy of the sample.

Answers

Answer 1
Answer:

Answer : The change in molar entropy of the sample is 10.651 J/K.mol

Explanation :

To calculate the change in molar entropy we use the formula:

\Delta S=n\int\limits^(T_f)_(T_i){(C_(p,m)dT)/(T)

where,

\Delta S = change in molar entropy

n = number of moles = 1.0 mol

T_f = final temperature = 300 K

T_i = initial temperature = 273 K

C_(p,m) = heat capacity of chloroform = 91.47+7.5* 10^(-2)(T/K)

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

\Delta S=1.0\int\limits^(300)_(273){((91.47+7.5* 10^(-2)(T/K))dT)/(T)

\Delta S=1.0* [91.47\ln T+7.5* 10^(-2)T]^(300)_(273)

\Delta S=1.0* 91.47\ln ((T_f)/(T_i))+7.5* 10^(-2)(T_f-T_i)

\Delta S=1.0* 91.47\ln ((300)/(273))+7.5* 10^(-2)(300-273)

\Delta S=8.626+2.025

\Delta S=10.651J/K.mol

Therefore, the change in molar entropy of the sample is 10.651 J/K.mol

Related Questions

Classify each of the following as a compound, a homogeneous mixture, or a heterogeneous mixture: (a) orange juice; (b) vegetable soup; (c) cement; (d) calcium sulfate; (e) teas
You have a stock solution of epinephrine at a concentration of 1 mg/mL. Knowing that the pipette you will use delivers 20 drops/mL, a. calculate the number of drops of the stock solution that must be added to a smooth muscle bath containing 25 mL of Locke’s solution so that the final concentration of epinephrine in the muscle bath will be 100 µg/mL, and
Ammonium nitrate dissociates in water according to the following equation:43() = 4+()+03−()When a student mixes 5.00 g of NH4NO3 with 50.0 mL of water in a coffee-cup calorimeter, the temperature of the resultant solution decreases from 22.0 °C to 16.5 °C. Assume the density of water is 1.00 g/ml and the specific heat capacity of the resultant solution is 4.18 J/g·°C.1) Calculate q for the reaction. You must show your work.2) Calculate the number of moles of NH4NO3(s) which reacted. You must show your work.3) Calculate ΔH for the reaction in kJ/mol. You must show your work.
A compound has a molecular formular of C12H24O6.What is the compound's empirical formula ​
What is the speed of a rocket in units of meter/ second if its travels at a speed of 1000 km/minutes?

Which description is not a property of a base? (2 points)a
pH lower than 7

b
Turns litmus paper blue

c
Bitter taste

d
Slimy feel

Answers

i think it’s A
because if the pH is lower than 7 than it is acidic

Answer:

the answer is a hope it helps.

Explanation:

Which two structures will provide a positive identification of a plant cell under a microscope? A.) Lysosomes, cell wall. B.) large central vacuole, cell wall. C.) large central vacuole ribosomes. D.)nucleoid, chloroplasts.

Answers

The Right Answer Is D.) Nucleoid chloroplasts. 
the answer is B large central vacuole and cell wall. they are the easiest/biggest things to see under a microscope to identify a plant cell

6.579 rounded to nearest hundreth

Answers

Answer:

6.58

Explanation:

At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressure of He would give a solubility of 0.730 M

Answers

We have that from the Question, it can be said that   The partial pressure of He would give a solubility of 0.730 M is

P_2=4.7atm

From the Question we are told

At a particular temperature, the solubility of He in water is 0.080 M when the partial pressure is 1.7 atm. What partial pressure of He would give a solubility of 0.730 M

Generally the equation for constant temperature  is mathematically given as

(C_2)/(C_1)=(P_2)/(P_1)\n\nTherefore\n\nP_2=(P_1C_1)/(C_1)\n\nP_2=(0.22*1.7)/(0.080)\n\nP_2=4.7atm\n\n

Therefore

The partial pressure of He would give a solubility of 0.730 M is

P_2=4.7atm

For more information on this visit

brainly.com/question/19007362?referrer=searchResults

Answer: Partial pressure of He that would give a solubility of 0.730 M is 15.5 atm

Explanation:

Henry's law states that the amount of gas dissolved or molar solubility of gas is directly proportional to the partial pressure of the liquid.

To calculate the molar solubility, we use the equation given by Henry's law, which is:

C_(He)=K_H* p_(liquid)

where,

K_H = Henry's constant =?

p_(He) = partial pressure = 1.7 atm

Putting values in above equation, we get:

0.080=K_H* 1.7atm\n\nK_H=0.047Matm^(-1)

To find partial pressure of He would give a solubility of 0.730 M

0.730=0.047Matm^(-1)* p_(liquid)

p_(liquid)=15.5atm

Thus partial pressure of He that would give a solubility of 0.730 M is 15.5 atm

A container was found in the home of the victim that contained 120 g of ethylene glycol in 550 g of liquid. How many drinks, each containing 100 g of this liquid, would a 85 kg victim need to consume to reach a toxic level of ethylene glycol

Answers

Answer:

0.432 drinks are toxic

Explanation:

The toxic dose of ethylene glycol is 0.1 mL per kg body weight (mL/kg). In grams (Density ethylene glycol = 1.11g/mL):

1.11g/mL * (0.1mL / kg) =  0.111g/kg

If the victim weighs 85kg, its letal dose is:

85kg * (0.111g/kg) = 9.435g of ethylene glycol

Using the concentration of ethylene glycol in the liquid:

9.435g of ethylene glycol * (550g liquid / 120g ethylene glycol) = 43.2g of liquid are toxic.

The drinks are:

43.2g of liquid * (1 drink / 100 g) =

0.432 drinks are toxic

Sulfonation of naphthalene, C10H8, results in two products. One product is kinetically favored and predominates in the beginning of the reaction. Because the reaction is reversible, eventually the kinetically slower but thermodynamically favored product predominates. Draw the structure of these two products. (The naphthalene ring is already drawn for you. Do not change the double bond configuration in the given structures.)

Answers

Explanation:

The sulfonation of the naphthalene yield 2 products under different conditions:

When the reaction is carried at 80 °C, 1-naphthalenesulfonic acid is the major product because it is kinetically favoured product as arenium ion formed in the transition state corresponding to 1-naphthalenesulfonic acid is more stable due to better resonance stabilization.

When the reaction is carried at 160 °C, 2-naphthalenesulfonic acid is the major product as it is more stable than 1-naphthalenesulfonic acid because of  steric interaction of the sulfonic acid group in 1-position and the hydrogen in 8-position.

The products are shown in image below.
Other Questions
How many lone pairs are on the central atom of BrF3?
I need soon this is due at 10:30 pm its curently 6:00 pmWhat is parasitism?
The standard free energy change for a reaction in an electrolytic cell is always:_________ a. Positive b. Negative c. Zero d. Impossible to determine
How many formula units are in a 7.3 x 10-3 gram sample of lithium chloride, LiCI?