Ask question

Ask question

All categories

faltersainse [42]

3 years ago

10

If you know that the change in entropy of a system where heat was added is 12 J/K, and that the temperature of the system is 250

K, what is the amount of heat added to the system? a)-5J b)-125J c)- 600 J d)-5000 J e)-8000 J

1 answer:

SVEN [57.7K]3 years ago

3 0

Solution:

Given:

Change in entropy of the system, ΔS = 12J/K

Temperature of the system, $T_{o}$ = 250K

Now, we know that the change in entropy of a system is given by the formula:

ΔS = $\frac{\Delta Q}{T_{o}}$

Amount of heat added, ΔQ = $\Delta S\times T_{o}$

ΔQ = 3000J

You might be interested in

During the reaction, 3.50 μmol of HCl are produced. Calculate the final pH of the reaction solution. Assume that the HCl is comp

lyudmila [28]

Answer:

The pH of the solution will be equal to 5.46

Explanation:

The dissociation reaction of HCl is equal to:

HCl → H+ + Cl-

To solve the exercise we must first convert the µmoles to moles using the following conversion factor:

3.5µmoles x $\frac{1 mol}{1x10^{6} umol}$ = 3.5x $10^{-6}$ moles

Assuming a liter of solution, we can calculate the molar concentration by:

M = $\frac{Number of moles}{Liter of solution}$

Replacing:

M = $\frac{3.5x10^{-6}moles }{1 L}$ = 3.5x $10^{-6}$ moles/L

As this acid dissociates completely, the concentration of protons and chloride will be equal to 3.5x $10^{-6}$ moles/L

The pH will be equal to:

pH = -log[H+]

Replacing:

pH = -log[3.5x $10^{-6}$ ] = 5.46

8 0

4 years ago

In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume t

gladu [14]

Answer:

F(x) = 0 ; x < 0

0.064 ; 0 ≤ x < 1

0.352 ; 1 ≤ x < 2

0.784 ; 2 ≤ x < 3

1 ; x ≥ 3

Explanation:

Each wafer is classified as pass or fail.

The wafers are independent.

Then, we can modelate X : ''Number of wafers that pass the test'' as a Binomial random variable.

X ~ Bi(n,p)

Where n = 3 and p = 0.6 is the success probability

The probatility function is given by :

$P(X=x)=f(x)=nCx.p^{x}.(1-p)^{n-x}$

Where $nCx$ is the combinatorial number

$nCx=\frac{n!}{x!(n-x)!}$

Let's calculate f(x) :

$f(0)=3C0.(0.6^{0}).(0.4^{3})=0.4^{3}=0.064$

$f(1)=3C1.(0.6^{1}).(0.4^{2})=0.288$

$f(2)=3C2.(0.6^{2}).(0.4^{1})=0.432$

$f(3)=3C3.(0.6^{3}).(0.4^{0})=0.6^{3}=0.216$

For the cumulative distribution function that we are looking for :

$P(X\leq x)=F(x)$

$F(0)=f(0)\\F(1)=f(0)+f(1)\\F(2)=f(0)+f(1)+f(2)\\F(3)=f(0)+f(1)+f(2)+f(3)=1$

$F(0)=0.064\\F(1)=0.064+0.288=0.352\\F(2)=0.064+0.288+0.432=0.784\\F(3)=0.064+0.288+0.432+0.216=1$

The cumulative distribution function for X is :

F(x) = 0 ; x < 0

0.064 ; 0 ≤ x < 1

0.352 ; 1 ≤ x < 2

0.784 ; 2 ≤ x < 3

1 ; x ≥ 3

5 0

3 years ago

The Review_c object has a lookup relationship up to the Job_Application_c object. The job_Application_c object has a master-deta

fomenos

Answer:

Utilize the Standard Controller for Position_c and a Controller Extension to query for Review_c data.

Explanation:

Data types such as individual, pedigree, sample, and marker can be displayed depending on the type of information needed, and the database fields will be recorded based on their levels with the appropriate data type placed in each field. The relationship that will be gotten will depend on the auto-populated defaults which can changed to suit our requirements.

6 0

3 years ago

Chlorine is one of the important commodity chemicals for the global economy. Before the advent of large scale

artcher [175]

The composition of gas in the feed, the percentage conversion and the

theoretical yield are combined to give the product stream composition.

Response:

The composition of gas in the product stream are;

HCl: 0.4 kmol/h, Cl₂: 1.6 kmol/h, H₂O: 1.6 kmol/h, O₂: 0.5 kmol/h

<h3>How can percentage conversion give the contents of the product stream?</h3>

The amount of oxygen used = 30% exceeding the theoretical amount

Number of moles of hydrochloric acid = 4 kmol/h

Percentage conversion = 80%

Required:

The composition of the gas in the product feed.

Solution;

The given reaction is; 4HCl + O₂ $\longrightarrow$ 2Cl₂ + 2H₂O

$Percentage \ conversion = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{Moles \ of \ limiting \ reactant \ supplied \ in \ the \, feed}}$

Which gives;

$80 \% = \mathbf{ \dfrac{Moles \ of \ limiting \ reactant \ reacted}{4 \, kmol/h}}$

Moles of limiting reactant reacted = 4 kmol/h × 0.80 = 3.6 kmol/h

Which gives;

Number of moles of HCl in the stream = 4 kmol/h - 3.6 kmol/h = 0.4 kmol/h

Number of moles of Cl₂ produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Similarly;

Number of moles of H₂O produced = 2 kmol/h × 0.8 = 1.6 kmol/h

Number of moles of O₂ in the product stream = 30% × 1 kmol/h + 20% × 1 kmol/h = 0.5 kmol/h

The composition of the production stream is therefore;

<u>HCl: 0.4 kmol/h</u>

<u>Cl₂: 1.6 kmol/h</u>

<u>H₂O: 1.6 kmol/h</u>

<u>O₂: 0.5 kmol/h</u>

Learn more about theoretical and actual yield here:

brainly.com/question/14668990

brainly.com/question/82989

7 0

3 years ago

A distillation column at 101 kPa is used to separate 350 kmol/h of a bubble point mixture of toluene and benzene into an overhea

AURORKA [14]

Answer:

A)

D = 158.42 kmol/h

B = 191.578 kmol/h

B) Rmin = 1.3095

Explanation:

<u>a) Determine the distillate and bottoms flow rates ( D and B ) </u>

F = D + B ----- ( 1 )

<em>Given data :</em>

F = 350 kmol/j

Xf = 0.45 mole

yD ( distillate comp ) = 0.97

yB ( bottom comp ) = 0.02

back to equation 1

350(0.45) = 0.97 D + 0.02 B ----- ( 2 )

where; B = F - D

Equation 2 becomes

350( 0.45 ) = 0.97 D + 0.02 ( 350 - D ) ------ 3

solving equation 3

D = 158.42 kmol/h

resolving equation 2

B = 191.578 kmol/h

<u>B) Determine the minimum reflux ratio Rmin</u>

The minimum reflux ratio occurs when the enriching line meets the q line in the VLE curve

first we calculate the value of the enriching line

Y =( Rm / R + 1 m ) x + ( 0.97 / Rm + 1 )

q - line ; y = ( 9 / 9-1 ) x - xf/9-1

therefore ; x = 0.45

Finally Rmin

= (( 0.97 / (Rm + 1 )) = 0.42

0.42 ( Rm + 1 ) = 0.97

∴ Rmin = 1.3095

<u />

<u />

<u />

<u />

<u />

8 0

3 years ago