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MAVERICK [17]
3 years ago
8

The production of ethylene glycol from ethylene chlorohydrin and sodium bicarbonate ↑ is carried out in a semibatch reactor. A 1

.5-molar solution of ethylene chlorohydrin is fed at a rate of 0.1 mole/minute to 1500 dm3 of a 0.75-molar solution of sodium bicarbonate. The reaction is elementary and carried out isothermally at 30°C where the specific reaction rate is 5.1 dm3/mol/h. Higher temperatures produce unwanted side reactions. The reactor can hold a maximum of 2500 dm3 of liquid. Assume constant density.

Chemistry
2 answers:
jarptica [38.1K]3 years ago
6 0

Answer:

The constant density decreases

Explanation:

As the temperature of a solvent increases, the solubility of any gas dissolved in that solvent decreases.

For example:

when the temperature of a river, lake or stream is raised high , due to discharge of hot water from some industrial process the solubility of the oxygen in the water is decreased .The fish and the other organisms that live in the water bodies such as rivers, ponds, lakes etc can survive only in the presence of oxygen and decrease in the concentration of the water due to increased temperature can lead to the death of the fish and this may in turn damage the ecosystem.

In the above example, water is considered as the solvent and the oxygen is considered as the solute. When the temperature of the solvent that is water increases, the solubility of the gas that is oxygen in the solvent decreases.

Therefore the answer is decreases

sergey [27]3 years ago
5 0

Answer:

See explanation.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

CH_2OHCH_2Cl+NaHCO_3\rightarrow (CH_2OH)_2+NaCl+CO_2\\A+B\rightarrow C+D+E

For which the differential equations in terms of the time variation are:

\frac{dn_A}{dt}= F_{A,0}+r_AV\\\frac{dn_B}{dt}=r_BV\\\frac{dn_C}{dt}=r_CV\\n_D=n_C\\\frac{dV}{dt}=v_0-\frac{r_CVM_{CO_2}}{\rho }

And the implicit equations:

F_{A,0}=0.1mol/min*60min/1h=6.0mol/h\\v_0=(6.0mol/h)/(1.5mol/dm^3)=4dm^3/h\\rho=1000g/L

r_A=-5.1C_AC_B\\r_B=r_A\\r_C=-r_A

C_A=\frac{n_A}{V} \\C_B=\frac{n_B}{V}

Thus, on the attached pictures you will find the conversion, concentrations, rate and volume profiles as well as the number of moles of ethylene glycol considering a maximum volume of 2500 dm³.

This is the code I used on matlab:

-

t=0;

na(1,1)=0;

nb(1,1)=1125;

nc(1,1)=0;

nd(1,1)=0;

V(1,1)=1500;

k=length(na);

C_A(1,1)=0;

C_B(1,1)=0.75;

x(1,1)=0;

t(1,1)=0;

r_A(k,1)=0;

while V(k,1)<2500

   na(k+1,1)=na(k,1)+0.01*(6+r_A(k,1)*V(k,1));

   nb(k+1,1)=nb(k,1)+0.01*(r_A(k,1)*V(k,1));

   nc(k+1,1)=nc(k,1)+0.01*(-r_A(k,1)*V(k,1));

   nd(k+1,1)=nd(k,1)+0.01*(-r_A(k,1)*V(k,1));

   V(k+1,1)=V(k,1)+0.01*(4-(-r_A(k,1))*V(k,1)*44/1000);

   r_A(k+1,1)=-5.1*C_A(k,1)*C_B(k,1);

   C_A(k+1,1)=na(k+1,1)/V(k+1,1);

   C_B(k+1,1)=nb(k+1,1)/V(k+1,1);

   C_C(k+1,1)=nc(k+1,1)/V(k+1,1);

   C_D(k+1,1)=nd(k+1,1)/V(k+1,1);

   x(k+1,1)=(nb(1,1)-nb(k+1,1))/nb(1,1);

   t(k+1,1)=t(k,1)+0.01;

   k=k+1;

end

-

Best regards.

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6 0
3 years ago
Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacc
Ira Lisetskai [31]

This an incomplete question, here is a complete question.

Calculate the empirical formula for each stimulant based on its elemental mass percent composition.

a. nicotine (found in tobacco leaves): C 74.03%, H 8.70%, N 17.27%

b. caffeine (found in coffee beans): C 49.48%, H 5.19 %, N 28.85% and O 16.48%

Answer:

(a) The empirical formula for the given compound is C_5H_7N

(b) The empirical formula for the given compound is C_4H_5N_2O

Explanation:

<u>Part A: nicotine </u>

We are given:

Percentage of C = 74.03 %

Percentage of H = 8.70 %

Percentage of N = 17.27 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 74.03 g

Mass of H = 8.70 g

Mass of N = 17.27 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{74.03g}{12g/mole}=6.17moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{8.70g}{1g/mole}=8.70moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{17.27g}{14g/mole}=1.23moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.23 moles.

For Carbon = \frac{6.17}{1.23}=5.01\approx 5

For Hydrogen  = \frac{8.70}{1.23}=7.07\approx 7

For Nitrogen = \frac{1.23}{1.23}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

The empirical formula for the given compound is C_5H_7N_1=C_5H_7N

<u>Part B: caffeine</u>

We are given:

Percentage of C = 49.48 %

Percentage of H = 5.19 %

Percentage of N = 28.85 %

Percentage of O = 16.48 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 49.48 g

Mass of H = 5.19 g

Mass of N = 28.85 g

Mass of O = 16.48 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = \frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{49.48g}{12g/mole}=4.12moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{5.19g}{1g/mole}=5.19moles

Moles of Nitrogen = \frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{28.85g}{14g/mole}=2.06moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{16.48g}{16g/mole}=1.03moles

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 1.03 moles.

For Carbon = \frac{4.12}{1.03}=4

For Hydrogen  = \frac{5.19}{1.03}=5.03\approx 5

For Nitrogen = \frac{2.06}{1.03}=2

For Nitrogen = \frac{1.03}{1.03}=1

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N : O = 4 : 5 : 2 : 1

The empirical formula for the given compound is C_4H_5N_2O_1=C_4H_5N_2O

6 0
3 years ago
A tank of oxygen with a pressure of 23 atm is moved from room temperature of 293K to a storage freezer at 239K. What is the fina
Gelneren [198K]

Answer:

18.76atm

Explanation:

Using the formula V1P1/T1 = V2P2/T2, from combined gas law. Volume is constant since we have not been given. Therefore the formula comes to be; P1/T1 = P2/T1

To get P2 = T2(P1/T1)

Where P2 is final pressure

P2 = 239K ( 23atm/293K)

=18.76atm

7 0
3 years ago
Predict the decay process and write the nuclear equation of Cf- 255
matrenka [14]

Answer:

curium

−

243

,

252

/

99

Es,  

251

/

98

Cf,  

214

/

82

Pb

Explanation: Im not very good with this but here ya go!

8 0
2 years ago
How many atoms of Na are in 1.89 mol of Na?
atroni [7]

Answer:

1.138158E24 atoms or 1.14 x 10^24 atoms

Explanation:

To find atoms/particles from moles you just want to convert using avogadro's number which is 6.022 x 10^23

1.89 mol x 6.022 • 10^23

———— = 1.138158E24 atoms

1 mol

so 1.138158E24 atoms or 1.14 x 10^24 for scientific notation

hope this helps :)

5 0
2 years ago
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