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2 years ago

Method used to separate ethanol from propanol

1 answer:

N76 [4]2 years ago

8 0

Fractional distillation is a method for separating a liquid from a mixture of two or more liquids. For example, liquid ethanol can be separated from a mixture of ethanol and water by fractional distillation. This method works because the liquids in the mixture have different boiling points.

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. A mixture of methane and air is capable of being ignited only if the mole percent of methane is between 5% and 15%. A mixture

postnew [5]

Answer:

A) Mass flow rate of air = 22.892 kmol/hr

B)percentage by mass of oxygen in the product gas = 22.52%

Explanation:

We are given that the mixture containing 9.0 mole% methane in air flowing.

Thus, we have 0.09 mole of methane(CH4) and the remaining will be the air which is (100% - 9%) = 91% = 0.91

Molar mass of CH4 = 12 + 1(4) = 16 g/mol

We are given the average molecular weight of air = 29 g/mol

Thus;

Average molar mass of air and methane mixture is;

M_avg = (0.09 × 16) + (0.91 × 29)

M_avg = 27.83 g/mol

We are told that air flowing at a rate of 7 × 10² kg/h = 700 kg/h

Thus;

Mass flow rate of CH4 in air mixture = 700kg/h × (0.09CH4)/1 mix × (1/27.83kg/kmol) = 2.264 kmol/hr

Mass flow rate of air in mixture = 2.264kmol/h × 0.91kmol air/0.09kmolCH4 = 22.892 kmol/hr

We are told that the mixture is capable of being ignited if the mole percent of methane is between 5% and 15%.

Thus, for 5% of methane, the air required will be;

2.264kmol/h × 0.95kmol air/0.05kmol CH4 = 43.016 kmol/hr

Now, the dilution air needed will be =

43.016 - 22.892 = 20.124 kmol/hr

Total mass flow rate of mixture =

700kg/hr + (20.124kmol/hr × 29kg/mol) = 1283.596 kg/hr

We are told that air consist of 21 mole% Oxygen (O2).

Molar mass of oxygen = 32

Thus;

Mass fraction of oxygen in the product gas = 43.016kmol/h × (0.21molO2/1mol air) × (32kg oxygen/1kmol oxygen) × (1/1283.596kg/h) = 0.2252

Thus, written in percentage form, we have; 22.52%

So, percentage by mass of oxygen in the product gas = 22.52%

3 0

2 years ago

If you compared 1 m solutions, was a 1 m nacl solution more or less hypertonic than a 1 m sucrose solution? what is your evidenc

igor_vitrenko [27]

When you say the solution is hypertonic, it means that the solution has a higher osmotic pressure. The formula for this is:

P = iMRT,
for strong electrolytes, i = number of ions.
for nonelectrolytes, i = 1

1. The P for sucrose solution which is a nonelectrolyte (assuming room temp):
P = (1)(1m)(8.314 J/mol-K)(298 K)
P = 2477.572 Pa

The P for NaCl solution, which is a strong electrolyte:
P = (2)(1 m)(8.314)(298 K)
P = 4955.144 Pa

<em>So, that means that NaCl is more hypertonic than the sucrose solution.</em>

2. For the second question, the P for the combination of 1 m glucose (nonelectrolyte) and 1 m sucrose is:
P = (1)(1 m)(8.314)(298 K) + (1)(1)(8.314)(298 K) = 4955.144 Pa
<em>In this case, the osmotic pressures are now equal. It is not hypertonic, but isotonic.</em>

7 0

2 years ago

A graduated cylinder holds 100 mL of water. A lead weight is dropped into the cylinder bringing the new volume up to 450 mL. If

dolphi86 [110]

11.43g/mL

Explanation:

Given parameters:

Volume of water in the graduated cylinder = 100mL

Volume of water + lead weight = 450mL

Mass of lead weight = 4000g

Unknown:

Density of the lead weight = ?

Solution:

Density is the mass per unit volume of a body.

Density = $\frac{mass}{volume}$

Volume of the lead weight = volume of water displaced

Volume of lead weight = 450 - 100 = 350mL

Density = $\frac{4000}{350}$ = 11.43g/mL

learn more:

Density brainly.com/question/2690299

#learnwithBrainly

6 0

2 years ago

Wet steam at 1100 kPa expands at constant enthalpy (as in a throttling process) to 101.33 kPa, where its temperature is 105°C. W

nikklg [1K]

Answer:

There are 3 steps of this problem.

Explanation:

Step 1.

Wet steam at 1100 kPa expands at constant enthalpy to 101.33 kPa, where its temperature is 105°C.

Step 2.

Enthalpy of saturated liquid Haq = 781.124 J/g

Enthalpy of saturated vapour Hvap = 2779.7 J/g

Enthalpy of steam at 101.33 kPa and 105°C is H2= 2686.1 J/g

Step 3.

In constant enthalpy process, H1=H2 which means inlet enthalpy is equal to outlet enthalpy

So, H1=H2

H2= (1-x)Haq+XHvap.........1

Putting the values in 1

2686.1(J/g) = {(1-x)x 781.124(J/g)} + {X x 2779.7 (J/g)}

= 781.124 (J/g) - x781.124 (J/g) = x2779.7 (J/g)

1904.976 (J/g) = x1998.576 (J/g)

x = 1904.976 (J/g)/1998.576 (J/g)

x = 0.953

So, the quality of the wet steam is 0.953

7 0

2 years ago

Read 2 more answers

Which one is balanced or unbalanced force 1. Change object motion 2. Push or pull 3. Have direction 4. Do not change objects mot

Margaret [11]

Answer:

Explanation:

1. Change object motion - unbalanced force

2. Push or pull - Balanced forces do not cause a change in motion. When balanced forces act on an object at rest, the object will not move. If you push against a wall, the wall pushes back with an equal but opposite force. Neither you nor the wall will move. Forces that cause a change in the motion of an object are unbalanced forces. So if object move it is unbalanced.

3. Have direction - unbalanced force

4. Do not change objects motion - balanced force

5. Net force does not =0 - unbalanced force

6.net force =0 - balanced force

3 0

2 years ago

Method used to separate ethanol from propanol​

Method used to separate ethanol from propanol