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bixtya [17]
3 years ago
13

Which of the following represents enthalpy on this graph?

Chemistry
1 answer:
Nina [5.8K]3 years ago
7 0

exothermic reaction is reactants accomphen by evolution of heat

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Tetrahydrofuran (THF) is a common organic solvent with a boiling point of 339 K. Calculate the total energy (q) required to conv
poizon [28]

Explanation:

For the given reaction, the temperature of liquid will rise from 298 K to 339 K. Hence, heat energy required will be calculated as follows.

             Q_{1} = mC_{1} \Delta T_{1}

Putting the given values into the above equation as follows.

           Q_{1} = mC_{1} \Delta T_{1}

                      = 27.3 g \times 1.70 J/g K \times 41

                      = 1902.81 J

Now, conversion of liquid to vapor at the boiling point (339 K) is calculated as follows.

           Q_{2} = energy required = mL_{v}

    L_{v} = latent heat of vaporization

Therefor, calculate the value of energy required as follows.

             Q_{2} = mL_{v}

                         = 27.3 \times 444

                         = 12121.2 J

Therefore, rise in temperature of vapor from 339 K to 373 K is calculated as follows.

            Q_{3} = mC_{2} \Delta T_{2}

Value of C_{2} = 1.06 J/g,    \Delta T_{2} = (373 -339) K = 34 K

Hence, putting the given values into the above formula as follows.

             Q_{3} = mC_{2} \Delta T_{2}

                       = 27.3 g \times 1.06 J/g \times 34 K

                       = 983.892 J

Therefore, net heat required will be calculated as follows.

            Q = Q_{1} + Q_{2} + Q_{3}

                = 1902.81 J + 12121.2 J + 983.892 J

                = 15007.902 J

Thus, we can conclude that total energy (q) required to convert 27.3 g of THF at 298 K to a vapor at 373 K is 15007.902 J.

5 0
3 years ago
Who discovered infrared light
pishuonlain [190]
<span>Sir Frederick William Herschel did.</span>
8 0
3 years ago
Read 2 more answers
A mixture of CO (g) and excess O2(g) is placed in a 1.0 L reaction vessel at 100.0C and a total pressure of 1.50 atm. The CO is
umka2103 [35]

Answer:

Partial pressure of of CO₂ in the product mixture is 0,20atm

Explanation:

The balance equation is:

2CO(g) + O₂(g) → 2CO₂(g)

Total pressure of CO(g) and O₂(g) gases before reaction at 100,0°C and 1,0L is 1,50 atm. You can say:

X₁ + Y₁  = 1,50atm <em>(1)</em>

Where X₁ is initial partial pressure CO and Y₁ is initial partial pressure of O₂

After reaction partial pressures are:

X₂ = X₁ - 2n = 0; <em>2n = X₁</em>

Y₂ = Y₁ - n

Z₂ = 2n

Where Z₂ is final partial pressure of CO₂

After reaction pressure at 100,0°C and 1,0L is 1,40 atm, that means:

1,40 atm = (Y₂ + Z₂)

1,40 atm = Y₁ - n + 2n

1,40atm = Y₁ + n

1,40 atm = Y₁ + X₁/2 <em>(2)</em>

Replacing (1) in (2)

1,40 atm = 1,50atm - X₁ + X₁/2

-0,10 atm = - X₁/2

<em>0,20 atm = X₁</em>.

As 2n = X₁; 2n =<em> Z₂ = 0,20 atm</em>

I hope it helps!

7 0
3 years ago
An important reaction sequence in the industrial production of nitric acid is the following:
Nataly_w [17]

Answer:

25.0 mol O₂ are required in the second reaction

Explanation:

N₂ (g) + 3H₂ (g)  →  2NH₃ (g)

4NH₃ (g)  +  5O₂ (g)  →  4NO (g) + 6 H₂O(l)

Molar ratio in first reaction is 1:2

For every mol of N₂. I make 2 moles of ammonia. If I have 20 moles of N₂, i'm going to get, 40 moles of ammonia.

In the second reaction, molar ratio between products is 4:5.

If I obtained 40 moles of ammonia in first step, let's prepare the rule of three.

4 moles of ammonia react with 5 moles of O₂

40 moles of ammonia react with ( 40.5) /4 = 25moles

5 0
4 years ago
True or false
Fittoniya [83]

Answer:1

Explanation: friction is showing us how in the past

7 0
3 years ago
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