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faltersainse [42]

3 years ago

10

ethylene and steam at 320 celsius and 1 atm are fed to a reaction process as an equimolar mixture. the process produces ethanol

which leaves the process as liquid at 25 celcius. what is the heat transfer associated with the process per mole of ethanol produced

1 answer:

Charra [1.4K]3 years ago

6 0

Answer:

Total amount of heat transfer, H_total = 115653.232 J/mol

Explanation:

Full explanation and step by step instructions is attached

You might be interested in

What types of materials dissolve in water

Nata [24]

Answer:

Things like salt, sugar and coffee dissolve in water. They are soluble. They usually dissolve faster and better in warm or hot water. Pepper and sand are insoluble, they will not dissolve even in hot water.

i hope this helped you, have a great day

3 0

2 years ago

One liter of N (g) at 2.1 bar and two liters of Ar(g) at 3.4 bar are mixed in a 4.0-L 2 flask to form an ideal-gas mixture. Calc

Vika [28.1K]

Answer:

Explanation:

From the information given:

Step 1:

Determine the partial pressure of each gas at total Volume (V) = 4.0 L

So, using:

$\text{The new partial pressure for }N_2 \ gas}$

$P_1V_1=P_2V_2$

$P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{2.1 \ bar \times 1\ L}{4.0 \ L} \\ \\ P_2 = 0.525 \ bar$

$\text{The new partial pressure for }Ar \ gas}$

$P_2=\dfrac{P_1V_1}{V_2} \\ \\ P_2=\dfrac{3.4 \ bar \times 2 \ L}{4.0 \ L} \\ \\ P_2 = 1.7 \ bar$

$Total pressure= P [N_2] + P[Ar] \ \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (0.525 + 1.7)Bar \\ \\ . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 2.225 \ Bar$

Now, to determine the final pressure using different temperature; to also achieve this, we need to determine the initial moles of each gas.

According to Ideal gas Law.

$2.1 \ bar = 2.07 \ atm \\ \\3.4 \ bar = 3.36 \ atm$

For moles N₂:

$PV = nRT \\ \\ n = \dfrac{PV}{RT}$

$n = \dfrac{2.07 \ atm \times 1 \ L }{0.08206 \ L .atm. per. mol. K \times 304 \ K}$

$n = 0.08297 \ mol \ N_2$

For moles of Ar:

$PV = nRT \\ \\ n = \dfrac{PV}{RT}$

$n = \dfrac{3.36 \ atm \times 1 \ 2L }{0.08206 \ L .atm. per. mol. K \times 377 \ K}$

$n = 0.2172 \ mol \ Ar$

$\mathtt{total \ moles = moles \ of \ N_2 + moles \ of \ Ar}$

$=0.08297 mol + 0.2037 mol \\ = 0.2867 mol gases$

Finally;

The final pressure of the mixture is:

$PV = nRT \\ \\ P = \dfrac{nRT}{V} \\ \\ P = \dfrac{0.2867 \ mol \times 0.08206 \ L .atm/mol .K\times 377 K}{4.0 \ L}$

P = 2.217 atm

P ≅ 2.24 bar

7 0

3 years ago

2H2(g)+2NO(g)→2H2O(g)+N2(g) Part A If the concentration of NO changed from 0.100 M to 0.025 M in the first 15 minutes of the rea

andrew11 [14]

Answer:

0.0025 M/min

Explanation:

The rate of a reaction can be calculated for an element, based on its stoichiometric coefficient. For a reaction:

aA + bB = cC + dD , the rate will be

r = -(1/a)x(Δ[A]/Δt) = -(1/b)x(Δ[B]/Δt) = (1/c)x(Δ[C]/Δt) = (1/d)x(Δ[D]/Δt)

Where Δ[X] is the variation of the concentration of the X compound, Δt is the time variation, and the signal of minus in the reagents compounds is because they are disappearing, so Δ[X] will be negative, and r must be positive. So, for the reaction given:

r = -(1/2)x(Δ[NO]/Δt)

r = -(1/2)x( (0.025 - 0.1)/15)

r = 0.0025 M/min

6 0

2 years ago

Sodium phosphate is added to a solution that contains 0.0030 M aluminum nitrate and 0.016 M calcium chloride. The concentration

gavmur [86]

Explanation:

It is given that aluminium nitrate and calcium chloride are mixed together with sodium phosphate.

And, $K_{sp} of AlPO_{4} = 9.84 \times 10^{-21}$

$K_{sp} of Ca_{3}(PO_{4})_{2} = 2.0 \times 10^{-29}$

Let us assume that the solubility be "s". And, the reaction equation is as follows.

$AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

$9.84 \times 10^{-21} = s \times s$

s = $9.92 \times 10^{-11}$

Also, $Ca_{3}(PO_{4})_{2} \rightleftharpoons 3Ca^{2+} + 2PO^{3-}_{4}$

$2 \times 10^{-29} = (3s)^{3} \times (2s)^{2}$

s = $7.14 \times 10^{-7}$

This means that first, aluminium phosphate will precipitate.

Now, we will calculate the concentration of phosphate when calcium phosphate starts to precipitate out using the $K_{sp}$ expression as follows.

$K_{sp} = [Ca^{2+}]^{3}[PO^{3-}_{4}]^{2}$

$2.0 \times 10^{-29} = (0.016)^{3}[PO^{3-}_{4}]^{2}$

$2.0 \times 10^{-29} = 4.096 \times 10^{-6} \times [PO^{3-}_{4}]^{2}$

$[PO^{3-}_{4}]^{2}$ = $4.88 \times 10^{-24}$

= $2.21 \times 10^{-12}$ M

Similarly, calculate the concentration of aluminium at this concentration of phosphate as follows.

$AlPO_{4} \rightleftharpoons Al^{3+} + PO^{3-}_{4}$

$K_{sp} = [Al^{3+}][PO^{3-}_{4}]$

$9.84 \times 10^{-21} = [Al^{3+}] \times 2.21 \times 10^{-12}$

$[Al^{3+}] = 4.45 \times 10^{-9}$ M

Thus, we can conclude that concentration of aluminium will be $4.45 \times 10^{-9}$ M when calcium begins to precipitate.

7 0

2 years ago

a 24.37mol of gas occupied 5.32ml vessel. what mole of the gas should be removed to get a volume of 2.56ml?

Pani-rosa [81]

Mole of the gas should be removed to get a volume of 2.56ml is 10.26 mol.

<h3>Equation :</h3>

Given data,

Volume₁ = 5.32ml

number of moles at 5.32ml n₁ = 24.37mol

Molarity = ?

Volume₂ = 2.56ml

number of moles = ?

First to find molarity of gas using formula,

M = nV

where,

M is molarity

n is numbers of mole

V is volume

So, putting the values of volume₁,

M = 24.37mol x 5.32ml

M = 129.65 mol/ml

So again to get moles of second volume in same formula we have,

M = nV

So,

n = M / V

n₂ = 129.65 / 2.56ml

n = 50.64mol

So,

n₂ - n₁ = 50.64mol - 24.37mol

n = 26.27

So,

mole of the gas removed = 26.27 / 2.56

mole of the gas removed = 10.26mol

To know more about volume :

brainly.com/question/1578538

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4 0

1 year ago