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BabaBlast [244]

3 years ago

13

Calculate the standard entropy of vaporization of ethanol at its boiling point 285 K. The standard molar enthalpy of vaporizatio

n of ethanol at its boiling point is 40.5 kJ/mol

2 answers:

kkurt [141]3 years ago

7 0

Answer:

The standard entropy of vaporization is 142 J/mol*K

Explanation:

Step 1: Data given

The boiling point = 285 Kelvin

The standard molar enthalpy of vaporization of ethanol at its boiling point is 40.5 kJ/mol

Step 2: Calculate the standard entropy of vaporization

ΔS = ΔH /T

⇒ with ΔS = the change in entropy

⇒ with ΔH = enthalpy of vaporization of ethanol = 40.5 kJ/mol = 40.5 *10³ J/mol

⇒ with T = the temperature =285 Kelvin

ΔS = (40.5 * 10³ J/mol)/ 285 Kelvin

ΔS = 142.1 ≈ 142 J/mol*K

The standard entropy of vaporization is 142 J/mol*K

(Note: The boiling point of ethanol is not 285K but 352 K)

Ivanshal [37]3 years ago

3 0

Answer:

standard entropy of vaporization of ethanol = 142.105 J/K-mol

Explanation:

given data

enthalpy of vaporization of ethanol = 40.5 kJ/mol = 40.5 × $10^{3}$ J/mol

entropy of vaporization of ethanol boiling point = 285 K

to find out

standard entropy of vaporization of ethanol

solution

we get here standard entropy of vaporization of ethanol that is expess as

standard entropy of vaporization of ethanol ΔS = $\frac{\Delta H}{T}$ .............1

here ΔH is enthalpy of vaporization of ethanol and T is temperature

put value in equation 1

standard entropy of vaporization of ethanol ΔS = $\frac{40.5*10^3}{285}$

standard entropy of vaporization of ethanol = 142.105 J/K-mol

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Calculate the standard enthalpy of formation of NOCl(g) at 25 ºC, knowing that the standard enthalpy of formation of NO(g) at th

stepan [7]

Answer:

The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol

Explanation:

The ∆H (heat of reaction) of the combustion reaction is the heat that accompanies the entire reaction. For its calculation you must make the total sum of all the heats of the products and of the reagents affected by their stoichiometric coefficient (number of molecules of each compound that participates in the reaction) and finally subtract them:

Enthalpy of the reaction= ΔH = ∑Hproducts - ∑Hreactants

In this case, you have: 2 NOCl(g) → 2 NO(g) + Cl₂(g)

So, ΔH= $2*H_{NO} +H_{Cl_{2} }-2*H_{NOCl}$

Knowing:

ΔH= 75.5 kJ/mol
$H_{NO}$ = 90.25 kJ/mol
$H_{Cl_{2} }$ = 0 (For the formation of one mole of a pure element the heat of formation is 0, in this caseyou have as a pure compound the chlorine Cl₂)
$H_{NOCl}$ =?

Replacing:

75.5 kJ/mol=2* 90.25 kJ/mol + 0 - $H_{NOCl}$

Solving

- $H_{NOCl}$ =75.5 kJ/mol - 2*90.25 kJ/mol

- $H_{NOCl}$ =-105 kJ/mol

$H_{NOCl}$ =105 kJ/mol

<u><em>The standard enthalpy of formation of NOCl(g) at 25 ºC is 105 kJ/mol</em></u>

8 0

3 years ago

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strojnjashka [21]

Answer:

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7 0

3 years ago

How many grams are in 3.14 moles of PI₃?

OverLord2011 [107]

Answer:

$\boxed {\boxed {\sf 1290 \ g \ PI_3}}$

Explanation:

We want to convert from moles to grams, so we must use the molar mass.

<h3>1. Molar Mass</h3>

The molar mass is the mass of 1 mole of a substance. It is the same as the atomic masses on the Periodic Table, but the units are grams per mole (g/mol) instead of atomic mass units (amu).

We are given the compound PI₃ or phosphorus triiodide. Look up the molar masses of the individual elements.

Phosphorus (P): 30.973762 g/mol
Iodine (I): 126.9045 g/mol

Note that there is a subscript of 3 after the I in the formula. This means there are 3 moles of iodine in 1 mole of the compound PI₃. We should multiply iodine's molar mass by 3, then add phosphorus's molar mass.

I₃: 126.9045 * 3=380.7135 g/mol

PI₃: 30.973762 + 380.7135 = 411.687262 g/mol

<h3>2. Convert Moles to Grams</h3>

Use the molar mass as a ratio.

$\frac {411.687262 \ g \ PI_3}{ 1 \ mol \ PI_3}$

We want to convert 3.14 moles to grams, so we multiply by that value.

$3.14 \ mol \ PI_3 *\frac {411.687262 \ g \ PI_3}{ 1 \ mol \ PI_3}$

The units of moles of PI₃ cancel.

$3.14 *\frac {411.687262 \ g \ PI_3}{ 1 }$

$1292.698 \ g\ PI_3$

<h3>3. Round</h3>

The original measurement of moles has 3 significant figures, so our answer must have the same. For the number we calculated, that is the tens place.

1292.698

The 2 in the ones place tells us to leave the 9.

$1290 \ g \ PI_3$

3.14 moles of phosphorous triiodide is approximately equal to <u>1290 grams of phosphorus triodide.</u>

4 0

3 years ago

Perform the calculation and record the answer with the correct number of significant figures.

kotykmax [81]

The question is incomplete, here is the complete question:

A. (6.5-6.10)/3.19

B. (34.123 + 9.60) / (98.7654 - 9.249)

<u>Answer:</u>

<u>For A:</u> The answer becomes 0.1

<u>For B:</u> The answer becomes 0.4884

<u>Explanation:</u>

Significant figures are defined as the figures present in a number that expresses the magnitude of a quantity to a specific degree of accuracy.

Rules for the identification of significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant. For example: 664, 6.64 and 66.4 all have three significant figures.
All zeros between the integers are always significant. For example: 5018, 5.018 and 50.18 all have four significant figures.
All zeros preceding the first integers are never significant. For example: 0.00058 has two significant figures.
All zeros after the decimal point are always significant. For example: 2.500, 25.00 and 250.0 all have four significant figures.
All zeroes used solely for spacing the decimal point are not significant. For example: 10000 has one significant figure.

<u>Rule applied for addition and subtraction:</u>

The least precise number present after the decimal point determines the number of significant figures in the answer.

<u>Rule applied for multiplication and division:</u>

In case of multiplication and division, the number of significant digits is taken from the value which has least precise significant digits

<u>For A:</u> (6.5-6.10)/3.19

This a a problem of subtraction and division.

First, the subtraction is carried out.

$\Rightarow \frac{6.5-6.10}{3.19}=\frac{0.4}{3.19}$

Here, the least precise number after decimal was 1.

$\Rightarrow \frac{0.4}{3.19}=0.125$

Here, the least precise number of significant digit is 1. So, the answer becomes 0.1

<u>For B:</u> (34.123 + 9.60) / (98.7654 - 9.249)

This a a problem of subtraction, addition and division.

First, the subtraction and addition is carried out.

$\Rightarow \frac{34.123+9.60}{98.7654-9.249}=\frac{43.723}{89.5164}=\frac{43.72}{89.516}$

Here, the least precise number after decimal in addition are 2 and in subtraction are 3

$\Rightarrow \frac{43.72}{89.516}=0.48840$

Here, the least precise number of significant digit are 4. So, the answer becomes 0.4884

6 0

3 years ago

Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of

lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

$U_{2}-U_{1}=Q-W$

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

$dw=Pdv$

The pressure is constant, so: $w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg}$ (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

$U_{2}-U_{1}=500-(3*75)=275kJ$

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

$P_{1}V_{1}=nRT_{1}$

$P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}$

Subtracting the first from the second:

$1.5P_{1}V_{1}=nR(T_{2}-T_{1})$

Isolating $T_{2}$ :

$T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}$

Assuming that it is water steam, n=0.1666 kmol

$V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}$

$T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76$ ºC

7 0

3 years ago