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My name is Ann [436]

3 years ago

6

A sample of benzene was vaporized at 25◦C. When 37.5 kJ of heat was supplied, 95.0 g of the liquid benzene vaporized. What is th

e enthalpy of vaporization of benzene at 25◦C?
1. 30.8 kJ/mol
2. 13.4 kJ/mol
3. 43.0 kJ/mol
4. 6.09 kJ/mol
5. 25.6 kJ/mol

1 answer:

grandymaker [24]3 years ago

8 0

Answer:

Enthalpy of vaporization = 30.8 kj/mol

Explanation:

Given data:

Mass of benzene = 95.0 g

Heat evolved = 37.5 KJ

Enthalpy of vaporization = ?

Solution:

Molar mass of benzene = 78 g/mol

Number of moles = mass/ molar mass

Number of moles = 95 g/ 78 g/mol

Number of moles = 1.218 mol

Enthalpy of vaporization = 37.5 KJ/1.218 mol

Enthalpy of vaporization = 30.8 kj/mol

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How many atoms altogether make up one molecule of glucose

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Complete and balance the following redox equation using the set of smallest whole– number coefficients. Now sum the coefficients

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Answer : The balanced chemical equation in a acidic solution is,

$BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)$

The sum of the coefficients is, 17

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion $(H^+)$ at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

$BrO_3^-(aq)+Sb^{3+}(aq)\rightarrow Br^-(aq)+Sb^{5+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-$

First balance the main element in the reaction.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-$

Now balance oxygen atom on both side.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-\rightarrow Br^-+3H_2O$

Now balance hydrogen atom on both side.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}$

Reduction : $BrO_3^-+6H^+\rightarrow Br^-+3H_2O$

Now balance the charge.

Oxidation : $Sb^{3+}\rightarrow Sb^{5+}+2e^-$

Reduction : $BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O$

The charges are not balanced. Now multiplying oxidation reaction by 3 and then adding both equation, we get the balanced redox reaction.

Oxidation : $3Sb^{3+}\rightarrow 3Sb^{5+}+6e^-$

Reduction : $BrO_3^-+6H^++6e^-\rightarrow Br^-+3H_2O$

The balanced chemical equation in acidic medium will be,

$BrO_3^-(aq)+6H^+(aq)+3Sb^{3+}(aq)\rightarrow Br^-(aq)+3H_2O(l)+3Sb^{5+}(aq)$

The sum of the coefficients = 1 + 6 + 3 + 1 + 3 + 3

The sum of the coefficients = 17

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3 years ago

Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll

makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature $T_1$ = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

$\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}$

where; B = - $152.5 \ cm^3 /mol$ C = -5800 $cm^6/mol^2$

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

$\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

$4.138*10^{-4} \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}$

Multiplying through with V² ; we have

$4.138*10^4 \ V ^3 = V^2 - 152.5 V - 5800 = 0$

$4.138*10^4 \ V ^3 - V^2 + 152.5 V + 5800 = 0$

V = 2250.06 cm³ mol⁻¹

Z = $\frac{PV}{RT}$

Z = $\frac{1800*2250.06}{8.314*10^3*523.15}$

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

$T_c = 647.1 \ K \\ \\ P_c = 22055 \ kPa \\ \\ \omega = 0.345$

$T__{\gamma}} = \frac{T}{T_c}$

$T__{\gamma}} = \frac{523.15}{647.1}$

$T__{\gamma}} = 0.808$

$P__{\gamma}} = \frac{P}{P_c}$

$P__{\gamma}} = \frac{1800}{22055}$

$P__{\gamma}} = 0.0816$

$B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}$

$B_o = 0.083 - \frac{0.422}{0.808^{1.6}}$

$B_o = 0.51$

$B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}$

$B_1 = -0.282$

The compressibility is calculated as:

$Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}$

$Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}$

Z = 0.9386

$V= \frac{ZRT}{P}$

$V= \frac{0.9386*8.314*10^3*523.15}{1800}$

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At $T_1 = 523.15 \ K \ and \ P = 1800 \ k Pa$

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V = 0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = $\frac{PV}{RT}$

Z = $\frac{1800*10^3 *0.1249}{729.77*523.15}$

Z = 0.588

3 0

3 years ago

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yaroslaw [1]

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8 0

3 years ago