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vodka [1.7K]
3 years ago
8

The normal boiling point of ethanol is 78.4 °C and 101.3 KPa. The heat of vaporization for ethanol is 42.32 kJ/mol. Determine th

e vapor pressure of ethanol at 95.0 °C. Use the Clasius Clapeyron Equation: Ln [P2/P1] = - [∆Hvap /R] * [1/T2 - 1/T1]

Chemistry
2 answers:
Mandarinka [93]3 years ago
7 0

Answer:

6.1×10^4Pa or 61KPa

Explanation:

The Clausius-Clapeyron equation is used to estimate the vapour pressure at different temperature, once the enthalpy of vaporization and the vapor pressure at another temperature is given in the question. The detailed solution is shown in the image attached. The temperatures were converted to kelvin and the energy value was converted from kilojoule to joule since the value of the gas constant was given in unit of joule per mole per kelvin. The fact that lnx=2.303logx was also applied in the solution.

nekit [7.7K]3 years ago
6 0

Answer:

169.4 kpa

Explanation:

Okay, we are given the following parameters from the question and they are; normal boiling point of ethanol = 78.4 °C( 351.4 k) , pressure= 101.3 kpa = 101.3 × 10^3 pa, heat of vaporization for ethanol = 42.32 kJ/mol = 42.32 ×10^3 J/mol, vapor pressure of ethanol at 95.0 °C(368 k)= ??(not given). The question asked us to use the Clapeyron Equation for this particular Question,all we have to do is to slot in the values given and evaluate, soft !.

The Clapeyron Equation is given below;

==> ln [P2/P1] = - [∆Hvap /R] × [1/T2 - 1/T1].

ln [ P2/ 101.3 × 10^3] = - [42.32 × 10^3/ 8.314] × [ 1/368 - 1/ 351.4].

==> ln P2 - ln 101.3 × 10^3 =

==> ln P2 - 11.53 = - 5090.21 × [0.0027 - 0.0028).

==>ln P2 = 0.51 + 11.53.

==> ln P2 = 12.04.

P2 = e^12.04.

P2= 169396.940460334134 pa.

P2 = 169.4 kpa.

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givi [52]

Answer:

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Explanation:

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3 molecules of BaCl_{2} produce 2 molecules of AlCl_{3}

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We know 1 mol of molecule = N_{A} number of molecules.

So 3 moles of BaCl_{2} produce 2 moles of AlCl_{3}.

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3 0
3 years ago
How many moles of calcium chloride, cacl2, can be added to 1.5 l of 0.020 m potassium sulfate, k2so4, before a precipitate is ex
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When CaSO4 → Ca2+ + SO4
So when we have Ksp = [Ca2+][SO4]

when Ksp = 4.93 x 10^-5
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4 0
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At a certain concentration of H2 and NH3, the initial rate of reaction is 0.120 M / s. What would the initial rate of the reacti
mel-nik [20]

The question is incomplete, here is the complete question:

The rate of certain reaction is given by the following rate law:

rate=k[H_2]^2[NH_3]

At a certain concentration of H_2 and [tex]I_2, the initial rate of reaction is 0.120 M/s. What would the initial rate of the reaction be if the concentration of [tex]H_2 were halved.Answer : The initial rate of the reaction will be, 0.03 M/sExplanation :Rate law expression for the reaction:[tex]rate=k[H_2]^2[NH_3]

As we are given that:

Initial rate = 0.120 M/s

Expression for rate law for first observation:

0.120=k[H_2]^2[NH_3] ....(1)

Expression for rate law for second observation:

R=k(\frac{[H_2]}{2})^2[NH_3] ....(2)

Dividing 2 by 1, we get:

\frac{R}{0.120}=\frac{k(\frac{[H_2]}{2})^2[NH_3]}{k[H_2]^2[NH_3]}

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3 years ago
Draw the product formed when 2-butanol undergoes reaction with TsCl and Et3N. CH3C6H4SO2Cl.
Pie

Answer:

Explanation:

In organic chemistry, the reaction between 2-butanol with TsCl and Et3N is known as the tosylation of the alcohol hydroxyl group. Alcohol is being changed to tosylate by the use of tosyl chloride under the influence of a base. Tosylation of alcohol is an example of a nucleophilic substitution reaction. From the image attached below, we will see how the reaction between 2-butanol proceed into the product by using tosyl chloride and a base(Et3N).

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Mandarinka [93]

Answer:

Weak electrolytes are

HgCl2 and NH₃

Explanation:

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  • Electrolytes may therefore be decomposed by passing electric current through them.
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