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2 years ago

13

Draw a structural formula for an alkyne of molecular formula c8h14 that yields 3-ethylhexane on catalytic hydrogenation. click t

he "draw structure" button to launch the drawing utility.

2 answers:

saveliy_v [14]2 years ago

6 0

Answer:

3-ethyl-1-hexyne or 4-ethyl-1-hexyne.

Explanation:

Hello,

By means of the more likely reagents to undergo the stated hydrogentation via catalysis with palladium or platinum, its structure is shown on the attached picture which is eligible for either 3-ethyl-1-hexyne or 4-ethyl-1-hexyne since they lead to the very same product. Do not forget that such hydrogenation covers up all the insaturations turning out into the specified alkane.

Best regards.

OverLord2011 [107]2 years ago

3 0

If it an alkyne and gives 3-ethylhexane on hydrogenation then the reaction will be

There will be three isomers of alkyne with given molecular formula

The reactions are shown in the attached figure

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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

2 years ago

A particular first-order reaction has a rate constant of 1.35 × 102 s-1 at 25.0°c. what is the magnitude of k at 75.0°c if ea =

IgorC [24]

According to this formula:
K= A*(e^(-Ea/RT) when we have K =1.35X10^2 & T= 25+273= 298K &R=0.0821
Ea= 85.6 KJ/mol So by subsitution we can get A:
1.35x10^2 = A*(e^(-85.6/0.0821*298))
1.35x10^2 = A * 0.03
A= 4333
by substitution with the new value of T(75+273) = 348K & A to get the new K
∴K= 4333*(e^(-85.6/0.0821*348)
= 2.16 x10^2

8 0

2 years ago

A precipitate forms when mixing solutions of sodium fluoride (NaF) and lead II nitrate (Pb(NO3)2). Complete and balance the net

Margarita [4]

Answer:

Pb^2+(aq) + 2F-(aq) → PbF2(s)

Explanation:

Step 1: Data given

sodium fluoride = NaF

lead(II)nitrate Pb(NO3)2

Step 2: The unbalanced equation

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + NaNO3(aq)

Step 3: Balancing the equation

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + NaNO3(aq)

On the left side we have 2x NO3 (in Pb(NO3)2), on the right side we have 1x NO3 (in NaNO3). To balance the amount of NO3 we hvae to multiply NaNO3 on the right side by 2.

NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + 2NaNO3(aq)

On the left side we have 1x Na (in NaF), on the right side we have 2x Na (in 2NaNO3). To balance the amount of Na we have to multiply NaF on the left side by 2. Now the equation is balanced.

2NaF(aq) + Pb(NO3)2(aq) → PbF2(s) + 2NaNO3(aq)

Step 4: Calculate net ionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

2Na+(aq) + 2F-(aq) + Pb^2+(aq) + 2NO3-(aq) → PbF2(s) + 2Na+(aq) + NO3-(aq)

Pb^2+(aq) + 2F-(aq) → PbF2(s)

4 0

2 years ago

Which choice(s) correctly rank(s) the bonds in terms of increasing polarity?

Marina CMI [18]

Answer:

(II) only correctly rank the bonds in terms of increasing polarity.

Explanation:

Bond polarity is proportional to difference in electronegativity between bonded atoms.

Atoms Electronegativity Bond Electronegativity difference

Cl 3.0 Cl-F 1.0

Br 2.8 Br-Cl 0.2

F 4.0 Cl-Cl 0

H 2.1 H-C 0.4

C 2.5 H-N 0.9

N 3.0 H-O 1.4

O 3.5 Br-F 1.2

I 2.7 I-F 1.3

Si 1.9 Cl-F 1.0

P 2.2 Si-Cl 1.1

Si-P 0.3

Si-C 0.6

Si-F 2.1

So, clearly, order of increasing polarity : O-H > N-H > C-H

So, (II) only correctly rank the bonds in terms of increasing polarity

4 0

2 years ago

If the speed of a vehicle increases by 22%, by what factor does its kinetic energy increase?

Mila [183]

Answer:

The kinetic increases by 48.84 %

Explanation:

The expression for the kinetic energy is:-

$K.E.=\frac{1}{2}\times mv^2$

Where, m is the mass of the object

v is the velocity of the object

Let the new velocity is:- v'

v is increased by 22 %. Thus, v' = 1.22 v

So, the new kinetic energy is:-

$K.E.'=\frac{1}{2}\times mv'^2=\frac{1}{2}\times m{1.22}^2=\frac{1}{2}\times mv^21.4884=1.4884K.E.$

<u>Thus, the kinetic increases by 48.84 %</u>

7 0

2 years ago

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