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2 years ago

If isomerization requires breaking the pi bond, what minimum energy is required for isomerization in j/mol?

Physics

1 answer:

aliina [53]2 years ago

6 0

<span>The minimum energy required for isomerization is 267 000 J/mol
</span>

The isomerization of cis-but-2-ene to trans-but-2-ene requires breaking of the π bond.

The bond energy of a C-C σ bond is 347 kJ/mol.

The bond energy of a C=C double bond (σ + π) is 614 kJ/mol.

So the bond energy of a π bond is (614 – 347) kJ/mol = 267 kJ/mol =
267 000 J/mol.

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--->Two aircraft P and Q are flying at the same speed. 300 m/s, The direction along which P is flying is at right angles to t

REY [17]

Answer:

The magnitude of the velocity of the aircraft P relative to aircraft Q is zero

Explanation:

The velocity of the two aircraft, P & Q, v = 300 m/s

The angle of the direction between them, Ф = 90°

The magnitude of the velocity of aircraft P relative to aircraft Q is given by the formula

Substituting the values in the above equation

v = 300 x cos 90°

= 300 x 0

= 0

Since the aircraft are at right angles, the velocity of one aircraft relative to the other is zero.

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2 years ago

Hippos spend much of their lives in water, but amazingly, they don’t swim. manatees, They have, like little very body fat. The d

kenny6666 [7]

Answer:

428.59 N

Explanation:

Buoyant force, $B=Vg\rho$ where V is volume, g is gravitational constant and \rho is density

$B+F_{upward}=mg$ where $F_{upward}$ is upward force

$Vg\rho_{w}+F_{upward}=mg$

$F_{upward}=mg- Vg\rho_{w}$

$F_{upward}=g(mg- V\rho_{w})=g(m-m\frac {\rho_{w}{\rho_{hippo}}$ where $\rho_{hippo}$ is the density of hippo

$F_{upward}=mg(1-\frac {\rho_{w}}{\rho_{hippo}})$

Using g as 9.81

$F_{upward}=1500*9.81*(1-1000/1030)= 428.5922 N$

Therefore, the upward force=428.59 N

3 0

2 years ago

How much energy is needed to change the speed of a 1600 kg sport utility vehicle from 15.0 m/s to 40.0 m/s?

podryga [215]

Answer:

Energy needed = 1100 kJ

Explanation:

Energy needed = Change in kinetic energy

Initial velocity = 15 m/s

Mass, m = 1600 kg

$\texttt{Initial kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 15^2=180000J$

Final velocity = 40 m/s

$\texttt{Final kinetic energy = }\frac{1}{2}mv^2=0.5\times 1600\times 40^2=1280000J$

Energy needed = Change in kinetic energy = 1280000-180000 = 1100000J

Energy needed = 1100 kJ

4 0

2 years ago

Drag each tile to the correct location.

In-s [12.5K]

Answer:

Look at the image please

Explanation:

5 0

2 years ago

100-ft-long horizontal pipeline transporting benzene develops a leak 43 ft from the high-pressure end. The diameter of the leak

Amanda [17]

Answer:

Explanation:

The mass flow rate of benzene from the leak in the pipeline containing benzene is:

$Q_m=AC_o\sqrt{2\rho g_cP_g}$

Here, $Q_m$ is the mass flow rate through the leak of the pipeline. $A$ is the area of the hole, $C_o$ is the discharge rate, $\rho$ is the fluid density, $g_c$ is the gravitational constant and $P_g$ is the constant gauge pressure within the process unit.

The diametre of the leak (d) is 0.1 in. Convert from in to ft.

$d=(0.1 in)(\frac{1ft}{12in})\\=8.33\times 10^{-3}ft$

Calculate the area (A) of the hole. The area of the hole is.

$A=\frac{\pi d^2}{4}$

Substitute 3.14 for $\pi$ and $8.33\times 10^{-3}ft$ for d and calculate A.

$A=\frac{\pi d^2}{4}\\\\\frac{(3.14)(8.33\times 10^{-3})^2}{4}\\\\5.45\times 10^{-5}ft^2$

The specific gravity of benzene is 0.8794. Specific gravity is the ratio of th density of a substance to the density of a reference substance.

Specific gravity of benzene = density of benzenee/denity of reference substance

Rewrite the expression in terms of density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

Take the reference substance as water. Density of water is $62.4\frac{Ib_m}{ft^3}$ . Calculate density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

$=(0.8794)(62.4\frac{Ib_m}{ft^3})\\\\54.9\frac{Ib_m}{ft^3}$

Calculate the pressure at the point of leak. The pressure is the average of the pressure of the high and low pressure end. Write the expression to calculate the average pressure.

Upstream x distance from upstream pressure end

$P_g$ =+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

Calculate the distance from the downstream pressure end. The distance from upstream pressure end is 43 ft. Total of the pipe is 100 ft.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

The distance from upstream pressure end is 43 ft. Total length of the pipe is 100 ft. Substitute the values in the equation.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

= 100ft - 43ft = 57 ft

Substitute 50 psig for upstream, 43 ft fr distance from the upstream pressure end, 40 psig for downstream pressure, 57 ft for distance from the downstream pressure end, and 100 ft for the total length of the horizontal pipeline and calculate $P_g$ .

Upstream x distance from upstream pressure end

$P_g$ =+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

$=\frac{(50psig\times 43ft)+(40psig \times 57ft)}{100ft}\\\\=44.3psig$

Convert the pressure from psig to $Ib_f/ft^2$

$P_g=(44.3psig)(\frac{1\frac{Ib_f}{ft^2}}{1psig})(144\frac{in^2}{ft^2})\\\\=6,379.2\frac{Ib_f}{ft^2}$

The leak is like a sharp orifice. Take the value of the discharge coefficient as 0.61.

Substitute $5.45\times 10^{-5}ft^2$ for A. 0.61 for $C_o$ , $54.9\frac{Ib_m}{ft^3}$ for $\rho$ , $32.17\frac{ft.Ib_m}{Ib_f.s^2}$ for $g_c$ , and $6,379.2\frac{Ib_f}{ft^2}$ for $P_g$ and calculate $Q_m$

$Q_m=AC_o\sqrt{2\rho g_cP_g}\\\\=(5.45\times 10^{-5}ft^2)(0.61)\sqrt{2(54.9\frac{Ib_m}{ft^3})(32.17\frac{ft.Ib_m}{Ib_f.s^2})(6,379.2\frac{Ib_f}{ft^2})}\\\\(3.3245\times 10^{-5}ft^2)\sqrt{22,533,031.21\frac{Ib^2_m}{ft^4.s^2}}\\\\=0.158\frac{Ib_m}{s}$

The mass flow rate of benzene through the leak in the pipeline is $0.158\frac{Ib_m}{s}$

8 0

2 years ago