Which type of investigation is lised to compare two subiects?

Complete the electron-pushing mechanism for the e1 reaction when 2-methylbutan-2-ol is treated with 20% sulfuric acid.

Sav [38]

E1 reaction works in the mechanism that the removal of an HX substituent results in the formation of a double bond. The E1 reaction for 2-methylbutan-2-ol is shown in the figure. This reaction is called acid-catalyzed dehydration of a tertiary alcohol.

The mechanism works in three major steps:
1. The OH group of the main reactant is hydrated by H2SO4 so it becomes H2O.
2. The H2O leaves taking electrons with it. This results to a carbocation intermediate on the carbon atom where it was attached.
3. Another H2O protonates the beta carbon. This is the carbon atom next to the carbocation. It will donate its electrons to the neighboring C-C bond, as indicated by the arrow. The carbons are rehybridized from sp3 to sp2, which is a pi bond. As a result, a double bond forms.

The product is 2-methyl-2-butene.

6 0

4 years ago

Read 2 more answers

A 37 kg object has an applied force of 85N [R] acting on it. The coefficient of

Sliva [168]

Answer:

Explanation:

This is quite tricky! You need to do 2 different equations to solve all the parts of this problem. First is finding the acceleration in one dimension, which has an equation of

F - f = ma

where F is the applied Friction,

f is the frictional force acting against F,

m is the mass of the object, and

a is the acceleration of the object (NOT the velocity!)

This is Newton's Second Law expanded on a bit. The sum of the forces working on an object is equal to the object's mass times its acceleration. We have F, but we need f which is found in the equation

f = μ $F_n$ which is the coefficient of kinetic friction times the weight of the object. Weight is found in the equation

w = mg where m is mass and g is the pull of gravity. Let's start there and work backwards:

w = 37(9.8) to 2 sig figs so

w = 360N. Now fill that in to find f:

f = (.17)(360) to 2 sig figs so

f = 61. Now for the final answer in the original equation way back up at the top:

85 - 61 = 37a and do the subtraction on the left side first:

24 = 37a and then we divide to 2 sig figs to get

a = .65 m/s/s

Since we are moving in a straight line (as opposed to on an angle) the displacement is found in

d = rt which simply says that the distance an object moves is equal to its rate times the time. Therefore,

d = 2.2(3.4) to 2 sig figs so

d = 7.5 m

6 0

3 years ago

The flywheel of a steam engine runs with a constant angular velocity of 190 rev/min. When steam is shut off, the friction of the

Ivanshal [37]

Answer:

(a) $\alpha = - 1.32\ rev/m^{2}$

(b) $\theta = 13674\ rev$

(c) $\alpha_{tan} = 8.75\times 10^{- 4}\ m/s^{2}$

(d) $a = 22.458\ m/s^{2}$

Solution:

As per the question:

Angular velocity, $\omega = 190\ rev/min$

Time taken by the wheel to stop, t = 2.4 h = $2.4\times 60 = 144\ min$

Distance from the axis, R = 38 cm = 0.38 m

Now,

(a) To calculate the constant angular velocity, suing Kinematic eqn for rotational motion:

$\omega' = \omega + \alpha t$

$omega'$ = final angular velocity

$\omega$ = initial angular velocity

$\alpha$ = angular acceleration

Now,

$0 = 190 + \alpha \times 144$

$\alpha = - 1.32\ rev/m^{2}$

Now,

(b) The no. of revolutions is given by:

$\omega'^{2} = \omega^{2} + 2\alpha \theta$

$0 = 190^{2} + 2\times (- 1.32) \theta$

$\theta = 13674\ rev$

(c) Tangential component does not depend on instantaneous angular velocity but depends on radius and angular acceleration:

$\alpha_{tan} = 0.38\times 1.32\times \frac{2\pi}{3600} = 8.75\times 10^{- 4}\ m/s^{2}$

(d) The radial acceleration is given by:

$\alpha_{R} = R\omega^{2} = 0.32(80\times \frac{2\pi}{60})^{2} = 22.45\ rad/s$

Linear acceleration is given by:

$a = \sqrt{\alpha_{R}^{2} + \alpha_{tan}^{2}}$

$a = \sqrt{22.45^{2} + (8.75\times 10^{- 4})^{2}} = 22.458\ m/s^{2}$

5 0

3 years ago

When two forces or more forces act on an object, the forces are

sashaice [31]

The forces are net forces. I think this is correct.

5 0

3 years ago

Two loudspeakers (A and B) are 3.20m apart and emitting a sound with a frequency of 400Hz. An observer is 2.10m directly in fron

TiliK225 [7]

Answer:

The observer hears a loud sound

Explanation:

In order to know if the observer hears a loud or a quiet sound, you need to know if there is a constructive or destructive interference between the sound waves of the loudspeakers.

You first calculate the distance between the observer and the loudspeakers.

The distances are given by:

d1: distance to loudspeaker A = 2.10m

d2: distance to loudspeaker B

$d_2=\sqrt{(3.20m)^2+(2.10m)^2}=3.827m$