Answer:
Explanation:
The equilibrium mechanism for the reversible acid is catalyzed by the isomerization of non conjugated β, γ- unsaturated ketones, like 3-cyclohexanone to their conjugated α, I²- unsaturated isomers.
Oxygen of the Carbonyl group in the ketone is protonated by the acid and this is followed by the abstraction of an α- hydrogen from the protonated 3-cyclo hexanone to yield ethanol
2-cyclo hexanone can be obtained by acid catalyzation of 3-cyclohexanone isomers through the formation of it's "enol".
We take the derivative of Ohm's law with respect to time: V = IR
Using the product rule:
dV/dt = I(dR/dt) + R(dI/dt)
We are given that voltage is decreasing at 0.03 V/s, resistance is increasing at 0.04 ohm/s, resistance itself is 200 ohms, and current is 0.04 A. Substituting:
-0.03 V/s = (0.04 A)(0.04 ohm/s) + (200 ohms)(dI/dt)
dI/dt = -0.000158 = -1.58 x 10^-4 A/s
The answer for this problem would be:
Assuming non-relativistic momentum, then you have:
ΔxΔp = mΔxΔv = h / (4)
Δv = h / (4πmΔx)
m ~ 1.67e-27 h ~ 6.62e-34,Δx = 4e-15 -->
Δv ~ 6.62e-34 / (4π * 1.67e-27 * 4e-15) ~ 7,886,270 m/s ~ 7.89e6 m/s
That's about 1% of the speed of light, the assumption that it's non-relativistic.
Answer:
hope helps
Explanation:
In a weightless environment a force of 5 Newtons is applied horizontally to the right on a rock with a mass of 1 kg and to a pebble with a mass of 0.1 kg.
The specific heat capacity of the substance is ![0.963 J/g^{\circ}C](https://tex.z-dn.net/?f=0.963%20J%2Fg%5E%7B%5Ccirc%7DC)
Explanation:
When an object of mass m is supplied with a certain amount of energy Q, its temperature increases according to the equation:
![Q=mC_s \Delta T](https://tex.z-dn.net/?f=Q%3DmC_s%20%5CDelta%20T)
where
m is the mass of the object
is its specific heat capacity
is the increase in temperature of the object
In this problem, we have
![Q=300 cal \cdot 4.814 = 1444.2 J](https://tex.z-dn.net/?f=Q%3D300%20cal%20%5Ccdot%204.814%20%3D%201444.2%20J)
m = 50 g
![\Delta T = 20C-(-10C)=30^{\circ}C](https://tex.z-dn.net/?f=%5CDelta%20T%20%3D%2020C-%28-10C%29%3D30%5E%7B%5Ccirc%7DC)
Therefore, we can solve for
to find its specific heat capacity:
![C_s = \frac{Q}{m\Delta T}=\frac{1444.2}{(50)(30)}=0.963 J/gC](https://tex.z-dn.net/?f=C_s%20%3D%20%5Cfrac%7BQ%7D%7Bm%5CDelta%20T%7D%3D%5Cfrac%7B1444.2%7D%7B%2850%29%2830%29%7D%3D0.963%20J%2FgC)
Learn more about specific heat capacity:
brainly.com/question/3032746
brainly.com/question/4759369
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