Answer:
Alcohols are usually named by the first procedure and are designated by an -ol suffix, as in ethanol, CH3CH2OH (note that a locator number is unnecessary on a two-carbon chain). On longer chains the location of the hydroxyl group determines chain numbering. For example: (CH3)2C=CHCH(OH)CH3 is 4-methyl-3-penten-2-ol.10
Explanation:
Reactivity is not a physical property because it's a characteristic of a chemical element or compound and is involved in chemical process.
A four cycle engine works with 4 basic steps to a successful rotation of the crankshaft: the intake, compression, power and exhaust stroke. Each engine cylinder has four openings for the intake, exhaust, spark plug and fuel injection. ... The compression makes the air-fuel combination volatile for easier ignition.
What kind of control like scientific?
Explanation:
Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.
a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement a is wrong.
b. Expression for second order reaction is as follows:
![\frac{1}{[A]} =\frac{1}{[A]_0} +kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%20%2Bkt)
the above equation can be written in the form of Y = mx + C
so, the plot between 1/[A] and t is linear. So the statement b is true.
c.
Expression for half life is as follows:
![t_{1/2}=\frac{1}{k[A]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BA%5D_0%7D)
As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.
d.
Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong