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
The specific heat capacity the substance is calculated using the below formula
Q(heat) = Mc delta T
Q =1560 cal
m(mass) 312 g
delta T (change in temperature ) = 15 c
C= specific heat capacity=?
by making c the subject of the formula
c=Q/m delta T
= 1560 cal/ 312g x 15 c = 0.33 cal/g/c (answer B)
Answer:
Yes.
Explanation:
It should be noted that the meaning of molarity is the ratio of moles of solute per liter of solution.
It should be understood that when determining or finding the molarity of an unknown compound ,the process should be performed or carried out at least 3 times. This is done to remove any form of doubt.
The first calculated value for the concentration of the compound will be regarded as rough value, while the second and the third will be regarded as the first and second values respectively.
In this case, the third value for the concentration of HCl will be calculated to for confirmation of other value, that is to be finally sure of its concentration.
Answer:
Scandium(III) fluoride, ScF3, is an ionic compound. It is slightly soluble in water but dissolves in the presence of excess fluoride to form the ScF63− anion.
hope it will help you......
Im pretty sure it would be d.
