Answer:
total time = 304.21 s
Explanation:
given data
y = 50% = 0.5
n = 1.1
t = 114 s
y = 1 - exp(-kt^n)
solution
first we get here k value by given equation
y = 1 -
...........1
put here value and we get
0.5 = 1 - e^{(-k(114)^{1.1})}
solve it we get
k = 0.003786 = 37.86 ×
so here
y = 1 -
1 - y =
take ln both side
ln(1-y) = -k ×
so
t =
.............2
now we will put the value of y = 87% in equation with k and find out t
t = ![\sqrt[1.1]{-\frac{ln(1-0.87)}{37.86*10^{-4}}}](https://tex.z-dn.net/?f=%5Csqrt%5B1.1%5D%7B-%5Cfrac%7Bln%281-0.87%29%7D%7B37.86%2A10%5E%7B-4%7D%7D%7D)
total time = 304.21 s
Castor oil is increasingly becoming an important bio-based raw material for industrial applications. The oil is non-edible and can be extracted from castor seeds from the castor plant belonging to the family Euphorbiaceae. The oil is a mixture of saturated and unsaturated fatty acid esters linked to a glycerol. The presence of hydroxyl group, a double bond, carboxylic group and a long chain hydrocarbon in ricinoleic acid (a major component of the oil), offer several possibilities of transforming it into variety of materials. The oil is thus a potential alternative to petroleum-based starting chemicals for the production of materials with variety of properties. Despite this huge potential, very little has recently been reviewed on the use of castor oil as a bio-resource in the production of functional materials. This review therefore highlights the potential of castor oil in the production of these diverse materials with their projected global market potential. The review gives the background information of castor oil and its geographical availability, the properties and its uses as bio-based resource for synthesis of various materials. The review further highlights on the use of castor oil or ricinoleic acid as a green capping agent in the synthesis of nanomaterials.

Answer:
See explanation
Explanation:
The complete question is
Write a SELECT statement that returns the same result set as this SELECT statement, but don't use a
join. Instead, use a subquery in a WHERE clause that uses the IN keyword.
SELECT DISTINCT category_name
FROM categories c JOIN products p
ON c.category_id = p.category_id
ORDER BY category_name
The join clause gives us all the registers of both tables, for which the category_id of categories coincides to the category_id in the table products. That is, we can first select the distinct category_id's from products and the see the categories from categories table, whose id is in that list. That means writing the following query:
Select distinct category_name
from categories c
where c.category_id in (
select distinct p.category_id
from products p)
order by category_name
Input: what is put in, taken in, or operated on by any process or system.
Output: the amount of something produced by a person, machine, or industry.