Polylactic acid is the correct answer
Answer:
The answer is "Option B"
Explanation:
From the query, the following knowledge is derived:
Yield in percentage = 47%
Performance of theory = 4860 g
Actual yield Rate =?
The percentage return is defined simply by the ratio between both the real return as well as the conceptual return multiplied by the 100. It's also represented as numerically:
Now We can obtain the percent yield as followed using the above formula:
The value of the Actual yield Rate =
The Actual yield Rate= 2284.2 g.
The half life equation is -->P(t) = Pi (0.5) ^ (t/c)
c is equal to the element to reach its half-life (5 seconds)t is equal to the duration of time the element is expose to (20 seconds)Pi is the initial amount (340)0.5 is the base of this exponential function to represent half-life.P(t) is the expression for the function of time
P(20) = 340 (0.5)^20/5P(20) = 340 (0.5) ^4P(20)= 21.25 grams
Fraction = P(t)/Pi = P(20)/Pi =21.25/340 =1/16
Therefore, when given 20 seconds, 340 grams of Fluorine-21 will degrade to 21.25 grams OR 1/16 of its original mass.
Hope this method helps! (This is my answer btw, I think you may have accidentally posted twice?)
Answer:
0.72 g of the lower oxide gave 0.8 g of higher oxide when oxidised. ... Thus, 90g of lower oxide contains as much metal as 100g of higher oxide, i.e., 80g (given). Hence, 80g of metal combines with 10g of oxygen in the lower oxide and 20g of oxygen in the higher oxide.
