The Triassic period is one of the geological periods and is the shortest of the Mesozoic era. Cambrian is the older era than the Triassic period.
<h3>What is Cambrian and triassic period?</h3>
The Triassic period existed 250 -200 million years ago and started after the vastest and most extreme devastation ever. It is the 1st period of the Mesozoic and is known for the movement of the Dinosaurs on the land, flying pterosaurs and swimming plesiosaurs and ichthyosaurs.
The Cambrian is the period of the Paleozoic era and is the most important period because of the appearance of many animals. In this period, the temperature raised on the planet and the ice sheets melted at a very dangerous rate leading to mass devastation.
Therefore, option A. <u>Cambrian</u> is older than the Triassic period.
Learn more about Cambrian and Triassic periods here:
brainly.com/question/11417834
Answer: Heyo Kenji Here! Here's your answer- In a fat molecule, the fatty acids are attached to each of the three carbons of the glycerol molecule with an ester bond through the oxygen atom. During the ester bond formation, three molecules are released. Since fats consist of three fatty acids and a glycerol, they are also called triacylglycerols or triglycerides.
Explanation: Hope this helps!
Have a nice day!
- Kenji ^^
C. a scientific approach to answering questions
A. NaCl(s) and O2(g)
B. 2NaClO3(s) —> 2NaCl(s) + 3O2(g)
C. moles NaClO3 = 100 g / 106.44 g/mol = 0.939 mol NaClO3
D. 0.939 mol NaCl (because the NaClO3 and NaCl are in a 1 to 1 ratio)
E. grams NaCl = 0.939 mol • 58.44 g/mol = 54.9 g NaCl
F. moles of O2 = 0.939 mol NaClO3 • (3 mol O2 / 2 mol NaClO3) = 1.41 mol O2
G. grams of O2 = 1.41 mol • 32 g/mol = 45.1 g O2
H. Percent yield = 10/45.1 • 100% = 22.2% yield
Answer: 0.20 M
Explanation:
According to the dilution law,

where,
= molarity of stock solution = 1.40 M
= volume of stock solution = 72.0 ml
= molarity of diluted solution = m
= volume of diluted solution = 248 ml


Now 124 mL portion of this prepared solution is diluted by adding 133 mL of water.
According to the dilution law,

where,
= molarity of stock solution = 0.41 M
= volume of stock solution = 124 ml
= molarity of diluted solution = m
= volume of diluted solution = (124 +133) ml = 257 ml


Thus the final concentration of the solution is 0.20 M.