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:
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Molarity=Moles of solute/Volume of solution in L
So
- 0.56M=moles/2.5L
- moles=0.56(2.5)
- moles of Iodine=1.4mol
Mads of Iodine
- Moles(Molar mass)
- 1.4(126.9)
- 177.66g
Answer:
3224 kJ/mol
Explanation:
The combustion of benzoic acid occurs as follows:
C₇H₆O₂ + 13/2O₂ → 7CO₂ + 3H₂O + dE
The change in temperature in the reaction is the change due the energy released, that is:
3.256K * (10.134kJ / K) = 33.00kJ are released when 1.250g reacts
To find the heat released per mole we have to find the moles of benzoic acid:
<em>Moles benzoic acid -Molar mass: 122.12g/mol-:</em>
1.250g * (1mol / 122.12g) = 0.0102 moles
<em />
The dE combustion per mole of benzoic acid is:
33.00kJ / 0.0102moles =
<em>3224 kJ/mol </em>
We're given the [OH⁻] as 8.34 × 10⁻¹² M. Using the formula pOH = -log[OH⁻], the pOH of this solution would be -log(8.34 × 10⁻¹²) ≈ 11.08.
The pOH is, for lack of a better term, the "opposite" of pH: A pOH of 7 is neutral; a pOH less than 7 is <em>basic</em>; and a pOH greater than 7 is <em>acidic</em>.
This follows from the relation, pH + pOH = 14. In this case, with a pOH of 11.08, our pH would be 14 - 11.08 = 2.92, which is acidic (pH < 7).
Thus, the correct answer choice is B.