Layer 2 and layer 9 are the same relative age.
Explanation:
The statement that best describes the rock layers is that Layer 2 and layer 9 are the same relative age..
The relative age is used in placing sedimentary rocks in order of their occurrence.
To do this, we apply the sedimentary laws.
The ones applicable here are:
- Principle of superposition states that in an undisturbed sequence, the oldest layer is at the base and youngest on top.
- Principle of cross cutting states that a fault and intrusion are younger than the rocks they cut through.
- Principle of fossil and fauna succession states that fossils and fauna succeed on another in a determinable form.
We see that layers 2 and 9 have the same fossil and are the same lithological units.
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Answer:
the system has infinitely many solutions.
Explanation:
The system is 2x + y = 1 and 4x + 2y = 2. Solutions to a system are the intersection points. Since these two lines are the same line they intersect everywhere. There are infinitely many solutions.
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Answer:
4190.22 L = 4.19 m³.
Explanation:
- For the balanced reaction:
<em>2P₂ + 5O₂ ⇄ 2P₂O₅. </em>
It is clear that 2 mol of P₂ react with <em>5 mol of O₂ </em>to produce <em>2 mol of P₂O₅.</em>
- Firstly, we need to calculate the no. of moles of 6.92 kilograms of P₂O₅ produced through the reaction:
no. of moles of P₂O₅ = mass/molar mass = (6920 g)/(283.88 g/mol) = 24.38 mol.
- Now, we can find the no. of moles of O₂ is needed to produce the proposed amount of P₂O₅:
<u><em>Using cross multiplication:</em></u>
5 mol of O₂ is needed to produce → 2 mol of P₂O₅, from stichiometry.
??? mol of O₂ is needed to produce → 24.38 mol of P₂O₅.
∴ The no. of moles of O₂ needed = (5 mol)(24.38 mol)/(2 mol) = 60.95 mol.
- Finally, we can get the volume of oxygen using the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 606.1 mm Hg/760 = 0.8 atm).
V is the volume of the gas in L (V = ??? L).
n is the no. of moles of the gas in mol (n = 60.95 mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (396.90°C + 273 = 669.9 K).
∴ V of oxygen needed = nRT/P = (60.95 mol)(0.0821 L.atm/mol.K)(669.9 K)/(0.8 atm) = 4190.22 L/1000 = 4.19 m³.
True.
A catalyst is a substancr that increases the rate of a chemical reaction without itself undergoing any permanent chemical change.
