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.
learn more:
Sedimentary rocks brainly.com/question/2740663
#learnwithBrainly
I will use [pV/T] in the state 1 = [pV/T] in the state 2.
State 1:
p = 1.0 atm
V = 25 liter
T = 100 + 273.15 = 373.15 K
State 2:
p = 19.71 mmHg * 1.atm / 760 mmHg = 0.0259atm
V= ?
T = 25 + 273.15 = 298.15 K
Application of the formula
1.0 atm * 25 liter / 373.15 k = 0.0259 atm * V / 298.15 K =>
V = [1.0atm * 25 liter / 373.15 K]*298.15K/0.0259atm = 771 liter
Acid Rain is formed when chemicals in the air get into rain and up the acidity levels!!!!!!
Hope this helps guys!
Answer:
See explanation below
Explanation:
The question is incomplete, cause you are not providing the structure. However, I found the question and it's attached in picture 1.
Now, according to this reaction and the product given, we can see that we have sustitution reaction. In the absence of sodium methoxide, the reaction it's no longer in basic medium, so the sustitution reaction that it's promoted here it's not an Sn2 reaction as part a), but instead a Sn1 reaction, and in this we can have the presence of carbocation. What happen here then?, well, the bromine leaves the molecule leaving a secondary carbocation there, but the neighbour carbon (The one in the cycle) has a more stable carbocation, so one atom of hydrogen from that carbon migrates to the carbon with the carbocation to stabilize that carbon, and the result is a tertiary carbocation. When this happens, the methanol can easily go there and form the product.
For question 6a, as it was stated before, the mechanism in that reaction is a Sn2, however, we can have conditions for an E2 reaction and form an alkene. This can be done, cause the extoxide can substract the atoms of hydrogens from either the carbon of the cycle or the terminal methyl of the molecule and will form two different products of elimination. The product formed in greater quantities will be the one where the negative charge is more stable, in this case, in the primary carbon of the methyl it's more stable there, so product 1 will be formed more (See picture 2)
For question 6b, same principle of 6a, when the hydrogen migrates to the 2nd carbocation to form a tertiary carbocation the methanol will promove an E1 reaction with the vecinal carbons and form two eliminations products. See picture 2 for mechanism of reaction.
Answer:
d.) It is a binary molecular compound.
Explanation:
The compound in question has a formula 
. The compound is not acidic in nature and the element 'M' is not a metal. This shows that the compound does not contain any metal. Based on the definition of a binary molecular compound as a compound comprising elements that are not metals. Therefore, the compound is obviously a binary molecular compound.
