Answer:
Igneous rocks do not contain any fossils.
Explanation:
This is because any fossils in the original rock will have melted when the rock melted to form magma. Sedimentary rocks are formed from the broken remains of other rocks that become joined together.
Answer:
4-oxopentanoic acid.
Explanation:
In this case, we must remember that the Grignard reaction is a reaction in which <u>carbanions</u> are produced. Carboanions have the ability to react with CO2 to generate a new C-C bond and a carboxylate ion. Finally, the acid medium will protonate the carboxylate to produce the <u>carboxylic acid group.
</u>
The molecules that can follow the mechanism described above are the molecules: p-methylbenzoic acid, cyclopentane carboxylic acid and 3-methylbutanoic acid. (See figure 1)
In the case of <u>4-oxopentanoic acid</u>, the possible carbanion <u>will attack the carbonyl group</u> to generate a cyclic structure and an alcohol group (1-methylcyclopropan-1-ol). Therefore, this molecule cannot be produced by this reaction. (See figure 2)
Atoms of one element are converted to atoms of another element through transmutation or nuclear transmutation.
Answer:
The correct answer is option D.
Explanation:
Step 1: 
Step 2: 
Overall reaction can be determined by adding all the reactions of mechanism:

Reactants in an overall reaction = 
Products in an overall reaction = 
According to question , the role of oxygen gas is product.
Answer:
a. Rate = k×[A]
b. k = 0.213s⁻¹
Explanation:
a. When you are studying the kinetics of a reaction such as:
A + B → Products.
General rate law must be like:
Rate = k×[A]ᵃ[B]ᵇ
You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.
If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1
Rate = k×[A]¹[B]ᵇ
In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]
Rate = k×[A][B]⁰
<h3>Rate = k×[A]</h3>
b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:
Rate = k×[A]
0.320M/s = k×[1.50M]
<h3>k = 0.213s⁻¹</h3>