Answer:
Photosynthesis
Explanation:
If thats one of the options, thats the answer. Let me know if its not, and I'll be glad to help :)
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:
Molar mass of Al2O3 = 101.961276 g/mol
This compound is also known as Aluminium Oxide.
Convert grams Al2O3 to moles or moles Al2O3 to grams
Molecular weight calculation:
26.981538*2 + 15.9994*3
Percent composition by element
Element Symbol Atomic Mass # of Atoms Mass Percent
Aluminium Al 26.981538 2 52.925%
Oxygen O 15.9994 3 47.075%
Explanation:
Percent composition by element
Element Symbol Mass Percent
Aluminium Al 52.925%
Oxygen O 47.075%
Answer:
0.9715 Fraction of Pu-239 will be remain after 1000 years.
Explanation:


Where:
= decay constant
=concentration left after time t
= Half life of the sample
Half life of Pu-239 =
[
![\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]](https://tex.z-dn.net/?f=%5Clambda%20%3D%5Cfrac%7B0.693%7D%7B24%2C000%20y%7D%3D2.8875%5Ctimes%2010%5E%7B-5%7D%20y%5E%7B-1%5D)
Let us say amount present of Pu-239 today = 
A = ?
![A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}](https://tex.z-dn.net/?f=A%3Dx%5Ctimes%20e%5E%7B-2.8875%5Ctimes%2010%5E%7B-5%7D%20y%5E%7B-1%5D%5Ctimes%201000%20y%7D)


0.9715 Fraction of Pu-239 will be remain after 1000 years.