Answer:
Mostly Para
Explanation:
First, let's assume that the molecule is the toluene (A benzene with a methyl group as radical).
Now the nitration reaction is a reaction in which the nitric acid in presence of sulfuric acid, react with the benzene molecule, to introduce the nitro group into the molecule. The nitro group is a relative strong deactiviting group and is metha director, so, further reactions that occur will be in the metha position.
Now, in this case, the methyl group is a weak activating group in the molecule of benzene, and is always ortho and para director for the simple fact that this molecule (The methyl group) is a donor of electrons instead of atracting group of electrons. Therefore for these two reasons, when the nitration occurs,it will go to the ortho or para position.
Now which position will prefer to go? it's true it can go either ortho or para, however, let's use the steric hindrance principle. Although the methyl group it's not a very voluminous and big molecule, it still exerts a little steric hindrance, and the nitro group would rather go to a position where no molecule is present so it can attach easily. It's like you have two doors that lead to the same place, but in one door you have a kid in the middle and the other door is free to go, you'll rather pass by the door which is free instead of the door with the kid in the middle even though you can pass for that door too. Same thing happens here. Therefore the correct option will be mostly para.
Answer:
Explanation:
1. Calculate the rate constant
The integrated rate law for first order decay is
where
A₀ and A_t are the amounts at t = 0 and t
k is the rate constant
2. Calculate the half-life
1) H
2) He
3) Li
4) Be
5) B
6) C
7) N
8) O
9) F
10) Ne
11) Na
12) Mg
13) Al
14) Si
15) P
16) S
17) Cl
18) Ar
19) K
It is true that substances that have a high melting point and conduct electricity in the liquid phase are ionic substances, and if one of your options is NaCl, then that is the correct answer.
Answer:
The factor that will change the volume of the diver's lungs upon reaching the surface is 4
Explanation:
Given data:
Pressure increases 1 atm = 101.325 kPa
34 ft = 10.3632 m
Depth of 102 ft = 31.0896 m
Question: What factor will the volume of the diver's lungs change upon arrival at the surface, V₂/V₁ = ?
The pressure at 31.0896 m:
The factor will the volume of the diver's lungs change upon arrival at the surface:
