Answer:
See explanation below
Explanation:
You forgot to put the picture to do so. In this case, I manage to find one, and I hope is the one you are looking for. If not, then post it again and I'll gladly help you out again.
According to the picture with the answer, we have a cyclohexane with 4 methyl groups there. Two of them are facing towards the molecule with a darker bond. This means that the alkyl bromide, should have a bromine in one of the bonds, and in order to produce an E2 reaction, this bromine should be facing in the opposite direction of the methyl groups which are facing towards. This is because an E2 reaction occurs with the less steric hindrance in the molecule. If the bromine is in the same direction as the methyl group, it will cause a lot more of work to do a reaction, and therefore, an E2 reaction. I will promote instead a E1 or a sustitution product.
Therefore the alkyl bromide should be like the one in the picture 2.
Answer: temperature and salinity.
Explanation:
Unequal heating of the atmosphere
Answer:
(a) Alkali metals: Francium (Fr)
(b) Chalcogens: Polonium (Po)
(c) Noble gases: Radon (Rn)
(d) Alkaline earth metals: Radium (Ra)
Explanation:
In the periodic table, the atomic mass increases down the group. Therefore, the last element of a group is the heaviest element of the group.
(a) alkali metals: The chemical elements that are present in group 1 of the periodic table, except hydrogen.
<u>The heaviest member of this group is francium (Fr)</u>
(b) chalcogens: The chemical elements that are present in group 16 of the periodic table
<u>The heaviest member of this group is polonium (Po)</u>
(c) noble gases: The chemical elements that are present in group 18 of the periodic table
<u>The heaviest member of this group is radon (Rn)</u>
(d) alkaline earth metals: The chemical elements that are present in group 2 of the periodic table.
<u>The heaviest member of this group is radium (Ra)</u>
Answer:
0.0468 g.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life time of the reaction (t1/2 = 1620 years).
∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).
a is the initial concentration (a = 12.0 g).
(a-x) is the remaining concentration.
∴ kt = lna/(a-x)
(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).
5.54688 = ln(12)/(a-x).
Taking e for the both sides:
256.34 = (12)/(a-x).
<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>
