Answer:
Albert Einstein is perhaps most famous for introducing the world to the equation E=mc2. In essence, he discovered that energy and mass are interchangeable, setting the stage for nuclear power—and atomic weapons. His part in the drama of nuclear war may have ended there if not for a simple refrigerator.
Explanation:
Albert Einstein is perhaps most famous for introducing the world to the equation E=mc2. In essence, he discovered that energy and mass are interchangeable, setting the stage for nuclear power—and atomic weapons. His part in the drama of nuclear war may have ended there if not for a simple refrigerator.
Anything asymmetrical is polar
Reduction <span>always results in a lowering of the oxidation number. The reaction of the system above is written as:
</span><span>Cu2+(aq) + Fe(s) --> Cu(s) + Fe2+(aq)
</span>
From the reaction, we see that copper goes from the +2 to a neutral charge. Lowering of the oxidation number happens so this is the element that is being reduced.
12.0g x 1 mol / 63.546g = 0.188839581mol
<span>So, for every 1 mole, we have 6.022 x 10^23 of whatever we're measuring. This gives us a conversion factor of (1 mole / 6.022 x 10^23 atoms) or (6.022 x 10^23 atoms / 1 mole).
</span>
0.188839581 mol x (6.022 x 10^23 atoms) / 1 mol = 1.137191955 x 10^23
<span>Remember from before that we are limited to 3 significant figures. Since our calculations are complete, we can now round down to: 1.14 x 10^23 </span>
<span>That should be your answer!
Hope it helps!
xo</span>
Answer:
Explanation:
If we look at the structure of 1-Bromopropane; we will see that it is a derivative of alkane family by the the substitution of an alkyl group. The position of the Bromine in the propane is 1, making 1-Bromopropane a primary alkyl-halide.
Primary alkyl - halide undergo SN2 mechanism. This nucleophilic reaction needs to be a strong alkyl halide , such as 1-Bromopropane used otherwise it will result to a reactive mechanism if a weak electrophile is used.
However, the critical and the main objective here is to Draw the major substitution product if the reaction proceeds in good yield. If no reaction is expected or yields will be poor, draw the starting material in the box. If a charged product is formed, be sure to draw the counterion.
The attached diagrams portraying this notions is shown in the attached file below.
