Answer:
no.
Explanation:
The reason this has
never happened is due to the source of magnetic fields: moving electric
charges. When electric charges (e.g. electrons) move in circles, they
produce a magnetic field. In a piece of iron, it is very easy to line up
these circles, getting all the little magnets to work together as one big
magnet.
For each of these circles, one side is the north pole and one side is the
south pole. Since each circle has two sides, each circle has a north and a
south pole. Even the smallest possible magnets (spinning electrons) have a
north and a south pole.
A is obviously out because it leads to a volume of 125.0 milliliters of the new solution and gives you a lower concentration than you were aiming for.
D is out because you are adding 75 milliliters of the stock solution, so your concentration would be too high. You only need 25.0 milometers of stock solution per 100 milliliters of the new solution.
C is also out because it leads to 50.0 milliliters stock solution per 100 milliliters of the new solution and hence the wrong concentration.
B is by default the correct answer. It also details the correct technique. First you add the stock solution (This you know from your calculations to be 25 milliliters.) then you add the water up to the volume you needed. (Because the calculations only tell you the total volume of water not what you need to add) You also add the water last so you can rinse the neck of the flask to make sure you also get all the stock solution residue into the stock solution.
I would add the final step of stirring, but B is the only answer that can be correct.
Answer:
A lower ph is always more acidic, due to the increased concentration of hydrogen ions in the solution/substance.
A ph of 3 is 100 times more acidic than a pH of 5, and this is due to the increments on the scale.
20.181 u
The average atomic mass of Ne is the <em>weighted average</em> of the atomic masses of its isotopes.
We multiply the atomic mass of each isotope by a number representing its relative importance (i.e., its % abundance).
Thus,
Avg. at. mass
= (0.904 83× 19.992 u) + (0.002 71 × 20.994) + (0.092 53× 21.991 u)
= 18.0894 u + 0.0569 u + 2.0348 u = 20.181 u
