I think it is aluminum oxide
Hi!
The correct option would be 3.85x10^(24)
To find the number of atoms in 250g of potassium, we need to first calculate the number of atoms in
1 mole of Potassium = 39g which contains 6.022x10^(23) atoms of K
<em>(Avogadro's constant value for the amount of molecules/atoms in one mole of any substance)</em>
<em>Solution</em>
So as 39g of Potassium contains 6.022x10^(23) K atoms
1g of Potassium would contain 6.022x10^(23) / 39 = 1.544 x10^(22) atoms
So 250g of Potassium would contain 1.544x10^(22) x 250 = 3.86x10^(24) atoms
The percent composition of this compound :
Mg = 72.182%
N = 27.818%
<h3>Further explanation</h3>
Given
9.03 g Mg
3.48 g N
Required
The percent composition
Solution
Proust stated the Comparative Law that compounds are formed from elements with the same Mass Comparison so that the compound has a fixed composition of elements
Total mass of the compound :
= 9.03 g + 3.48 g
= 12.51 g
The percent composition :
Mg : 9.03/ 12.51 g x 100% = 72.182%
N : 3.48 / 12.51 g x 100% = 27.818%
Answer:
we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf
Explanation:
when we run a TLC plate in a 50/50 mixture of hexanes and ethyl acetate we will except an increase in the polarity of the system and this will cause the Non-polar spot to be near the solvent front, while the polar spot will run at an approximate speed of 0.5 Rf
The speed of the polar spot depends largely on the level of polarity, an increase in the polarity will see both spots of Neat hexane run when we run a TLC plate in a 50/50 mixture of hexanes and ethyl acetate
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.
