Answer:
F2 is the limiting reactant
27.6 grams of NaF is produced.
Explanation:
Balance the equation first.
2Na+ F2 ---> 2NaF
To find the limiting reactant, solve for how much NaF can be produced with Na and F2
12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)
=0.658 moles NaF
16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)
=0.705 moles NaF
Since F2 produced the least NaF, F2 is the limiting reactant.
Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.
0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF
27.6 moles of NaF would be theoretically produced.
Explanation:
-Filter help — delete some big unreacted, undesirable species (norit is probably from what you are sorting through, its only carbon which cleans up things)
— extract with DCM because you are probably in an aqueous phase, and some butanoate is in it
- Anhydrous sodium absorbs excess of water (dries the material)
-evaporation in the hood to clear the DCM and crystallize the product.
They would most likely get sick. Or suffer from organ failure. When one part of your body doesn’t work it causes a weakness, making you more susceptible to sicknesses.
H2SO3 or sulfurous acid is actually a strong acid. We know
for a fact that strong acids completely dissociate into its component ions in a
solution, that is:
<span>H2SO3 --> 2H+ + SO3-</span>
<span>So from the equation above, there are 2 moles of H+</span>
Answer:
nine
There are nine orbitals in the n = 3 shell. There is one orbital in the 3s subshell and three orbitals in the 3p subshell. The n = 3 shell, however, also includes 3d orbitals. The five different orientations of orbitals in the 3d subshell are shown in the figure below.
Explanation:
