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.
Let's start with the amount given in percent. Let our basis be 100 grams of compound. So, that means that in this amount, 57.1 g is oxygen and 100-57.1=42.9 g is carbon. Since there is 1:1 atom ratio, it also means that moles oxygen = moles carbon.
Moles = Mass/Relative Mass
Let x be the relative mass of oxygen
57.1/16 = 42.9/x
Solving for x,
<em>x = 12.02 amu</em>
Answer:
pKa = 3.675
Explanation:
∴ <em>C</em> X-281 = 0.079 M
∴ pH = 2.40
let X-281 a weak acid ( HA ):
∴ HA ↔ H+ + A-
⇒ Ka = [H+] * [A-] / [HA]
mass balance:
⇒<em> C</em> HA = 0.079 M = [HA] + [A-]
⇒ [HA] = 0.079 - [A-]
charge balance:
⇒ [H+] = [A-] + [OH-]... [OH-] is negligible; it comes from to water
⇒ [H+] = [A-]
∴ pH = - log [H+] = 2.40
⇒ [H+] = 3.981 E-3 M
replacing in Ka:
⇒ Ka = [H+]² / ( 0.079 - [H+] )
⇒ Ka = ( 3.981 E-3 )² / ( 0.079 - 3.981 E-3 )
⇒ Ka = 2.113 E-4
⇒ pKa = - Log ( 2.113 E-4 )
⇒ pKa = 3.675
Answer:
Π = iMRT ½
Explanation:
111 g
= 0.033 mol
0.033 mol CaCl2
0.09632 kg solvent
= 0.0344 m
13.7 g x 1 mol C3H7OH
60.10 g
0.5 L
(0.0821 L.atm/K.mol) (300.15K
