Answer:
The answer to your question is 242 ml
Explanation:
Data
HI 0.211 M Volume = x
KMnO₄ 0.354 M Volume = 24 ml
Balanced Chemical reaction
12HI + 2KMnO₄ + 2H₂SO₄ → 6I₂ + Mn₂SO₄ + K₂SO₄ + 8H₂O
Process
1.- Calculate the moles of KMnO₄ 0.354 M in 24 ml
Molarity = moles / volume (L)
moles = Molarity x volume (L)
moles = 0.354 x 0.024
moles = 0.0085
2.- From the balanced chemical reaction we know that HI and KMnO₄ react in the proportion 12 to 2. Then,
12 moles of HI --------------- 2 moles of KMnO₄
x --------------- 0.0085 moles of KMnO₄
x = (0.0085 x 12)/2
x = 0.051 moles of HI
3.- Calculate the milliliters of HI 0.211 M
Molarity = moles/volume
Volume = moles/molarity
Volume = 0.051/0.211
Volume = 0.242 L or Volume = 242 ml
Answer:
Primer postulado:
Así Bohr asumió que el átomo de hidrógeno puede existir solo en ciertos estados discretos, los cuales son denominados estados estacionarios del átomo. En el átomo no hay emisión de radiación electromagnética mientras el electrón no cambia de órbita.
Explanation:
Because there is less friction on the marble floor.
Answer:
what each cells job is divided
Explanation:
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.
