19372)19292
Hahaha ss
Qowieuww
Answer:
Explanation:
Approx.
425
⋅
g
Explanation:
2
A
l
(
s
)
+
3
C
l
2
(
g
)
→
2
A
l
C
l
3
(
s
)
You have given a stoichiometrically balanced equation, so bravo.
The equation explicitly tells us that
54
⋅
g
of aluminum metal reacts with
6
×
35.45
⋅
g
C
l
2
gas to give
266.7
⋅
g
of
aluminum trichloride
hope this helps
Answer:
The mole is important because it allows chemist to work with a subatomic world with macro world units and amount. Atoms molecules and formula units are very small and very difficult to work with usually. However the mole allows a chemist to work with amount large enough to use.
Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.


There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
![[NaOH]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6