Answer:
Explanation:
We're asked to calculate the number of atoms of
Ca
in
153
g Ca
.
What we must first do is convert the given mass of calcium to moles of calcium, using its molar mass (referring to a periodic table, this is
40.08
g
mol
):
153
g Ca
(
1
mol Ca
40.08
g Ca
)
=
3.82
mol Ca
Using Avogadro's number,
6.022
×
10
23
particles
mol
, we can calculate the number of atoms present:
3.82
mol Ca
(
6.022
×
10
23
atoms Ca
1
mol Ca
)
=
2.30
×
10
24
atoms Ca
The reaction between the reactants would be:
CH₃NH₂ + HCl ↔ CH₃NH₃⁺ + Cl⁻
Let the conjugate acid undergo hydrolysis. Then, apply the ICE approach.
CH₃NH₃⁺ + H₂O → H₃O⁺ + CH₃NH₂
I 0.11 0 0
C -x +x +x
E 0.11 - x x x
Ka = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
Since the given information is Kb, let's find Ka in terms of Kb.
Ka = Kw/Kb, where Kw = 10⁻¹⁴
So,
Ka = 10⁻¹⁴/5×10⁻⁴ = 2×10⁻¹¹ = [H₃O⁺][CH₃NH₂]/[CH₃NH₃⁺]
2×10⁻¹¹ = [x][x]/[0.11-x]
Solving for x,
x = 1.483×10⁻⁶ = [H₃O⁺]
Since pH = -log[H₃O⁺],
pH = -log(1.483×10⁻⁶)
<em>pH = 5.83</em>
Answer:
its not cear to answer this question
The balanced reaction for combustion is as follows ;
2C₂H₅OH + 6O₂ ---> 4CO₂ + 6H₂O
the stoichiometry of C₂H₅OH to O₂ is 2:6
that means 2 mol of C₂H₅OH reacts with 6 mol of O₂.
when 1 mol of C₂H₅OH reacts with 6/2 mol of O₂,
then 0.3020 mol of C₂H₅OH reacts with - 6/2 x 0.3020
therefore number of O₂ moles reacted = 0.91 mol
