Answer:
[Br₂] = 1.25M
Explanation:
2NO (g) + Br₂ (g) ⇄ 2NOBr (g)
Eq 0.80M ? 0.80M
That's the situation told, in the statement.
Let's make the expression for Kc
Kc = [NOBr]² / [Br₂] . [NO]²
Kc = 0.80² / [Br₂] . [0.80]²
0.80 = 1 / [Br₂]
[Br₂] = 1 / 0.80 → 1.25
Burette is a very accurate measuring instrument when adding solutions and has a measurement error of 0.05 mL.
Small volumes of solutions can be transferred from the burette at a controllable rate.
In this instance NaOH is in the burette.
Initial reading of NaOH is 0.20 mL
end point is the point at which the chemical reaction reaches completion. In acid base reactions, end point is when all the H⁺ ions have reacted with OH⁻ ions.
final reading of NaOH is 24.10 mL
to find the volume of NaOH dispensed we have to find the difference between final reading and initial reading
volume of NaOH added = 24.10 mL - 0.20 mL = 23.90 mL
volume of NaOH dispensed is 23.90 mL
Answer:
1.13 × 10⁶ g
Explanation:
Let's consider the reduction of aluminum (III) from Al₂O₃ to pure aluminum.
Al³⁺ + 3 e⁻ → Al
We can establish the following relations:
- 1 Ampere = 1 Coulomb / second
- The charge of 1 mole of electrons is 96,468 c (Faraday's constant)
- 1 mole of Al is produced when 3 moles of electrons circulate
- The molar mass of Al is 26.98 g/mol.
The mass of aluminum produced under these conditions is:
<em>V ≈ 18.5 cm³</em>
<em>Hi there !</em>
<em>density formula</em>
<em>d = m/V => V = m/d</em>
<em>V = 50g/2.7g/cm³</em>
<em>V = 18.(518) cm³</em>
<em>V ≈ 18.5 cm³</em>
<em>Good luck !</em>
