<span>This change is called BOILING.</span>
Draw eight circles and nine more in a big circle = this represents your nucleus
draw eight electrons (since electrons = protons in neutral elements) outside of it (2 in one ring, then 6 in a second ring)
Bronsted Lowry Acids are donates protons
Therefor it is <span>NH4+</span>
Molality is defined as the number of moles of solute in 1 kg of solvent.
molality of solution to be prepared is 0.50 molal
this means that in 1000 g of water there should be 0.50 mol of NaCl
if 1000 g of water should contain - 0.50 mol
then 750.0 g of water requires - 0.50 mol/kg x 0.750 kg = 0.375 mol
mass of NaCl in 0.375 mol - 58.5 g/mol x 0.375 mol = 21.9 g
therefore a mass of 21.9 g of NaCl is required
Answer:
11.12 → pH
Explanation:
This is a titration of a weak base and a strong acid.
In the first step we did not add any acid, so our solution is totally ammonia.
Equation of neutralization is:
NH₃ + HCl → NH₄Cl
Equilibrium for ammonia is:
NH₃ + H₂O ⇄ NH₄⁺ + OH⁻ Kb = 1.8×10⁻⁵
Initially we have 50 mL . 0.10M = 5 mmoles of ammonia
Our molar concentration is 0.1 M
X amount has reacted.
In the equilibrium we have (0.1 - x) moles of ammonia and we produced x amount of ammonium and hydroxides.
Expression for Kb is : x² / (0.1 - x) = 1.8×10⁻⁵
As Kb is so small, we can avoid the x to solve a quadratic equation.
1.8×10⁻⁵ = x² / 0.1
1.8×10⁻⁵ . 0.1 = x²
1.8×10⁻⁶ = x²
√1.8×10⁻⁶ = x → 1.34×10⁻³
That's the value for [OH⁻] so:
1×10⁻¹⁴ = [OH⁻] . [H⁺]
1×10⁻¹⁴ / 1.34×10⁻³ = [H⁺] → 7.45×10⁻¹²
- log [H⁺] = pH
- log 7.45×10⁻¹² = 11.12 → pH
