Answer:
54 grams ammonium chloride and 40 grams sodium hydroxide
Explanation:
A buffer is a solution that contains either a weak acid and its salt or a weak base and its salt, the solution is resistant to changes in pH. This means that, a buffer is an aqueous solution of either a weak acid and its conjugate base or a weak base and its conjugate acid.
A Buffer is used to maintain a stable pH in a solution, buffers can neutralize small quantities of additional acid of base. For any buffer solution, there is always a working pH range and a set amount of acid or base that can be neutralized before the pH will change. The amount of acid or base that can be added to a buffer before changing its pH is called its buffer capacity.
A good buffer mixture is supposed to have about equal concentrations of its both components. It is a rule of thumb therefore, that a buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other component.
The implication of this is that the ammonium chloride and sodium hydroxide should be of approximately the same concentration. If the masses are dissolved as shown in the answer, then we will have 1molL-1 of each component of the buffer in accordance with the rule of thumb stated above.
Answer:
Hands-free eyewash stations, sand bucket, fire blankets, fire extinguishers, fire alarm system and first aid kit.
Explanation:
Working in laboratories has many risks, therefore, preventive measures that should be incorporated to avoid the occurrence of any laboratory accidents.
Some of the important emergency equipment that should be available in laboratories are: hands-free eyewash stations, sand bucket, fire blankets, fire extinguishers, fire alarm system, chemical storage cabinet, first aid kits and fume hood.
Some of the personal protective equipment include lab coats, goggles, safety gloves and face shield.
Answer:
1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹
2. 0.58 mol
Explanation:
1.Given ΔO₂/Δt…
2H₂O₂ ⟶ 2H₂O + O₂
-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt
d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹
d[H₂O]/dt = 2d[O₂]/dt = 2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = 6.6 × 10⁻³mol·L⁻¹s⁻¹
2. Moles of O₂
(a) Initial moles of H₂O₂
(b) Final moles of H₂O₂
The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.
(c) Moles of H₂O₂ reacted
Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol
(d) Moles of O₂ formed
Answer:
V = 80.65L
Explanation:
Volume = ?
Number of moles n = 5 mol
Temperature (T) = 393.15K
Pressure = 1520mmHg
Ideal gas constant (R) = 62.363mmHg.L/mol.K
According to ideal gas law,
PV = nRT
P = pressure of the ideal gas
V = volume the gas occupies
n = number of moles of the gas
R = ideal gas constant (note this can varies depending on the unit of your variables)
T = temperature of the ideal gas
PV = nRT
Solve for V,
V = nRT / P
V = (5 * 62.363 * 393.15) / 1520
V = 80.65L
The volume the gas occupies is 80.65L
