Answer:
-21 kJ·mol⁻¹
Explanation:
Data:
H₃O⁺ + OH⁻ ⟶ 2H₂O
V/mL: 50 50
c/mol·dm⁻³: 1.0 1.0
ΔT = 4.5 °C
C = 4.184 J·°C⁻¹g⁻¹
C_cal = 50 J·°C⁻¹
Calculations:
(a) Moles of acid
So, we have 0.050 mol of reaction
(b) Volume of solution
V = 50 dm³ + 50 dm³ = 100 dm³
(c) Mass of solution
(d) Calorimetry
There are three energy flows in this reaction.
q₁ = heat from reaction
q₂ = heat to warm the water
q₃ = heat to warm the calorimeter
q₁ + q₂ + q₃ = 0
nΔH + mCΔT + C_calΔT = 0
0.050ΔH + 100×4.184×4.5 + 50×4.5 = 0
0.050ΔH + 1883 + 225 = 0
0.050ΔH + 2108 = 0
0.050ΔH = -2108
ΔH = -2108/0.0500
= -42 000 J/mol
= -42 kJ/mol
This is the heat of reaction for the formation of 2 mol of water
The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.
Answer:
3.47 x 10⁻⁹.
Explanation:
- The concentration of OH⁻ = 0.16 M.
∵ [OH⁻] = √(Kb)(C).
Kb for ammonia = 1.8 x 10⁻⁵.
∴ [OH⁻] = √(Kb)(C) = √(1.8 x 10⁻⁵)(0.16 M) = 2.88 x 10⁻⁶ M.
<em>∵ [H⁺][OH⁻] = 10⁻¹⁴.</em>
∴ [H⁺] = 10⁻¹⁴/[OH⁻] = 10⁻¹⁴/(2.88 x 10⁻⁶) = 3.47 x 10⁻⁹ M.
<em>∵ pH = - log[H⁺]</em> = - log(3.47 x 10⁻⁹) = <em>8.459 ≅ 8.46.</em>
Answer:
Please show a file so I can answer your question or tell me which experiment.
Explanation:
Definition:The control group (sometimes called a comparison group) is used in an experiment as a way to ensure that your experiment actually works. It's a way to make sure that the treatment you are giving is causing the experimental results, and not something outside the experiment.