What mass of carbon dioxide occupies a volume of 81.3 L at 204 kPa and temperature of 95.0 C
1 answer:
The mass of CO₂ present is 238.52g
<u>Explanation:</u>
Given:
Volume, V = 81.3 L
Pressure, P = 204 kPa = 204000 Pa
Temperature, T = 95°C
T = 95 + 273 K
T = 368 K
Mass of CO₂, x = ?
Molecular weight of CO₂ = 44 g/mol
According to gas law:
PV = nRT
where,
n = number of moles
n = given mass/ molecular mass
R = gas constant
R = 8.314 X 10³ L Pa K⁻¹ mol⁻¹
Substituting the value:
Therefore, the mass of CO₂ present is 238.52g
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Answer:
1) After adding 15.0 mL of the HCl solution, the mixture is before the equivalence point on the titration curve.
2) The pH of the solution after adding HCl is 12.6
Explanation:
10.0 mL of 0.25 M NaOH(aq) react with 15.0 mL of 0.10 M HCl(aq). Let's calculate the moles of each reactant.
There is an excess of NaOH so the mixture is before the equivalence point. When HCl completely reacts, we can calculate the moles in excess of NaOH.
NaOH + HCl ⇒ NaCl + H₂O
Initial 2.5 × 10⁻³ 1.5 × 10⁻³ 0 0
Reaction -1.5 × 10⁻³ -1.5 × 10⁻³ 1.5 × 10⁻³ 1.5 × 10⁻³
Final 1.0 × 10⁻³ 0 1.5 × 10⁻³ 1.5 × 10⁻³
The concentration of NaOH is:
NaOH is a strong base so [OH⁻] = [NaOH].
Finally, we can calculate pOH and pH.
pOH = -log [OH⁻] = -log 0.040 = 1.4
pH = 14 - pOH = 14 - 1.4 = 12.6