Answer:
Explanation:
Given : Density - 2.41 g/liter
Temperature - 25° C
Pressure : 770 mm Hg
R = 0.0821 L atm mol-¹K-¹
Find : Molecular mass of gas
Solution : Ideal gas equation with respect to density will be : PM = dRT. In the formula, P is pressure, M is molecular mass, d is density, R is gas constant and T is temperature.
Keeping the values in equation-
Pressure : 770 mm Hg = 1 atm
Temperature : 273 + 25 = 298 K
M = dRT/P
M = (2.41*0.0821*298)/1
M = 58.96 gram/mol
Thus, the molecular mass of gas is 58.96 gram/mol.
"The solubility of gases decreases as temperature rises" statements about trends in solubility is accurate.
<u>Option: D</u>
<u>Explanation:</u>
A substance's solubility is the quantity of that component that is needed at a defined degree of temperature to produce a saturated solution in any set quantity of solvent. Some compounds like hydrochloric acid, ammonia, etc have solubility that reduces with rising temperature. They are both standard-pressure gases.
When heating a solvent with a gas absorbed in it, both the solvent and the solute spike in the kinetic energy.When the gaseous solute's kinetic energy rises, the molecules have a higher propensity to overcome the solvent molecules' connection and migrate to the gas phase. Thus, a gas's solubility reduces with rising temperature.
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]=\frac{1.0 \times 10^{-3} mol }{25.0 \times 10^{-3} L} =0.040M](https://tex.z-dn.net/?f=%5BNaOH%5D%3D%5Cfrac%7B1.0%20%5Ctimes%2010%5E%7B-3%7D%20mol%20%7D%7B25.0%20%5Ctimes%2010%5E%7B-3%7D%20L%7D%20%3D0.040M)
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
