We can solve this problem when we use the conditions of a gas at standard temperature and pressure. It has been established that at STP where the temperature is 0 degrees Celsius and the pressure is 101.325 kPa, the volume of 1 mole of gas is 22.4 L. We will use this data for the calculations.
68.5 L ( 1 mol O2 / 22.4 L O2 ) = 3.06 mol O2
Answer:
The final temperature of the setup = 36.6°C
Explanation:
Let the final temperature of the setup be T
Heat lost by the copper tubing = Heat gained by water and the vessel
Heat lost by the copper tubing = mC ΔT = 455 × 0.387 × (89.5 - T) = (15759.61 - 176.1T) J
Heat gained by water = mC ΔT = 159 × 4.186 × (T - 22.8) = (665.6T - 15175.1) J
Heat gained by vessel = c ΔT = 10 × (T - 22.8) = (10T - 228) J
Heat lost by the copper tubing = Heat gained by water and the vessel
(15759.61 - 176.1T) = (665.6T - 15175.1) + (10T - 228)
851.7 T = 31162.71
T = 36.6°C
Boiling point, melting point, Appearance, mass, and taste.<span />
Answer:
X-ray photon is a quantum of electromagnetic energy.
Answer:
the quantity required can go from 117 ml (for maximum concentration) up to 2900 ml ( if the concentrated solution has molarity =0.420 M)
Explanation:
the amount of water required to dilute a solution V₁ liters of Molarity M₁ to V₂ liters of M₂
moles of hydrochloric acid = M₁ * V₁= M₂ * V₂
V₁ = V₂ * M₂/M₁
where
M₂ = 0.420 M
V₂ =2.90 L
Since the hydrochloric acid can be concentrated up to 38% p/V ( higher concentrations are possible but the evaporation rate is so high that handling and storage require extra precautions, like cooling and pressurisation)
maximum M₁ =38% p/V = 38 gr/ 0.1 L / 36.5 gr/mol = 10.41 M
then
min V₁ = V₂ * M₂/ max M₁ = 2.90 L* 0.420 M/ 10.41 M= 0.117 L = 117 ml
then the quantity required can go from 117 ml up to 2900 ml ( if M₁ = M₂)
