To solve the problem, we assume the sample to be ideal. Then, we use the ideal gas equation which is expressed as PV = nRT. From the first condition of the nitrogen gas sample, we calculate the number of moles.
n = PV / RT
n = (98.7x 10^3 Pa x 0.01 m^3) / (8.314 Pa m^3/ mol K) x 298.15 K
n = 0.40 mol N2
At the second condition, the number of moles stays the same however pressure and temperature was changed. So, the new volume is calculated as follows:
V = nRT / P
V = 0.40 x 8.314 x 293.15 / 102.7 x 10^3
V = 9.49 x 10^-3 m^3 or 9.49 L
Answer:
339kJ
Explanation:
Given parameters:
Mass of steam = 150g = 0.15kg
Initial temperature of steam = 100°C
Final temperature of water = 100°C
Unknown:
Quantity of heat that must be removed to condense the steam = ?
Solution:
The heat involved here is a latent heat because there is no change temperature. The process is just a phase change.
H = mL
m is the mass
L is the latent heat of vaporization = 2,260 kJ/kg
Insert the parameters and solve;
H = 0.15kg x 2,260 kJ/kg
H = 339kJ
Answer:
See explanation
Explanation:
Electrons transition between energy levels in an atom due to gain or loss of energy. An electron may gain energy and move from its ground state to one of the accessible excited states. The electron quickly returns to ground state, emitting the energy previously absorbed as a photon of light. The wavelength of light emitted is measured using powerful spectrometers.
Atoms can be excited thermally or by irradiation with light of appropriate frequency.
Burette is a very accurate measuring instrument when adding solutions and has a measurement error of 0.05 mL.
Small volumes of solutions can be transferred from the burette at a controllable rate.
In this instance NaOH is in the burette.
Initial reading of NaOH is 0.20 mL
end point is the point at which the chemical reaction reaches completion. In acid base reactions, end point is when all the H⁺ ions have reacted with OH⁻ ions.
final reading of NaOH is 24.10 mL
to find the volume of NaOH dispensed we have to find the difference between final reading and initial reading
volume of NaOH added = 24.10 mL - 0.20 mL = 23.90 mL
volume of NaOH dispensed is 23.90 mL
