Pressure of argon = 546.8 kPa
Conversion factor: 1 atm = 101.325 kPa
Pressure of argon = 546.8 kPa x 1 atm/101.325 kPa = 5.4 atm
Moles of argon = 15.82
Volume of argon = 75.0 L
According to Ideal gas law,
PV = nRT
where P is the pressure, V is the volume , n is the number of moles, R is the universal gas constant, and T is the temperature
T = PV/nR = (5.4 atm x 75.0 L) / (15.82 x 0.0821 L.atm.mol⁻¹K⁻¹)
T = 311.82 K
Hence the temperature of the canister is 311.82 K.
Answer:
Option B
Explanation:
We will check the solubility graph for potassium nitrate, KNO
3. Based on the graph it can be said that the temperature of solution when 130 grams of KNO3 dissolves in 100 grams of water is near to 65 degree Celsius. Now if three grams of solute is increased then the temperature of the solution will increase by a degree or so and hence the most probable temperature would be 68 degree Celsius.
Hence, option B is correct
the actual yield is the amount of Na₂CO₃ formed after carrying out the experiment
theoretical yield is the amount of Na₂CO₃ that is expected to be formed from the calculations
we need to first find the theoretical yield
2Na₂O₂ + 2CO₂ ---> 2Na₂CO₃ + O₂
molar ratio of Na₂O₂ to Na₂CO₃ is 2:2
number of Na₂O₂ moles reacted is equal to the number of Na₂CO₃ moles formed
number of Na₂O₂ moles reacted is - 7.80 g / 78 g/mol = 0.10 mol
therefore number of Na₂CO₃ moles formed is - 0.10 mol
mass of Na₂CO₃ expected to be formed is - 0.10 mol x 106 g/mol = 10.6 g
therefore theoretical yield is 10.6 g
percent yield = actual yield / theoretical yield x 100%
81.0 % = actual yield / 10.6 g x 100 %
actual yield = 10.6 x 0.81
actual yield = 8.59 g
therefore actual yield is 8.59 g
Answer:
There is None
Explanation:
This is because it is a derived function dependent on other factors.
q = mCΔT
The correct specific heat capacity of water is <em>4.187 kJ/(kg.K)</em>.
ΔT = q/mC = 87 kJ/[648.00 kg x 4.187 kJ/(kg.K)] = 87 kJ/(2713 kJ/K) = 0.032 K
Tf = Ti + ΔT = 298 K + 0.032 K = 298.032 K
