21. _____ Which of the following entities has the greatest mass?
1 answer:
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Answer:
Explanation:
Given:
Pressure = 745 mm Hg
Also, P (mm Hg) = P (atm) / 760
Pressure = 745 / 760 = 0.9803 atm
Temperature = 19 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (19 + 273.15) K = 292.15 K
Volume = 0.200 L
Using ideal gas equation as:
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.9803 atm × 0.200 L = n × 0.0821 L.atm/K.mol × 292.15 K
⇒n = 0.008174 moles
From the reaction shown below:-
1 mole of react with 2 moles of
0.008174 mole of react with 2*0.008174 moles of
Moles of = 0.016348 moles
Volume = 13.4 mL = 0.0134 L ( 1 mL = 0.001 L)
So,
Answer : The value of work done for the system is 1144.69 J
Explanation :
First we have to calculate the moles of
Molar mass of = 65.01 g/mole
Now we have to calculate the moles of nitrogen gas.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 mole of react to give 3 mole of
So, 0.311 moles of react to give
moles of
Now we have to calculate the volume of nitrogen gas.
Using ideal gas equation:
where,
P = Pressure of gas = 1.00 atm
V = Volume of gas = ?
n = number of moles = 0.466 mole
R = Gas constant =
T = Temperature of gas =
Putting values in above equation, we get:
As initially no nitrogen was present. So,
Volume expanded = Volume of nitrogen evolved
Thus,
Expansion work = Pressure × Volume
Expansion work = 1.00 atm × 11.3 L
Expansion work = 11.3 L.atm
Conversion used : (1 L.atm = 101.3 J)
Expansion work = 11.3 × 101.3 = 1144.69 J
Therefore, the value of work done for the system is 1144.69 J