Consider the following reaction: 2 H2(g) + 2 NO(g) → 2 H2O(g) + N2(g) If the concentration of NO changed from 0.100 M to 0.025 M
1 answer:
Answer: The average rate of the reaction during this time interval is, 0.005 M/s
Explanation:
The given chemical reaction is:
The expression will be:
where,
= final concentration of
= 0.025 M
= initial concentration of
= 0.100 M
= final time = 15 minutes
= initial time = 0 minutes
Putting values in above equation, we get:
Hence, the average rate of the reaction during this time interval is, 0.005 M/s
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Answer:
a) for adiabatic reversible, ΔU(internal energy is constant) = 0, ΔH = 0(no heat is entering or leaving the surrounding)
workdone (w) = -8442.6 J ≈ -8.443 KJ
heat transferred (q) of the ideal gas = - w
q = 8.443 KJ
b) For ideal gas at adiabatic reversible, Internal energy (ΔU) = 0 and enthalpy (ΔH) = 0
the workdone(w) in the ideal gas= - 4567.5 J ≈ - 4.57 KJ
the heat transfer (q) of an ideal gas = 4.5675 KJ
Explanation:
given
mole of an ideal gas(n) = 2.5 mol
Temperature (T) = 20°C
= (20°C + 273) K = 293 K
Initial pressure of the ideal gas(P₁) = 20 atm
Final pressure of the ideal gas(P₂) = 5 atm.
2) (a)for adiabatic reversible process,
note: adiabatic process is a process by which no heat or mass is transferred between the system and its surrounding.
Work done (w) = nRT ln
= 2.5 mol × 8.314 J/mol K × 293 K × ln
= 6090.01 J × [-1.3863]
= -8442.6 J ≈ -8.443 KJ
So, the work done (w) of ideal gas = -8.443 KJ
For ideal gas at adiabatic reversible, Internal energy (U) = 0 and Enthalpy (H) = 0
From first law of thermodynamics:-
U = q + w
0 = q + w
q = - w
q = - (-8.443 KJ)
q = 8.443 KJ
heat transfer (q) of the ideal gas = 8.443 KJ
(b) For adiabatic irreversible, the temperature T remains constant because the internal energy U depends only on temperature T. Since at constant temperature, the entropy is proportional to the volume, therefore, entropy will increase.
Work done (w) = -nRT(1 - ln
)
= - 2.5 mol × 8.314 J / mol K× 293 K × [1- (5 atm /20 atm)]
= - 6090.01 J × 0.75
= - 4567.5 J ≈ - 4.57 KJ
∴work done(w) of an ideal gas = - 4.57 KJ
For ideal gas at adiabatic Irreversible, Internal energy (U) = 0 and Enthalpy (H) = 0
From first law of thermodynamics:-
U = q + w
0 = q + w
q = - w
q = - (-4.5675 KJ)
q = 4.5675 KJ
the heat transfer (q) of an ideal gas = 4.5675 KJ
Answer:
Explanation:
From the 1 g sample you have:
0.546 grams of cobalt chloride
1-0.546=0.454 grams of water
Now:
1) The salt
Of the 0.546 g, 0.248 g are cobalt (Mr=58.9) and the rest id Cl (Mr=35.45):
Dividing:
So the molecular formula will be:
2) The water
The water's molecular weight is M=18 :
Bonding with the Co:
The complete formula of the hydrate: