Answer:
Solution:-
The gas is in the standard temperature and pressure condition i.e. at S.T.P
Therefore,
V
i
=22.4dm
3
V
f
=?
As given that the expansion is isothermal and reversible
∴ΔU=0
Now from first law of thermodynamics,
ΔU=q+w
∵ΔU=0
∴q=–w
Given that the heat is absorbed.
∴q=1000cal
⇒w=−q=−1000cal
Now,
Work done in a reversible isothermal expansion is given by-
w=−nRTln(
V
i
V
f
)
Given:-
T=0℃=273K
n=1 mol
∴1000=−nRTln(
V
i
V
f
)
⇒1000=−1×2.303×2×273×log(
22.4
V
f
)
Explanation:
<span>2 HCl + Ba(OH)2 = BaCl2 + 2 H2O is the answer.</span>
Answer:
<h2>
= (
1.08 /
2.2
) 100% = 49%</h2>
Explanation:
Balanced Equation: 2CH₃CH₃(g) + 5O₂(g) → 2CO₂(g) + 6H₂O(g)
Calculate moles of CH₃CH₃ and O₂
1.2 ₃₃ (
1 ₃₃/
30.0694 ₃₃
) = 0.040 ₃₃
8.6 ₂ (
1 2/
31.998 ₂
) = 0.27 ₃₃
Find limiting reagent 0.040 ₃₃ (
5 ₂/
2 ₃₃
) = 0.10 ₂
CH₃CH₃ is the limiting Reagent
CH₃CH₃ (L.R.) O₂ CO₂ H₂O
Initial (mol) 0.040 0.27 0 0
Change
(mol)
-2x=-0 -5x=
-0.10 +2x=+0.040 +6x=+0.12
Final (mol) 0 0.117 0.040 0.12
0.040 − 2 = 0 = 0.020
Determine percent yield
0.12 ₂ (
18.0148 ₂
/1 ₂
) = 2.2 ₂
= (
1.08 /
2.2
) 100% = 49%
