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:
Your protons are correct but it’s 28 neutrons not 27!
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We will use the formula for freezing point depression :
but first, we need to get the molality m of the solution:
- molality m = moles of C2H5OH / mass of water Kg
when moles of C2H5OH = mass of C2H5OH/ molar mass of C2H5OH
= 11.85 g / 46 g/mol
= 0.258 moles
and when we have the mass of water Kg = 0.368 Kg
so, by substitution on the molality formula:
∴ molality m = 0.258 moles / 0.368 Kg
= 0.7 mol/Kg
and when C2H5OH is a weak acid so, there is no dissociation ∴ i = 1
and when Kf is given = 1.86 C/m
so by substitution on ΔTf formula:
when ΔTf = i Kf m
∴ ΔTf = 1 * 1.86C/m * 0.7mol/Kg
= 1.302 °C
The answer is B. 6CO2+H2O yields C6H12O6+ 6H20.
