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:
A. experiment 1 i believe its the best answer
When 1.00 L of a 0.500 M LiBr solution is tripled by dilution with water . the number of moles of lithium in new solution is 0.5 moles.
given that :
volume = 1.00 L
molarity = 0.500 M
The solution is tripled by dilution but there will be no effect in number of mole by the dilution.
the number of moles = molarity × volume in L
the number of moles = 1 × 0.500
the number of moles = 0.5 mol
Thus, When 1.00 L of a 0.500 M LiBr solution is tripled by dilution with water . the number of moles of lithium in new solution is 0.5 moles.
To learn more about number of moles here
brainly.com/question/14919968
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Well if you were asking it was a true or false questions the answer is: True
Nonane (b) has the highest melting point.
-----------------------------
A caveat: I'm assuming that we're dealing with the straight-chain isomers of these alkanes (specifically pentane and nonane). The straight-chain isomer of pentane (<em>n</em>-pentane, CH3-[CH2]3-CH3) has a melting point of -129.8 °C; the straight-chain isomer of nonane (<em>n-</em>nonane, CH3-[CH2]7-CH3) has a melting point of -53.5 °C. The pattern holds as you go down (or up): The more carbon atoms, the higher the melting point. So, in decreasing order of melting points here, you'd have the following: nonane > pentane > butane > ethane.
However, one structural isomer of pentane, neopentane, has a melting point of -16.4 °C, which is <em>higher </em>that the melting point of <em>n</em>-nonane despite neopentane having the same molecular formula as its straight-chain isomer. Of course, you're not to blame for coming up with this question; this is just some extra info to keep in mind.
