ΔS= nΔHvap/T,
Where, ΔS = Change in entropy, n = moles of water = 39.3/18 = 2.188 moles, ΔHvap = 40.67kJ/mol = 40670 J/mol, T = Temperature (K) = 100+272.15 = 373.15 K
Therefore,
ΔS = (2.188*40670)/373.15 = 237.96 J/K
<h2>
Answer:</h2>
-310J
<h2>
Explanation:</h2>
The change in internal energy (ΔE) of a system is the sum of the heat (Q) and work (W) done on or by the system. i.e
ΔE = Q + W ----------------------(i)
If heat is released by the system, Q is negative. Else it is positive.
If work is done on the system, W is positive. Else it is negative.
<em>In this case, the system is the balloon and;</em>
Q = -0.659kJ = -695J [Q is negative because heat is removed from the system(balloon)]
W = +385J [W is positive because work is done on the system (balloon)]
<em>Substitute these values into equation (i) as follows;</em>
ΔE = -695 + 385
ΔE = -310J
Therefore, the change in internal energy is -310J
<em>PS: The negative value indicates that the system(balloon) has lost energy to its surrounding, thereby making the process exothermic.</em>
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I would say light. Hope this helps
To calculate the strength of the eye lens, we use the formula
( thin lens formula)
Here, u is the object distance and v is image distance.
Given, 
and 
Therefore,
Thus, strength of the eye lens is
