The 18o-labeled methanol (CH3O*H) will appear in the products side at position b.
<h3>
Position of 18o-labeled methanol in the products</h3>
The 18O label will appear at position b in the product as indicated in the image.
This methoxy group in the product formed in position b comes from the 18O-labeled methanol (CH3OH).
While the oxygens at positions a and c in the product come from the unlabeled hemiacetal.
Thus, the 18o-labeled methanol (CH3O*H) will appear in the products side at position b.
Learn more about methanol here: brainly.com/question/17048792
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Ice floats because it is about 9% less dense than liquid water. ... The heavier water displaces the lighter ice, so ice floats to the top. One consequence of this is that lakes and rivers freeze from top to bottom, allowing fish to survive even when the surface of a lake has frozen over
There are 3 sig figs. 8 and 2 are obvious. The ending 0 is significant because it is at the end of a decimal number, and doesn't have to be measured.
<span>First - you need the empirical formula.
So, assume you have 100 g of the compound.
If so, you'll have 54.53 gram of C, 9.15 g of H and 36.32 g of O. Find the number of moles of each.
54.53 g C (1 mole C / 12.01 g C) = 4.540
9.15 g H (1 mole H / 1.008 g H) = 9.077
36.32 g O (1 mole O / 15.9994 g O) = 2.270
Take the smallest number found and divide the others by it to get the empirical formula.
4.540/2.270 = 2.
9.077/2.270 = 4.
2.270/2.270 =1.
So, that gives you the empirical formula of C2H4O.
Find the weight of this compound. C = 12, H = 1, O = 16. So, C2H4O is 44 amu.
132/44 = 3.
So, 3 (C2 H4 O) = C6H12O3 = molecular formula.</span>
Given:
ΔV = 5 - 1.48 L = 3.52 L = 0.00352 m³, th chane in volume
p = 0.818 atm = 0.819*101325 Pa = 82985.2 Pa
By definition, the work done is

Answer: 292.1 J