Answer: 92mg of ethyl alcohol
2 milli molar of ethyl alcohol means 2 milli mole glucose for 1 liter of water. Since we want to make 1L of solution, then the amount of ethyl alcohol needed is: 2 milli mole/l x 1l= = 2 x 10^-3 mole
Ethyl alcohol molecular weight is 46g/mole, then 2 mili mole of glucose is= 2x 10^-3 mole x 46 grams/mole= 92 grams x 10^-3= 92mg
Answer:
<h3>The man's average body temp. will fall by 0.6°C to 38.4°C</h3>
Explanation:
The enthalpy (heat content) of the water, using a datum of 0°C, is
Hw = Mw kg x Cp,w (specific heat capacity kJ/kg °C) x Tw °C
= 1 kg x 4.18 kJ/kg°C x 3°C = 12.5 kJ
Hman (pre drink) = 68 kg x 3.6 kJ/kg/°C x 39°C = 9547 kJ
So Hman (post drink) = H (pre drink) + Hw = 9547 + 12.5 = 9559.5 kJ because no heat is lost immediately.
But Hman (after drink) has mass 68 + 1 = 69 kg
Also his new Cp will be approx (3.6 x 68/69) + (4.18 x 1/69) = 3.608 kJ/kg°C
So Hman = 9559.5 kJ (from above) = 69 kg x 3.608 kJ/kg°C x Tnew
Therefore the man's new overall temp. = 38.4°C which is a drop of 0.6°C
453.87 (whatever label you use, most highschools are different but i dont want to give you the wrong label!)