Answer:
1. C + O₂ → CO₂
2. C + CO₂ → 2 CO
3. Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂
<u>Answer:</u> For the given amount of sweat lost, the amount of energy required will be 692,899 Joules.
<u>Explanation:</u>
We are given:
Heat of vaporization for water = 2257 J/g
Amount of sweat lost = 307 grams
Applying unitary method:
For 1 g of sweat lost, the energy required is 2257 Joules
So, for 307 grams of sweat lost, the energy required will be = ![\frac{2257J}{1g}\times 307g=692,899J](https://tex.z-dn.net/?f=%5Cfrac%7B2257J%7D%7B1g%7D%5Ctimes%20307g%3D692%2C899J)
Hence, for the given amount of sweat lost, the amount of energy required will be 692,899 Joules.
1 atm = 760 mm Hg
Where 760 mm is the height of Mercury for 1 atm.
For 0.900 atm,
Height of mercury = 0.900*760 = 684 mm Hg