This is probably a little late lol but the answer is the mouth, because when you chew food, it releases nutrients into your body.
Answer:
A) 2.27 grams of tungsten trioxide must be needed to prepare 1.80 g of tungsten.
B)To prepare 1.80 g of tungsten we will need 0.0587 grams of hydrogen gas.
Explanation:
![WO_3(s)+3H_2(g)\rightarrow W(s)+3H_2O(g)](https://tex.z-dn.net/?f=%20WO_3%28s%29%2B3H_2%28g%29%5Crightarrow%20W%28s%29%2B3H_2O%28g%29)
A) Mass of tungsten prepared = 1.80 g
Moles of tungsten =
According to reaction, 1 mol of tungsten is obtained from 1 mole of tungsten trioxide.
Then 0.009783 moles of tungsten will be obtained from:
of tungsten trioxide
Mass of 0.009783 moles of tungsten trioxide :
0.009783 mol × 231.8 g/mol = 2.27 g
2.27 grams of tungsten trioxide must be needed to prepare 1.80 g of tungsten.
B) According to reaction,1 mol of tungsten is produced by 3 moles of hydrogen gas
Then 0.009783 moles of tungsten are produced by :
![\frac{3}{1}\times 0.009783 mol=0.02935 mol](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7B1%7D%5Ctimes%200.009783%20mol%3D0.02935%20mol)
Mass of 0.02935 moles of hydrogen gas:
0.02935 mol × 2 g/mol =0.0587 g
To prepare 1.80 g of tungsten we will need 0.0587 grams of hydrogen gas.
It is endothermic reaction ΔH>0 (sign is +).
Because it is spontaneous reaction ΔG<span><0 (Gibbs free energy)
</span>ΔG=ΔH-TΔS, so must be TΔS>ΔH and ΔS<span>>0 (sign +).</span>
The right answer for the question that is being asked and shown above is that: "-deltaH." The change that will always be true for an exothermic process is a <span>-deltaH. The change that it exhibits is always true for an exothermic process.</span>