Balanced equation: 2Fe + 3H2O → Fe2O3 +3H2
Convert g to mols:
285/55.845 = 5.1034 mols
Mole ratio of Iron and Iron (III) Oxide: 2:1
5.1034/2 = 2.5517 mols
Answer:
The boiling point of HF is <u><em>higher than</em></u> the boiling point of H2, and it is <u><em>higher than</em></u> the boiling point of F2.
Explanation:
In HF, inter- molecule forces will be present between the hydrogen and fluorine atoms. There will be hydrogen bonding present among the hydrogen and fluorine atoms. Hydrogen bonds are strong bonds and hence the boiling point for HF would be high as much energy will be required to break these bonds.
H2 and F2 will only have intra-molecular attractions and there will be no hydrogen bonds present in them. As a result, their boiling point will be lower.
<span>1.75 mol H2O x (6.022x10^23 molecules H2O / 1 mol H2O) = 1.05x10^24 molecules H2O if you need a further example let me know </span>
Answer:
In a semiconductor, the bonding molecular orbitals that contain electrons are referred to as the valence band, while the antibonding orbitals that are completely empty are referred to as the conduction band.
The conduction band occupies a higher energy level than the valence band. The band gap is what separates the two orbitals.
Answer:
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to solve this problem by firstly setting up the equilibrium expression for the given reaction, in agreement to the law of mass action:
![K=\frac{[NO]^2}{[N_2][O_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO%5D%5E2%7D%7B%5BN_2%5D%5BO_2%5D%7D)
Next, we plug in the given concentrations on the data table to obtain:
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