First, find the number of moles for each element. The molar
mass for nitrogen is 14 g/mol and that of hydrogen is 1 g/mol.
1.40 g N / 14 g/mol = 0.1 mol N
0.20 g H / 1 g/mol = 0.2 mol H
Find the mole ratio. Divide both numbers with the much lower
value. In this case, it is 0.1 mol N.
For N: 0.1 ÷ 0.1 = 1
For H: 0.2÷0.1 = 2
Thus, the empirical formula is
.
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Answer:
A. ![K=\frac{[N_2O]^2}{[N_2]^2[O_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BN_2O%5D%5E2%7D%7B%5BN_2%5D%5E2%5BO_2%5D%7D)
Explanation:
Hello there!
In this case, for us to figure out the appropriate equilibrium expression, it will be firstly necessary for us to recall the law of conservation of mass which states that the equilibrium constant of an equilibrium chemical reaction is written by dividing the products and reactants and including the stoichiometric coefficients as exponents. In such a way, for the given reaction, we will have:
![K=\frac{[N_2O]^2}{[N_2]^2[O_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BN_2O%5D%5E2%7D%7B%5BN_2%5D%5E2%5BO_2%5D%7D)
As N2O is the product whereas N2 and O2 are reactants; thus, the equilibrium expression will be A.
Regards!
Mass of Li = 237.38 g
<h3>Further explanation</h3>
The mole itself is the number of particles contained in a substance amounting to 6.02.10²³

<h3>Known</h3>
Moles of Li = 34.2
Molar mass(MW) of Li = 6.941 g/mol
then mass of Lithium (Li) :
