Answer:
1.55×10²² molecules.
Explanation:
We'll begin by calculating the number of mole in 5.32 g of pure lead (Pb). This can be obtained as follow:
Mass of Pb = 5.32 g
Molar mass of Pb = 207 g/mol
Mole of Pb =?
Mole = mass /molar mass
Mole of Pb = 5.32/207
Mole of Pb = 0.0257 mole
Finally, we shall determine the number of molecules in 0.0257 mole of Pb. This can be obtained as follow:
From Avogadro's hypothesis,
I mole of Pb contains 6.02×10²³ molecules.
Therefore, 0.0257 mole will contain = 0.0257 × 6.02×10²³ = 1.55×10²² molecules.
Therefore, 5.32 g of pure lead (Pb) contains 1.55×10²² molecules.
This is categorized as a combustion reaction.
Answer:
At equilibrium, reactants predominate.
Explanation:
For every reaction, the equilibrium constant is defined as the ratio between the concentration of products and reactants. Thus, for the reaction N2 (g) + O2 (g) ⇌ 2NO the expression of its equilibrium constant is:
![Keq = \frac{[NO]^{2}}{[O_{2} ][N_{2}]}](https://tex.z-dn.net/?f=Keq%20%3D%20%5Cfrac%7B%5BNO%5D%5E%7B2%7D%7D%7B%5BO_%7B2%7D%20%5D%5BN_%7B2%7D%5D%7D)
Since the equilibrium constant is Keq = 4.20x10-31 the concentration of reactants O2 and N2 must be much higher than products to obtain such a small number as 4.20x10-31 at the equilibrium. Hence, at equilibrium reactants predominate.