Answer:
Explanation:
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Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:
![K=\frac{[NO_2]^2}{[NO]^2[O_2]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BNO%5D%5E2%5BO_2%5D%7D)
Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:
In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.
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<span>The choices are as follows:
h2o + 2o2 = h2o2
fe2o3 + 3h2 = 2fe + 3h2o
al + 3br2 = albr3
caco3 = </span><span>cao + co2
The correct answers would be the second and the last option. The equations that are correctly balanced are:
</span> fe2o3 + 3h2 = 2fe + 3h2o
caco3 = cao + co2
To balance, it should be that the number of atoms of each element in the reactant and the product side is equal.
<span>NaCl is poster-compound for ionic bonding. The bonds in NaCl have about 70% ionic character, making the bond highly polar. its overstatement to state that there is actual ion in NaCl with +1 and -1 charge but actual charge of Na and Cl is +1 and -1 ion, since Nacl exist as a network of highly charged particle and not discrete molecule, NaCl particle does not exhibit intermolecular forces.
Water molecule on other hand exhibit London dispersion force, keesom force, and hydrogen bonding.
The polar water molecule are attracted to the polarized Na and Cl atoms. This is what allow NaCl(s) to dissolve and ionize in water. Therefore type of attraction in NaCl is ion-dipole attraction.</span>
Mass number is apart of (what ever subject it is)
