It could get bigger or it could have something a lot different to it than anything else
For many (but not all) problems, you can simply round the atomic weights and the molar mass to the nearest 0.1 g/mole. HOWEVER, make sure that you use at least as many significant figures in your molar mass as the measurement with the fewest significant figures. In other words, never let your molar mass be the measured value that determines how many signficant figures to use in your answer!
Let's rewrite the reaction for clarity:
2 SO₂(g) + O₂(g) ⇆ 2 SO₃(g) δhºrxn = –198 kj/mol
The equilibrium constant of a reaction is the ratio of the concentration its products to its reactants which are raised to their respective stoichiometric coefficients. For this reaction, the K would be
K = [SO₃]²/[SO₂]²[O₂]
To get a larger K, the products must be greater than the reactants. This means that the forward reaction must be favored to yield more of the product SO₃. There are different ways to do this: by manipulating the pressure, concentration or temperature.
For the concentration, you should add more amounts of the reactants. For the pressure, we should increase it. This is because the product side has only 2 moles of gas compared to 3 moles of gas in the reactants. So, it wall have more room for the product even at a higher pressure. Lastly, since the reaction is exothermic manifested by the negative sign of δhºrxn , the reaction would favor the forward reaction at high temperatures.
Answer :
The correct answer is %IC = 10 % and bond is covalent bond with slight polarity.
<u>Percent Ionic Character :</u>
It is defined as percent of ionic character present in a polar covalent bond . The formula of % ionic character (%IC) is given as follows :
Where Xa = Electronegativity of A atom and Xb = Electronegativity of B atom
Given : Molecule is TiAl₃
Electronegativity of Ti = 2.0
Electronegativity of Al = 1.6 ( From image shared )
Plug the value in above formula :
Value of e⁻¹ = 0.90
Percent ionic character = 1 - 0.90 * 100
Percent Ionic character = 10 %
<u>Since the % IC is 10 % , which is very less comparatively , hence the bond is covalent and very less polar .</u>
