Explanation:
Carbon dioxide is a polar molecule whose positive center is on the carbon atom: This positive center is able to attract (and accept) the lone electron pairs present on the oxide ion (O2-). carbon dioxide is acts as a Lewis acid
A Lewis acid can accept a pair of electrons from a Lewis base. The boron in BF3 is electron poor and has an empty orbital, so it can accept a pair of electrons, making it a Lewis acid. A Lewis acid is defined as an electron-pair acceptor.
In CO molecule, there is a lone pair on both carbon and oxygen. The substance which can donate an electron pair are called Lewis base. It is clear that CO molecule can donate an electron pair and hence, it is a Lewis base. Also, CO can be BOTH a Lewis acid and base.
Oxygen is a Lewis base (that too a weak one), not a Lewis acid. REASON: It has lone pair of electrons, which can be donated to electron-deficient species (Lewis acids).
Methane is Neither a Lewis Acid or Lewis Base.
Answer:
The equilbrium constant is 179.6
Explanation:
To solve this question we can use the equation:
ΔG = -RTlnK
<em>Where ΔG is Gibbs free energy = 12.86kJ/mol</em>
<em>R is gas constant = 8.314x10⁻³kJ/molK</em>
<em>T is absolute temperature = 298K</em>
<em>And K is equilibrium constant.</em>
Replacing:
12.86kJ/mol = -8.314x10⁻³kJ/molK*298K lnK
5.19 = lnK
e^5.19 = K
179.6 = K
<h3>The equilbrium constant is 179.6</h3>
0.625 moles of ge X 6.02x10^23 atoms/ 1 mol of ge equal to 3.76x10^23 atoms of ge, just times with Avogadro.
The amount of grams that are in 2.3 moles of N = 32.223 or 32/100
Because there are 14.01 grams per mile of nitrogen atoms.
So…
14.01 x 2.3= 32.223
Hope this helps :)
Answer: The correct answer is -297 kJ.
Explanation:
To solve this problem, we want to modify each of the equations given to get the equation at the bottom of the photo. To do this, we realize that we need SO2 on the right side of the equation (as a product). This lets us know that we must reverse the first equation. This gives us:
2SO3 —> O2 + 2SO2 (196 kJ)
Remember that we take the opposite of the enthalpy change (reverse the sign) when we reverse the equation.
Now, both equations have double the coefficients that we would like (for example, there is 2S in the second equation when we need only S). This means we should multiply each equation (and their enthalpy changes) by 1/2. This gives us:
SO3 —>1/2O2 + SO2 (98 kJ)
S + 3/2O2 —> SO3 (-395 kJ)
Now, we add the two equations together. Notice that the SO3 in the reactants in the first equation and the SO3 in the products of the second equation cancel. Also note that O2 is present on both sides of the equation, so we must subtract 3/2 - 1/2, giving us a net 1O2 on the left side of the equation.
S + O2 —> SO2
Now, we must add the enthalpies together to get our final answer.
-395 kJ + 98 kJ = -297 kJ
Hope this helps!
