Answer:
Noble gases
Explanation:
The noble gases are non-metals that requires the highest amount of energy to remove an electron from their shells.
The reason for this difficult is that their electronic configuration confers a stable configuration them.
- The ionization energy is the energy required to remove the most loosely held electrons in an atom.
- Due to the special stability of noble gases, it is very difficult to remove electrons from an atom of noble gases.
Answer:
a) Mo the electron configuration: 42Mo: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d4
Mo3+ - is Paramagnetic
b) Au - [Xe] 4f14 5d10 6s1
For Au+ is not paramagnetic
c) Mn - [Ar] 3d5 4s2
Mn2+ is paramagnetic
d)Hf -[Xe] 4f¹⁴ 5d² 6s²
Hf2+ is not paramagnetic
Explanation:
An atom becomes positively charged when it looses an electron.
Diamagnetism in atom occurs whenever two electrons in an orbital paired equalises with a total spin of 0.
Paramagnetism in atom occurs whenever at least one orbital of an atom has a net spin of electron. That is a paramagnetic electron is just an unpaired electron in the atom.
Here is a twist even if an atom have ten diamagnetic electrons, the presence of at least one paramagnetic electron, makes it to be considered as a paramagnetic atom.
Simply put paramagnetic elements are one that have unpaired electrons, whereas diamagnetic elements do have paired electron.
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!
Answer is: concentratio of H₃O⁺ ions is 4.2·10⁻³ M.<span>
Chemical reaction: HCOOH(aq) + H</span>₂O(l) ⇄ HCOO⁻(aq) + H₃O⁺(aq).<span>
c(HCOOH) = 0,1 M.
[</span>H₃O⁺] = [HCOO⁻] = x.<span>
[HCOOH] = 0,1 M - x.
</span>Ka = [H₃O⁺] · [HCOO⁻] / [HCOOH].
0,00018 = x² / (0,1 M - x).<span>
Solve quadratic equation: x = </span>[H₃O⁺] = 0,0042 M.
Answer:
N2 + 3H2 → 2NH3. 14 moles a. If 6 mol of H2 are consumed, how many moles of NH3 are produced? ... a 3 mol H₂. 4. How many moles of nitrogen are needed to make 11 moles of NH3? Il mol NH₃ x Imol Na = 15.5
