50g U ( 1mol/ 238.03g) = .2 moles
238.03 is the molar mass of uranium- can be found on periodic table
Answer:
See explanation
Explanation:
We define the formal charge on an atom in a molecule as the charge it carries assuming that electrons in all chemical bonds of the molecule were shared equally between atoms irrespective of the electronegativity of each atom.
The formula for calculating the formal charge on an atom in a molecule is;
Formal Charge = [number of valence electrons on neutral atom] – [(number of lone electron pairs) + (½ number of bonding electrons)] ·
The formal charge on the two nitrogen atoms in diazomethane is obtained as follows;
Middle nitrogen atom = 5 – 8/2 – 0 = +1
Last nitrogen atom = 5 – 4/2 – 4 = –1
The Lewis structure of the molecule is shown in the image attached.
When we replace neutral ligands with negative ligands then the value of C.Δ will decrease.
In coordination chemistry, a ligand is an ion or molecule that binds to a significant steel atom to shape coordination complicated. The bonding with the metal generally includes a formal donation of 1 or more of the ligand's electron pairs regularly through Lewis Bases.
A ligand is defined as any molecule or atom that irreversibly binds to a receiving protein molecule, otherwise known as a receptor. whilst a ligand binds to its respective receptor, the shape and/or interest of the ligand is altered to provoke several special styles of cellular responses.
Learn more about ligands here:-brainly.com/question/1869211
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A.Δ will remain constant
B.Δ will increase
C.Δ will decrease
D.not enough information
The –OH+ group is most acidic proton in ln-OH as shown in figure (a). The proton is circled in the figure.
The stabilisation of the conjugate base produced is stabilises due to resonance factor. The possible resonance structures are shown in figure (b).
The acidity of a protonated molecule depends upon the stabilisation of the conjugate base produced upon deprotonation. The conjugate base of ln-OH is shown in figure (a).
The possible resonance structures are shown in figure (b). As the number of resonance structures of the conjugate base increases the stabilisation increases. Here the unstable quinoid (unstable) form get benzenoid (highly stable) form due to the resonance which make the conjugate base highly stabilise.
Thus the most acidic proton is assigned in ln-OH and the stability of the conjugate base is explained.