Visual representation of covalent bonding indicating the valence shell electrons in the molecule, lines represents the shared pair of electron and pair of electrons that are not involved in bonding are represented as dots(lone pairs) are known as Lewis structures.
Compound formation takes place in order to complete the octet of each element that is according to octet rule, each atom forms bond with other atom in order to complete their octet that is to get eight electrons in its valence shell and attain stability.
An organic compound of the form
is known as ketene.
The given ketene is
.
The number of valence electron of:



The number of valence electrons in
= 
2 electrons are involved in each single bond between carbon and hydrogen and 4 electrons are involved in each double bond formed between carbon-carbon and carbon-oxygen. Hence, the total number of electrons involved in bond formation are 12 and rest 2 pair of electrons are present on oxygen as lone pair of electrons.
Therefore, the attached image is the Lewis structure of
.
Answer:
The given statement is false.
Explanation:
- However, if the pluton exists beneath the ground, this could be conveniently shown in the illustration something from the peak such pluton appears convex in form resembling a lopolith and perhaps diapir, which would be a particular form of statistically significant pluton recognized as the sill.
- Mostly from the figure it could also be shown that subsurface sheets are lined or curved, throughout the pluton mold. And therefore it is inferred that such a pluton creates adjustment to something like the ground atmosphere by altering the form of the levels above it.
So that the given is incorrect.
a) (NH4)2SO4 --- 1 mole of it contains 2 moles of N, 8 moles of H, 1 mole of S, and 4 moles of O.
MM = (2 moles N x 14.0 g/mole) + (8 moles H x 1.01 g/mole) + (1 mole S x 32.1 g/mole) + (4 moles O x 16.0 g/mole) = 132 g/mole.
6.60 g (NH4)2SO4 x (1 mole (NH4)2SO4 / 132 g (NH4)2SO4) = 0.0500 moles (NH4)2SO4
b) The molar mass for Ca(OH)2 = 74.0 g/mole, calculated like (NH4)2SO4 above.
4.5 kg Ca(OH)2 x (1000 g / 1 kg) x (1 mole Ca(OH)2 / 74.0 g Ca(OH)2) = 60.8 moles Ca(OH)2
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