A cable raises a mass of 158.0 kg with an acceleration of 1.8 m/s2. What force (in N) of tension is in the cable?
1 answer:
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This problem is asking for an explanation of what we need to know about double and triple bonds to successfully predict molecular geometries in molecules. At the end, one comes to the conclusion that double and triple bonds contribute to the degree in which an atom is bonded and they also determine the lone pairs, which, at the same time, define the molecular geometry.
<h3>Molecular geometry:</h3>In chemistry, molecules are not necessarily flat arrangements of atoms, yet they have specific bond angles, orientations and shapes, which define the molecular geometry. In such a way, we can use the VSEPR theory in order to know the molecular geometry of a molecule; however, we first need its Lewis structure or at least the number and type of bonds to do so.
Consider water and carbon dioxide; the former has two hydrogen to oxygen bonds (O-H) and 2 lone pairs because O has six valence electrons but just 2 are bonded to complete the octet, so 4 unpaired electrons lead to two lone pairs. On the other hand, the latter has two double bonds (C=O) and 0 lone pairs because carbon has four valence electrons and they are all bonded to complete the octet.
In such a way, one can see how the double bond affected the bonding in CO2 in contrast to the H2O; situation that also applies to triple bonds, because CO2 has a linear molecular geometry whereas water has a bent one (see attached picture)
Hence, one comes to the conclusion that double and triple bonds contribute to the degree in which an atom is bonded and they also determine the lone pairs, which, at the same time, define the molecular geometry.
Learn more about molecular geometry: brainly.com/question/7558603
Learn more about the VSEPR theory: brainly.com/question/14225705
Answer:
Ether
SN1 mechanism
Explanation:
The nucleophile in this reaction is CH3OH. It is a poor nucleopile. We already know that a poor nucleophile reacting with a tertiary alkyl halide often leads to the substitution product as the major product.
Also, the iodide ion is a good leaving group. This makes the SN1 substitution more likely yielding the ether as the major product as shown in the image attached.
