What pressure will be exerted by 0.650 mol of a gas at 30.0C if it is contained in a 0.700 L vessel?
1 answer:
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Option (a) is correct.
A reducing agent is the one which loses electrons to other substance and an oxidizing agent is one which accepts electrons.
Here, In
, Cr has oxidation number 6+ in the L.H.S of the equation, but on R.H.S its oxidation number is 0 i.e. it Cr has gained electrons such that total charge is 0.
And the oxidation state of Al in the left-hand side of equation is 0 and in right-hand side, it is +6.i.e. it has donated its electrons to Cr.
Hence, Cr is the oxidizing agent and Al is the reducing agent.
A reducing agent is the one which loses electrons to other substance and an oxidizing agent is one which accepts electrons.
Here, In
And the oxidation state of Al in the left-hand side of equation is 0 and in right-hand side, it is +6.i.e. it has donated its electrons to Cr.
Hence, Cr is the oxidizing agent and Al is the reducing agent.
Explanation:
- As it is given that boiling point of propanamide is very high. So, reason for this is that easy formation of hydrogen bonds which are strong enough that we have to provide large amount of heat to break it.
As in , the hydrogen atoms which are present are positive in nature. Due to this they are able to form hydrogen bonds with the neighboring oxygen atom.
Hence, these bonds are so strong that high heat needs to given to break them.
- A propanoic acid contain carboxylic group as the functional group. So, this group is also able to form hydrogen bonding as it forms a hydrogen bond between an acid group and hydroxyl group of neighboring molecule.
Hence, it will also require high heat to break the bond due to which there will be increase in boiling point.
- In propanal, there is presence of aldehyde functional group and three carbon atoms chain which will not form strong bonding with the hydrogen atom of CHO. Due to this there will exist weak Vander waal's force that is not at all strong enough.
As a result, less energy will be needed to break the bonds in propanal. Hence, it has very low boiling point.
Answer:
The energy profile for rotation about the C-C bond in ethane is shown in the image, along with the Newman projections of the corresponding ethane conformer.
Explanation:
If you see the ethane molecule (second image) from the C-C bond axis (third image), as in the Newman projections, it's easy to draw an angle between one of the hydrogen atoms of the visible carbon, the carbon itself, and one of the hydrogens of the hidden carbon.
When you make a rotation about the C-C bond, the angle between those hydrogens will change. If you start with an eclipsed conformation, with each hydrogen of the hidden C exactly behind the hydrogens of the visible C, the angle will be 0°, or also 120° or 240°, as this rotations will be equivalent. On the other hand, if the angle is 60° (or 180°, or 300°), you will have a staggered conformation. The eclipsed conformation is less stable than the staggered one, because the interactions between hydrogens will be bigger (the repulsion between their electrons), and because of that the eclipsed conformations will be found in the maxima, while the staggered one will be found in the minima.
