<span>Now consider a low pressure area on a disk as shown below.A parcel of air at point A would move toward the center of the low pressure area. That movement would take it farther away from the center of the disk and therefore it would move to the west. A parcel of air at B would move toward the center of the low pressure area which would also take it closer to the center of the spinning disk where its speed is greater than the surrounding points. It would appear to move to the east. With A moving to the west and B moving to the east the line from A to B is rotating counterclockwise.</span>
Answer:
ΔG°rxn = +50.8 kJ/mol
Explanation:
It is possible to obtain ΔG°rxn of a reaction at certain temperature from ΔH°rxn and S°rxn, thus:
<em>ΔG°rxn = ΔH°rxn - T×S°rxn (1)</em>
In the reaction:
2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l)
ΔH°rxn = 3×ΔHfNO2 + ΔHfH2O - (2×ΔHfHNO3 + ΔHfNO)
ΔH°rxn = 3×33.2kJ/mol + (-285.8kJ/mol) - (2×-207.0kJ/mol + 91.3kJ/mol)}
ΔH°rxn = 136.5kJ/mol
And S°:
S°rxn = 3×S°NO2 + S°H2O - (2×S°HNO3 + S°NO)
ΔH°rxn = 3×0.2401kJ/molK + (0.0700kJ/molK) - (2×0.146kJ/molK + 0.2108kJ/molK)
ΔH°rxn = 0.2875kJ/molK
And replacing in (1) at 298K:
ΔG°rxn = 136.5kJ/mol - 298K×0.2875kJ/molK
<em>ΔG°rxn = +50.8 kJ/mol</em>
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Answer:
The formation of a meander. As the river erodes laterally, to the right side then the left side, it forms large bends, and then horseshoe-like loops called meanders . The formation of meanders is due to both deposition and erosion and meanders gradually migrate downstream.
The cations has positive charges that are metals while the anions have negative charges that are non-metals. Upon reaction, there is an exchange in charges that are reflected in the subscripts of the atoms. In this case, compound AX2 must have a cation, A belonging to group 2 A with +2 charge and anion, X belonging to Group 7A with -1 charge. Answer is D.
