Answer:
- <u><em>g) Neither plant should increase by 1 cm in height.</em></u>
Explanation:
See the graph for this question on the figure attached.
The growing of the <em>plant A</em> is represented by the line that goes above the other. At start, that line has a slope that rises about 0.75 cm ( height increase) in 1 day. From the day 2 and forward the slope of the line decreases. The line reaches its highest point about at day 4 and seems to start decreasing. Thus, you should predict that on the day six it <em>most likely </em>does not increase in height.
The growing of the <em>plant B</em> is represented by the line drawn below the other. As for the plant B, the growing decreases with the number of days. Between the days 4 and 5 the line is almost flat, which means that <em>most likely</em> this plant will not grow on the day six or grow less than 0.5 cm.
Thus, for both plants you can say that <em>on day six, most likley, neither should increase by 1 cm in height (</em>option g).
The elements of the "Noble" gases group is nonreactive. The reason for this is that noble gases are always or most of the time at room temperature.<span />
Explanation:



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<h2>yes can you. do this</h2>
Answer : The correct option is, (b) +115 J/mol.K
Explanation :
Formula used :

where,
= change in entropy
= change in enthalpy of vaporization = 40.5 kJ/mol
= boiling point temperature = 352 K
Now put all the given values in the above formula, we get:



Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K
Descriptive or Correlational. hope this helps