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).
Volume is the amount of space the matter takes up, density is mass OF the volume
<span>The </span>ΔH value is -22kcal
Since, we have the reaction as,
2Li(s) + F2(g) --> 2LiF(s)
we are only concerned with the limiting reactants. We calculate for the amount of product that can be produced with the given amount of reactants.
a. 1 g Li(1 mol / 6.941 g of Li)(2 mol LiF/2 mol Li) = 0.144 mol LiF2
1 g F2(1 mol/38 g)(2 mol LiF2/1 mol F2) = 0.052 mol LiF2
Answer: 1 g of F2
b. 10.5 g Li(1 mol/6.941 g of Li)(2 mol LiF/2 mol Li) = 1.512 mol LiF2
37.2 g F2(1 mol/38 g)(2 mol LiF2/1 mol F2) = 1.958 mol LiF2
Answer: 10.5 g of Li
c. (2.85 x 10^3 g Li)(1 mol/6.941 g of Li)(2 mol LiF/2 mol Li) = 410.60 mol LiF2
(6.79 x 10^3 g F2)(1 mol/38 g)(2 mol LiF2/1 mol F2) = 357.368 mol of LiF2
Answer: 6.79 x 10^3 g F2