Based on Gay-Lussac's law, which of the following statements is true for an ideal gas at constant volume and mole amount? (4 poi
1 answer:
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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).
Answer:
<u><em>Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.</em></u>
A chemistry student weighs out 0.0941 g of hypochlorous acid (HClo) into a 250. ml. volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.2000 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the equivalence point. Round your answer to 3 significant digits mL.
Explanation:
1 mole HClO = 74.44g
0.0941g = = 0.00126 moles
Concentration = no. of moles/volume in L
Hence, Concentration of HClO = 0.00126/ 0.250L
= 0.005M.
C1V1 =C2V2
0.005 × 250 mL = 0.2 × V2
<u><em>Volume of NaOH, aka V2 = 6.32 mL to 3 sig. fig.</em></u>