An anterior pituitary gland
Answer:
1.2 mL
Explanation:
<em>This is a problem of simple dilution. The dilution principle simply agrees that the number of moles before dilution must be equal to the number of moles after dilution.</em>
Recall that: number of moles = mass/molar mass or molarity x volume.
Hence, for the dilution principle:
initial molarity x initial volume = final molarity x final volume.
In this case, initial molarity of NaOH = 1 M, initial volume = ?, final molarity = 0.1 M, final volume = 12.0 mL.
Initial volume = final molarity x final volume/initial molarity
= 0.1 x 12/1 = 1.2 mL
It thus means that 1.2 mL of 1 M NaOH would be taken and then diluted up to 12.0 mL mark by the addition of distilled water in order to produce 12.0 mL, 0.10 M NaOH solution.
I think its 4 hope that helped
No. The recorded drop in water in the pipet of the potometer would have been used not only for plant transpiration, but also to return the plant cells to a turgid state. Turgid means to expand<span> due to high fluid content. Plants are required to maintain turgid cells to maintain an upright position as loss of turgidity results in plant wilting.</span>
