Answer:
Approximately 0.39 g or 0.4 g if you're rounding up
Explanation:
15/3.82 = 3.92
Let's round that up to 4
That means 15 days is around 4 half lives
4 half lives means 1/16 of the original mass will be left
25/16 = 0.390625
Answer:
B) CH3(CH2)10CO2H
Explanation:
Hello,
In this case, we define amphipathic as a compound having both a hydrophilic and a hydrophobic part, for that reason, the hydrophilic part will be water-soluble (polar) whereas the hydrophobic does not (nonpolar). In such a way, some functional groups such as hydroxyl and carboxyl tend to be polar by cause of the presence of O-H bonds whereas long-carboned chains tend to be nonpolar by cause of the presence of C-H and C-C bonds.
Therefore, since A) CH3CH2OH (ethanol) and D) CH3CO2H (acetic acid) are short-carboned chains with polar groups they are largely hydrophilic whereas C) CH3(CH2)10CH3 (dodecane) is highly nonpolar, we sum up that only B) CH3(CH2)10CO2H (dodecanoic acid or lauric acid) is amphoteric as it has a long-charboned part (nonpolar) and a water-soluble part (polar).
Best regards.
Answer:
E. Fracking
Explanation:
The process involves the high-pressure injection of 'fracking fluid' into a wellbore to create cracks in the deep-rock formations through which natural gas, petroleum, and brine will flow more freely.
Correct Answer: option <span>(1) Mn(s)
Reason:
The </span><span>spontaneity of electrochemical cell, depends on the it's Eo value. Electrochemical cells with positve Eo are spontanous and vice-versa.
</span>
In present case, the Eo of half-cell of interest are as follows:
Eo Zn2+/Zn = <span>-0.763v
</span>Eo Mg2+/Mg = 2.37v
Eo Mn2+/Mn = -1.18v
Therefore, Eo cell (with Zn as one of the half-cell) = Eo Zn2+/Zn - Eo Mn2+/Mn
= -0.763 - (-1.18)
= 0.417v
On other hand, Eo cell (with Mg as one of the half-cell) = Eo Mg2+/Zn - Eo Mn2+/Mn
= -2.37 - (-1.18)
= -1.19v
Thus, Mn(s) <span>metal will spontaneously react with Zn2+(aq), but will not spontaneously react with Mg2+(aq)</span>