Chemistry, 25.09.2019 00:20 latdoz0952

Aliquid mixture contains water (h2o, mw = 18.0), ethanol (c2h5oh, mw = 46.0) and methanol (ch3oh, mw = 32.0). using two different analytical techniques to analyze the mixture, it was determined that the water mole fraction was 0.250 while the water mass fraction was 0.134. determine the mole fraction ethanol (c2h5oh) and the mole fraction methanol (ch3oh) in the solution. report the values to the correct number of significant figures.

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Answer from: Angelav3

Mole fraction of ethanol is 0.363.

Mole fraction of methanol is 0.387.

Explanation:

Mole fraction of water = $\chi_1=0.250$

Mole fraction of ethanol = $\chi_2$

Mole fraction of methanol = $\chi_3$

$\chi_1+\chi_2+\chi_3=1$

$\chi_2+\chi_3=1-\chi_1=0.750$

$\chi_2+\chi_3=0.750$

$\chi_1=\frac{n_1}{n_1+n_2+n_3}$

$\chi_2=\frac{n_2}{n_1+n_2+n_3}$

$\chi_3=\frac{n_3}{n_1+n_2+n_3}$

n_1+n_2+n_3=1

Moles of water = n_1=0.250 mol

Moles of ethanol= n_2

Moles of methanol= n_3

n_2+n_3=0.750 mol

Mass of the mixture = M

Mass of water, m_1

$=n_1\times 18.0 g/mol=0.250 mol\times 18.0 g/mol=4.5 g$

Fraction of water by mass = 0.134

$\frac{n_1\times 18.0 g/mol}{M}=0.134$

$M=\frac{0.250 mol\times 18.0 g/mol}{0.134}=33.58 g$

Mass of ethanol = m_2

Mass of methanol = m_3

m_1+m_2+m_3=M

4.5 g+m_2+m_3=33.58 g

m_2+m_3=29.08 g ..[1]

$\frac{m_2}{46.0 g/mol}+\frac{m_3}{32.0 g/mol}=0.750 mol$

16m_1+23m_3=552 ..[2]

On solving [1] and [2]:

m_2 = 16.70, m_3= 12.38 g

Mole fraction of ethanol = chi_2

$\chi_2=\frac{n_2}{n_1+n_2+n_3}$

$=\frac{\frac{16.70 g}{46.0 g/mol}}{1 mol}= 0.363$

Mole fraction of methanol = chi_3

$\chi_3=\frac{n_3}{n_1+n_2+n_3}$

$=\frac{\frac{12.38 g}{32.0 g/mol}}{1 mol}= 0.387$

Answer from: Quest

Answer: abeexplanation:

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.

Answer from: Quest

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence, the electronegativity (or the ability to attract electrons) is lessened."

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Aliquid mixture contains water (h2o, mw = 18.0), ethanol (c2h5oh, mw = 46.0) and methanol (ch3oh, mw...

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