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Iteru [2.4K]
3 years ago
12

How do scientists predict the polarity of molecules, the distribution of non-polar and polar molecules?

Chemistry
2 answers:
EastWind [94]3 years ago
8 0

Answer:

Explanation:

Polar molecules interact with other molecules of similar polarity to form solutions. Non-polar molecules do not interact the same way.

andre [41]3 years ago
5 0

<u>Answer:</u>

The scientists predict the polarity of molecules and the distribution of non polar and polar molecules by finding out the net dipole moment of the molecular structure.

<u>Explanation:</u>

The net value of dipole moment is found out with the help of finding the bond angle between the atoms of the given molecule along with the distance between the atoms and the charge on the atoms of the molecule.

This will be applied in the appropriate formula which tells whether the molecule is polar or non polar.

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A solution is diluted by adding more _____, which means the _____ of the solution increases but the amount (moles) of solute sta
Rudik [331]

Answer:solvent ; volume

Explanation:

6 0
2 years ago
A mixture containing 20 mole % butane, 35 mole % pentane and rest
notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

0.9=\frac{y_bD}{z_bF}

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

8 0
3 years ago
Read 2 more answers
An unknown amount of acid can often be determined by adding an excess of base and then back-titrating the excess. A 0.3471−g sam
MAXImum [283]

Explanation:

The given data is as follows.

       Mass of mixture = 0.3471 g

As the mixture contains oxalic acid and benzoic acid. So, oxalic acid will have two protons and benzoic acid has one proton.

This means oxalic acid will react with 2 moles of NaOH and benzoic acid will react with 1 mole of NaOH.

Hence,   moles of NaOH in 97 ml = \frac{0.1090 \times 97}{1000}

                                                       = 10.573 \times 10^{-3} mol

Moles of HCl in 21.00 ml = \frac{0.2060 \times 21}{1000}

                                         = 4.326 \times 10^{-3} mol

Therefore, total moles of NaOH that reacted are as follows.

           10.573 \times 10^{-3} mol - 4.326 \times 10^{-3} mol      

                = 6.247 \times 10^{-3} mol

So, total 3 mole of NaOH will react with 1 mole of mixture. Therefore, number of moles of NaOH reacted with benzoic acid is as follows.

                  \frac{6.247 \times 10^{-3}}{3}

                    = 2.082 \times 10^{-3} mol

Since, molar mass of NaOH is 40 g/mol. Therefore, calculate the mass of NaOH as follows.

                    2.082 \times 10^{-3} mol \times 40 g/mol

                         = 83.293 \times 10^{-3} g

                         = 0.0832 g

Whereas molar mass of benzoic acid is 122 g/mol.

Therefore,       40 g NaOH = 122 g benzoic acid

So,            0.0832 g NaOH = \frac{122 g}{40 g} \times 0.0832 g

                                             = 0.253 g

Hence, calculate the % mass of benzoic acid as follows.

                      \frac{0.253 g}{0.3471 g} \times 100

                           = 73.10%

Thus, we can conclude that mass % of benzoic acid is 73.10%.

4 0
3 years ago
Three chemistry students measured the length of a copper bar. The recorded lengths were 5.05 cm, 5 cm , and 5.1 cm, What is the
Salsk061 [2.6K]
5.05 + 5 + 5.1 = 15.15cm Then you just divide it by the amount of measurements you had like this:15.15 ÷ 3 = 5.04999971cm Then you can just round it to the 3rd figure: 5.05cm < And that's the mean/average length of the bar. :) (Or the one above if you want all of the decimals too) 
3 0
4 years ago
A unicellular organism has how many cells?<br> O two<br> O many<br> O one<br> O<br> three
Flauer [41]
The answer is many.
5 0
3 years ago
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