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Kaylis [27]
3 years ago
10

Dipole-induced dipole attractions exist between molecules of water and molecules of gasoline, yet these two substances do not mi

x because water has such a strong attraction for itself. which compound might best help these two substances mix into a single liquid phase
Chemistry
1 answer:
lana [24]3 years ago
3 0

Gasoline is predominantly octane, C8H18.  Something like soap would be a great homogenizer.  Soap is composed of a long hydrocarbon chain with a tiny, highly polar tip on one end.  Usually, the soap is the anion of a salt, NaX.  This allows the polar end of the soap to stick to water, while the nonpolar end sticks to the oil.

\frac{kg*m*m}{m^3*s*\mu}=1=\frac{kg}{m*s*\mu}

\mu=\frac{kg}{m*s}

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Describe the relationship between pure chemistry and applied chemistry
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Pure chemistry: gain knowledge for own pleasure
Applied chemistry: gain knowledge to know how to use it 
7 0
3 years ago
What is the concentration of a solution containing 1.11 g sugar (sucrose, C12H22O11, MW = 342.3 g/mol, d = 1.587 g/cm3) in 432 m
djyliett [7]

Answer:

0.0075 M

0.0060 m

Explanation:

Our strategy here is to use the definition of molarity and molality to solve this question.

The molarity is the number of moles of solute, sucrose in this case, per liter of solution.

The molality is the number of moles of solute per kilogram of solvent.

So the molarity of the  solution is

M = moles of solute/ V solution

As we see we need the volume of solution since we are only given the volume of solvent, but this will be easy to compute since we have the density of  sucrose.

So determine the moles of sucrose , and the volume of solution:

Moles sucrose = 1.11 g/342.3 g/mol = 3.24 x 10⁻³ M

Volume of solution = Vol Sucrose + Vol glycerine

d = m/V ⇒ Vsucrose = m / d = 1.11 g/ 1.587 g/cm³ = 0.70 cm³

Vol solution = 432 mL + 0.70 mL = 432.7 mL  (1cm³  = 1 mL)

Vol solution = 432.7 mL x 1 L / 1000 mL = 0.4327 L

⇒ M = 3.2 x 10⁻³  mol / 0.4327 L = 0.0075  M

For the molarity what we need is to first calculate the kilograms of glycerine from the given density:

d = m/v ⇒ m = d x v = 1.261 g/cm³ x  432 cm³ = 544.75 g

Converting to Kg:

544.75 g x 1 Kg/ 1000 g = 0.544 kg

Now the molality is

m = mol sucrose/ kg solvent = 3.24 x 10⁻³ mol / 0.544 Kg = 0.0060 m

Note: In the calculation for  volume of solution we could have approximated it to that of just glycerine, but since the density of sucrose was given we calculated the total volume of solution to be more rigorous.

8 0
3 years ago
With no difference in temperatures and 100% humidity, will it rain? Why or Why not?
Artyom0805 [142]
Not necessarily, rain needs some mechanism such as instability of vertical air movement..
7 0
3 years ago
The specific heat for liquid ethanol is 2.46 J/(g•°C). When 210 g of ethanol is cooled from
avanturin [10]

Answer:

a) heat it from 23.0 to 78.3

q = (50.0 g) (55.3 °C) (2.46 J/g·°C) =  

b) boil it at 78.3

(39.3 kJ/mol) (50.0 g / 46.0684 g/mol) =  

c) sum up the answers from the two calculations above. Be sure to change the J from the first calc into kJ

Explanation:

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nasty-shy [4]

Answer:

Ionic bond

Explanation:

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