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1 year ago

For which of the following species are the dispersion forces strongest?

1 answer:

5 0

In order to see which species has the strongest dispersion forces, you need to calculate their molar mass, because the higher the molar mass, the stronger the dispersion forces.
Since E. C8H18 has the highest molar mass, its dispersion forces are also the strongest ones.

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A sample of vinegar was found to have an acetic acid concentration of 0.8846 m. What is the acetic acid % by mass? Assume the de

jenyasd209 [6]

Answer:

5.3%

Explanation:

Let the volume be 1 L

volume , V = 1 L

use:

number of mol,

n = Molarity * Volume

= 0.8846*1

= 0.8846 mol

Molar mass of CH3COOH,

MM = 2*MM(C) + 4*MM(H) + 2*MM(O)

= 2*12.01 + 4*1.008 + 2*16.0

= 60.052 g/mol

use:

mass of CH3COOH,

m = number of mol * molar mass

= 0.8846 mol * 60.05 g/mol

= 53.12 g

volume of solution = 1 L = 1000 mL

density of solution = 1.00 g/mL

Use:

mass of solution = density * volume

= 1.00 g/mL * 1000 mL

= 1000 g

Now use:

mass % of acetic acid = mass of acetic acid * 100 / mass of solution

= 53.12 * 100 / 1000

= 5.312 %

≅ 5.3%

3 0

2 years ago

How many sodium atoms are in 0.1310 g of sodium

Lana71 [14]

To convert grams to atoms, we first need to find the moles and then multiply by Avogadro's constant.

0.1310 g * (1 mol/22.99 g) * (6.022*10^23 atom/1 mol) = 3.431 *10^21 atoms.

3 0

2 years ago

Read 2 more answers

A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600. g of water at 20.0∘C. If the heat capacity of the bomb c

Nata [24]

Answer:

22.7

Explanation:

First, find the energy released by the mass of the sample. The heat of combustion is the heat per mole of the fuel:

ΔHC=qrxnn

We can rearrange the equation to solve for qrxn, remembering to convert the mass of sample into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:

qrxn=−(qwater+qbomb)

The heat absorbed by the water can be calculated using the specific heat of water:

qwater=mcΔT

The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:

qbomb=CΔT

Combine both equations into the first equation and substitute the known values, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute the terms of each multiplication and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)–(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)–(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)–50208 J+(420. J∘C×Tfinal)–8400 J]

Add the like terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, solve for Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

The answer should have three significant figures, so round to 22.7∘C.

8 0

2 years ago

In a laboratory experiment, the freezing point of an aqueous solution of glucose is found to be -0.325°C, What is the molal conc

Hatshy [7]

0.17 M is the is the molal concentration of this solution

Explanation:

Data given:

freezing point of glucose solution = -0.325 degree celsius

molal concentration of the solution =?

solution is of glucose=?

atomic mass of glucose = 180.01 grams/mole

freezing point of glucose = 146 degrees

freezing point of water = 0 degrees

Kf of glucose = 1.86 °C

ΔT = (freezing point of solvent) - (freezing point of solution)

ΔT = 0.325 degree celsius

molality =?

ΔT = Kfm

rearranging the equation:

m = $\frac{0.325}{1.86}$

m= 0.17 M

molal concentration of the glucose solution is 0.17 M

3 0

1 year ago

If you add 25.0 mL of water to 125 mL of a 0.150 M LiOH solution, what will be the molarity of the resulting diluted solution?

Alborosie

Concentration is the number of moles of solute in a fixed volume of solution

Concentration(c) = number of moles of solute(n) / volume of solution (v)

25.0 mL of water is added to 125 mL of a 0.150 M LiOH solution and solution becomes more diluted.

original solution molarity - 0.150 M

number of moles of LiOH in 1 L - 0.150 mol

number of LiOH moles in 0.125 L - 0.150 mol/ L x 0.125 L = 0.01875 mol

when 25.0 mL is added the number of moles of LiOH will remain constant but volume of the solution increases

new volume - 125 mL + 25 mL = 150 mL

therefore new molarity is

c = 0.01875 mol / 0.150 L = 0.125 M

answer is 0.125 M

7 0

2 years ago