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Wittaler [7]
2 years ago
8

The mole fraction of hcl in a solution prepared by dissolving 5.5 g of hcl in 200 g of c2h6o is ________. the density of the sol

ution is 0.79 g/ml.
Chemistry
1 answer:
Thepotemich [5.8K]2 years ago
5 0
The answer:
all that we search for is the number of mole of HCl and the number of mole of C2H6O

M(HCl) = 5.5g/ mole of HCl , so mole of HCl = 5.5/M(HCl), where M(HCl) is the molar mass.
M(HCl) = 1+ 36.5= 37.5 

moles of HCl = 5.5/37.5=0.14 

M(C2H6O) = 200g / moles of C2H6O, so moles of C2H6O=200g / M(C2H6O)

M(C2H6O)= 2x12+ 6 + 16=46,

moles of C2H6O=200g / 46 =<span>4.35 </span><span> moles 
</span>
the sum of the moles is    0.14 + <span>4.35 </span> = 4.501 moles

finally,  <span>The mole fraction of hcl in a solution prepared by dissolving 5.5 g of hcl in 200 g of c2h6o is 0.031
</span>
because it can be found by  0.14 / 4.501= 0.031

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Safrole was once used as a flavoring in root beer, until it was banned in 1960. what is the vapor pressure of a solution prepare
BartSMP [9]
Answer is: 48,25 torr.
Raoult's Law: p = x(solv) · p(solv)
p - <span>vapour pressure of a solution.
</span>x(solv) - <span>mole fraction of the solvent.
</span>p(solv) - <span>vapour pressure of the pure solvent.
</span>n(ethanol) = 950g ÷ 46,07g/mol = 20,62 mol.
x(solv) = moles of solvent ÷ total number of moles
x(solv) = 20,62 ÷ 21,77 = 0,965.
p = 0,965 ·50,0 torr = 48,25 torr.

4 0
2 years ago
Read 2 more answers
A 23.0g sample of a compound contains 12.0g of C, 3.0g of H, and 8.0g of O.What the empirical formula of the compound
Kryger [21]

Answer:

The empirical formula of compound is C₂H₆O.

Explanation:

Given data:

Mass of carbon = 12 g

Mass of hydrogen = 3 g

Mass of oxygen = 8 g

Empirical formula of compound = ?

Solution:

First of all we will calculate the gram atom of each elements.

no of gram atom of carbon = 12 g / 12 g/mol = 1 g atoms

no of gram atom of hydrogen = 3 g / 1 g/mol = 3 g atoms

no of gram atom of oxygen = 8 g / 16 g/mol = 0.5 g atoms

Now we will calculate the atomic ratio by dividing the gram atoms with the 0.5 because it is the smallest number among these three.

          C:H:O  =     1/0.5  :   3/0.5  :   0.5/0.5

          C:H:O  =     2      :     6      :     1

The empirical formula of compound will be C₂H₆O

5 0
2 years ago
A large cylindrical tank contains 0.750 m3 of nitrogen gas at 27°C and 7.50 * 103 Pa (absolute pressure). The tank has a tight‐f
Nuetrik [128]

Answer: The answer is 68142.4 Pa

Explanation:

Given that the initial properties of the cylindrical tank are :

Volume V1= 0.750m3

Temperature T1= 27C

Pressure P1 =7.5*10^3 Pa= 7500Pa

Final properties of the tank after decrease in volume and increase in temperature :

Volume V2 =0.480m3

Temperature T2 = 157C

Pressure P2 =?

Applying the gas law equation (Charles and Boyle's laws combined)

P1V1/T1 = P2V2/T2

(7500 * 0.750)/27 =( P2 * 0.480)/157

P2 =(7500 * 0.750* 157) / (0.480 *27)

P2 = 883125/12.96

P2 = 68142.4Pa

Therefore the pressure of the cylindrical tank after decrease in volume and increase in temperature is 68142.4Pa

8 0
2 years ago
Jane made this picture to represent a chemical reaction: Two circles, one white and the other gray are shown on the left. A smal
7nadin3 [17]

It represents a decomposition reaction because one reactant breaks apart and forms two products.

Explanation:

The type of chemical reaction depicted by Jane represents a decomposition reaction because on reactant breaks apart and forms two products.

In this reaction XY breaks apart to form X and Y. This is a decomposition reaction.

                     XY   →   X + Y

  • Decomposition or cracking is the formation of two or more products from a single reactant.
  • The break down of a compound into individual atoms or molecules falls into this category.
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      Find the image attached for more explanation:

Learn more:

Thermal decomposition brainly.com/question/11181911

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8 0
2 years ago
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Lithium has an atomic mass of 6.941 amu. Lithium has two common isotopes. The one isotope has a mass of 6.015 amu and a relative
koban [17]

Answer:

The atomic mass of second isotope is 7.016

Explanation:

Given data:

Average Atomic mass of lithium = 6.941 amu

Atomic mass of first isotope = 6.015 amu

Relative abundance of first isotope = 7.49%

Abundance of second isotope = ?

Atomic mass of other isotope = ?

Solution:

Total abundance = 100%

100 - 7.49 = 92.51%

percentage abundance of second isotope = 92.51%

Now we will calculate the mass if second isotope.

Average atomic mass of lithium = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass)  / 100

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6.941 =  45.05235 + (x92.51) / 100

6.941×100 = 45.05235 + (x92.51)

694.1 - 45.05235   = (x92.51)

649.04765 = x 92.51

x = 485.583 /92.51

x = 7.016

The atomic mass of second isotope is 7.016

3 0
2 years ago
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