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Wittaler [7]
3 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]3 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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vaieri [72.5K]

Answer:

m = 3.4126 g

Explanation:

First, the question is incomplete but I already put in the comments the rest of the question.

Let's solve the first two questions, and then the actual question you are asking here to give you a better explanation of how to do it.

1) We need the volume of the pool, in this case is easy. Assuming the pool is rectangular, we use the volume of a parallelepiped which is the following:

V = h * d * w

We have the data, but first we will convert the feet to centimeter. This is because is easier to work the volume in cm³ than in feet.

So the height, width and depth of the pool in centimeter are:

h = 32 * 30.48 = 975.36 cm

w = 18 * 30.48 = 548.64 cm

d = 5.3 * 30.48 = 161.54 cm

Now the volume:

V = 975.36 * 548.64 * 161.54

V = 86,443,528.79 cm³ or 86,443,528.79 mL or 86,443.5 L

2) If the pool has a pH of 6.4, the concentration of H+ can be calculated with the following expression:

[H+] = antlog(-pH) or 10^(-pH)

Replacing we have:

[H+] = 10^(-6.4)

[H+] = 3.98x10^-7 M

3) Finally the question you are asking for.

According to the reaction:

Na2CO3 + H+ → 2Na+ + HCO3−

We can see that there is ratio of 1:1 between the H+ and the Na2CO3, so, if we have initially a concentration of 3.98x10^-7 M, the difference between the new concentration of H+ and the innitial, will give the concentration to be added to the pool to raise the pH. Then, with the molecular weight of Na2CO3 (105.98 g/mol) we can know the mass needed.

The new concentration of [H+] is:

[H+] = 10^(-7.6) = 2.58x10^-8 M

The difference of both [H+] will give concentration of Na2CO3 used:

3.98x10^-7 - 2.58x10^-8 = 3.73x10^-7 M

The moles:

moles = 3.73x10^-7 * 86,443.5 = 0.0322 moles

Finally the mass:

m = 0.0322 * 105.98

m = 3.4126 g    

5 0
3 years ago
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dybincka [34]

Answer:

132.17 g

Explanation:

The reaction given , in the question is -

    CH₄ (g ) + 4 S ( g ) ---> CS₂ ( g ) + 2H₂S  ( g )

From the reaction , 4 mole of S is required for the production of 1 mole of  CS₂ .

since ,

Moles of  CS₂  = given mass of CS₂ / Molecular weight  of  CS₂

Since ,

the Molecular weight of  CS₂ = 76

Given ,  mass of CS₂ =  72.57 g

Moles of  CS₂ = 72.57  / 76 = 0.95 mol

Since ,

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2. Isomers.

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6. Electronegativity.

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Explanation:

1. A tetrahedral carbon is Sp^3 hybridized while a linear carbon is Sp hybridized. A tetrahedral carbon typically comprises of four bonds that are 109. 5° apart while a linear carbon atom comprises of two (2) bonds that are 180° apart.

2. Two different compounds that have the same molecular formula are known as isomer. For example Butane and Isobutane, Methoxyethane and Propanol have the same molecular formula (numbers of hydrogen and carbon atoms) but different structural formula.

3. Pi (π) bonds are generally weaker than sigma (σ) bonds. This is because the orbital paths of Pi bonds are parallel thereby causing an overlap.

4. Hybridization is the combination of two or more atomic orbitals to form the same number of hybrid orbitals, each having the same shape and energy.

5. A Pi bond is formed by side-by-side overlap of two p orbitals.

6. The electronegativity is a measure of an atom’s attraction for electrons in a bond and indicates how much a particular atom "wants" electrons.

7. Two Lewis structures that have the same atomic placement and σ structure but a different arrangement of π electrons are called resonance structures.

8. All single bonds are Sigma bonds.

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3 years ago
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Two is wrong the rest are right
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