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bixtya [17]
3 years ago
15

At a certain temperature the vapor pressure of pure heptane C7H16 is measured to be 454.mmHg. Suppose a solution is prepared by

mixing 102.g of heptane and 135.g of chloroform CHCl3.
Calculate the partial pressure of heptane vapor above this solution. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
OlgaM077 [116]3 years ago
5 0

Answer:

Mass of heptane = 102g

Vapor pressure of heptane = 454mmHg

Molar mass of heptane = 100.21

No of mole of heptane = mass/molar mass = 102/100.21

No of mole of heptane = 1.0179

Therefore the partial pressure of heptane = no of mole heptane *Vapor pressure of heptane

Partial pressure of heptane = 1.0179*454mmHg

Partial pressure of heptane = 462.1096 = 462mmHg

the partial pressure of heptane vapor above this solution = 462mmHg

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A balloon contains 2.0 L of air at 101.5 kPa . You squeeze the balloon to a volume of 0.25 L.
8_murik_8 [283]
<h3>Answer:</h3>

812 kPa

<h3>Explanation:</h3>
  • According to Boyle's law pressure and volume of a fixed mass are inversely proportional at constant absolute temperature.
  • Mathematically, P\alpha \frac{1}{V}

At varying pressure and volume;

P1V1=P2V2

In this case;

Initial volume, V1 = 2.0 L

Initial pressure, P1 = 101.5 kPa

Final volume, V1 = 0.25 L

We are required to determine the new pressure;

P2=\frac{P1V1}{V2}

Replacing the known variables with the values;

P2=\frac{(101.5)(2.0L)}{0.25L}

           = 812 kPa

Thus, the pressure of air inside the balloon after squeezing is 812 kPa

8 0
3 years ago
If this decay has a half life of 2.60 years, what mass of 72.5 g of sodium 22 will remain after 15.6 years
Vlad1618 [11]

Sodium-22 remain : 1.13 g

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.  

Usually, radioactive elements have an unstable atomic nucleus.  

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{T/t\frac{1}{2} }}}

T = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

half-life = t 1/2=2.6 years

T=15.6 years

No=72.5 g

\tt Nt=72.5.\dfrac{1}{2}^{15.6/2.6}\\\\Nt=72.5.\dfrac{1}{2}^6\\\\Nt=1.13~g

8 0
3 years ago
a) What substances are present in an aqueous buffer composed of HC2H3O2 and C2H3O2 - ?b) What happens when LiOH is added to a bu
Alex17521 [72]

Answer:

a) HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻

b) HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

Explanation:

a) In a HC₂H₃O₂/C₂H₃O₂⁻ buffer system, the following reactions take place:

HC₂H₃O₂ + H₂O ⇄ C₂H₃O₂⁻ + H₃O⁺

C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻

Thus, the species present are: HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻.

b) When LiOH is added to the buffer system, it is partially neutralized according to the following equation.

HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) When HBr is added to the buffer system, it is partially neutralized according to the following equation.

C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

3 0
3 years ago
Which determines the reactivity of an alkali metal?
adelina 88 [10]

Answer:

its ability to lose electron

3 0
3 years ago
Read 2 more answers
Sighting along the C2-C3 bond of 2-methylbutane, the least stable conformation (Newman projection) has a total energy strain of
natima [27]

Answer:

21 KJ/mol

Explanation:

For this question, we have to start with the <u>linear structure</u> of 2-methylbutane. With the linear structure, we can start to propose all the <u>Newman projections</u> keep it in mind that the point of view is between carbons 2 and 3 (see figure 1).

Additionally, we have several <u>energy values for each interaction</u> present in the Newman structures:

-) Methyl-methyl <em>gauche: 3.8 KJ/mol</em>

-) Methyl-H <em>eclipse: 6.0 KJ/mol</em>

-) Methyl-methyl <em>eclipse: 11.0 KJ/mol</em>

-) H-H <em>eclipse:</em> 4.0 KJ/mol

Now, we can calculate the energy for each molecule.

<u>Molecule A</u>

In this molecule, we have 2 Methyl-methyl <em>gauche </em>interactions only, so:

(3.8x2) = 7.6 KJ/mol

<u>Molecule B</u>

In this molecule, we have a Methyl-methyl <em>eclipse </em>interaction a Methyl-H <em>eclipse </em>interaction and an H-H <em>eclipse</em> interaction, so:

(11)+(6)+(4) = 21 KJ/mol

<u>Molecule C</u>

In this molecule, we have 1 Methyl-methyl <em>gauche </em>interaction only, so:

3.8 KJ/mol

<u>Molecule D</u>

In this molecule, we have three Methyl-H <em>eclipse </em>interaction, so:

(6*3) = 18 KJ/mol

<u>Molecule E</u>

In this molecule, we have 1 Methyl-methyl <em>gauche </em>interaction only, so:

3.8 KJ/mol

<u>Molecule F</u>

In this molecule, we have a Methyl-methyl <em>eclipse </em>interaction a Methyl-H <em>eclipse </em>interaction and an H-H <em>eclipse</em> interaction, so:

(11)+(6)+(4) = 21 KJ/mol

The structures with higher energies would be less stable. In this case, structures B and F with an energy value of 21 KJ/mol (see figure 2).

I hope it helps!

3 0
3 years ago
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