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3 years ago

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Acenapthalene has the empirical formula C6H5. A solution of 0.515 g of acenapthalene in 15.0 g CHCl3 boils at 62.5oC. The normal

boiling point of CHCl3 is 61.7oC; the Kb is 3.63 oC/molal. What is the molecular formula of acenapthalene? (Atomic weights: C = 12.01, H = 1.008, Cl = 35.45).

1 answer:

german3 years ago

4 0

Answer:

The molecular formula of an ascenapthalene is $C_{12}H_{10}$

Explanation:

$\Delta T_b=K_b\times m$

$\Delta T_b=K_b\times \frac{\text{Mass of acenapthalene}}{\text{Molar mass of acenapthalene}\times \text{Mass of chloroform in Kg}}$

where,

$\Delta T_f$ =Elevation in boiling point = $(62.5-61.7)^oC=0.8^oC$

Mass of acenapthalene = 0.515 g

Mass of $CHCl_3$ = 15.0 g = 0.015 kg (1 kg = 1000 g)

$K_b$ = boiling point constant = 3.63 °C/m

m = molality

Now put all the given values in this formula, we get

$0.8^0C=3.67 ^oC/m\times \frac{0.515}{\text{Molar mass of acenapthalene}\times 0.015kg}$

$\text{Molar mass of acenapthalene}=155.7875 g/mol$

Let the molecule formula of the Acenapthalene be $C_{6n]H_{5n}$

$6n\times 12 g/mol+5n\times 1 g/mol=155.7875 g/mol$

n = 2.0

The molecular formula of an ascenapthalene is $C_{12}H_{10}$

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what is the diffrence between displacement ans double displacement reactions write equations for these reactions

Lina20 [59]

Answer:

Single Displacement reaction

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Double displacement reaction

In a double displacement reaction, two atoms or a group of atoms switch places to form new compounds.

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8 0

3 years ago

17. Which of the following is NOT an empirical formula? *

Aloiza [94]

Answer: 17) d. $C_2H_6$

18. c. The empirical formula of a compound can be twice the molecular formula.

Explanation:

Molecular formula is the chemical formula which depicts the actual number of atoms of each element present in the compound.

Empirical formula is the simplest chemical formula which depicts the whole number of atoms of each element present in the compound.

To calculate the molecular formula, we need to find the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

The empirical mass can be calculated from empirical formula and molar mass must be known.

17. Thus the empirical formula of $C_2H_6$ should be $CH_3$

18. The molecular formula will either be same as empirical formula or is a whole number multiple of empirical formula. Thus the empirical formula of a compound can never be twice the molecular formula.

3 0

3 years ago

Read 2 more answers

I performed an experiment and mixed copper nitrate and potassium iodide. When they reacted, they formed a precipitate, even thou

Alexus [3.1K]

Answer:

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

Explanation:

The reaction performed in the experiment is;

2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2

The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.

The oxidation - reduction equation is as follows;

2Cu2^+ + 2e ----> 2Cu^+ reduction half equation

2I^- ----> I2 + 2e. Oxidation half equation

Balanced redox reaction equation;

2Cu2^+ + 2I^- ----> 2Cu^+ + I2

3 0

3 years ago

The combustion of 1.5011.501 g of fructose, C6H12O6(s)C6H12O6(s) , in a bomb calorimeter with a heat capacity of 5.205.20 kJ/°C

avanturin [10]

Answer : The internal energy change is -2805.8 kJ/mol

Explanation :

First we have to calculate the heat gained by the calorimeter.

$q=c\times (T_{final}-T_{initial})$

where,

q = heat gained = ?

c = specific heat = $5.20kJ/^oC$

$T_{final}$ = final temperature = $27.43^oC$

$T_{initial}$ = initial temperature = $22.93^oC$

Now put all the given values in the above formula, we get:

$q=5.20kJ/^oC\times (27.43-22.93)^oC$

$q=23.4kJ$

Now we have to calculate the enthalpy change during the reaction.

$\Delta H=-\frac{q}{n}$

where,

$\Delta H$ = enthalpy change = ?

q = heat gained = 23.4 kJ

n = number of moles fructose = $\frac{\text{Mass of fructose}}{\text{Molar mass of fructose}}=\frac{1.501g}{180g/mol}=0.00834mole$

$\Delta H=-\frac{23.4kJ}{0.00834mole}=-2805.8kJ/mole$

Therefore, the enthalpy change during the reaction is -2805.8 kJ/mole

Now we have to calculate the internal energy change for the combustion of 1.501 g of fructose.

Formula used :

$\Delta H=\Delta U+\Delta n_gRT$

or,

$\Delta U=\Delta H-\Delta n_gRT$

where,

$\Delta H$ = change in enthalpy = $-2805.8kJ/mol$

$\Delta U$ = change in internal energy = ?

$\Delta n_g$ = change in moles = 0 (from the reaction)

R = gas constant = 8.314 J/mol.K

T = temperature = $27.43^oC=273+27.43=300.43K$

Now put all the given values in the above formula, we get:

$\Delta U=\Delta H-\Delta n_gRT$

$\Delta U=(-2805.8kJ/mol)-[0mol\times 8.314J/mol.K\times 300.43K$

$\Delta U=-2805.8kJ/mol-0$

$\Delta U=-2805.8kJ/mol$

Therefore, the internal energy change is -2805.8 kJ/mol

5 0

3 years ago

The second-order rate constants for the reaction of oxygen atoms witharomatic hydrocarbons have been measured (R. Atkinson and J

Pepsi [2]

Answer:

A = 1,13x10¹⁰

Ea = 16,7 kJ/mol

Explanation:

Using Arrhenius law:

ln k = -Ea/R × 1/T + ln(A)

You can graph ln rate constant in x vs 1/T in y to obtain slope: -Ea/R and intercept is ln(A).

Using the values you will obtain:

y = -2006,9 x +23,147

As R = 8,314472x10⁻³ kJ/molK:

-Ea/8,314472x10⁻³ kJ/molK = -2006,9 K⁻¹

<em>Ea = 16,7 kJ/mol</em>

Pre-exponential factor is:

ln A = 23,147

A = e^23,147

<em>A = 1,13x10¹⁰</em>

<em></em>

I hope it helps!

6 0

3 years ago