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2 years ago

Name the method that used to separate the components of petroleum oil.

Chemistry

1 answer:

maw [93]2 years ago

6 0

Answer:

fractional distillation

Explanation:

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The rate constant of a reaction is measured at different temperatures. A plot of the natural log of the rate constant as a funct

Wittaler [7]

Answer:

The activation energy is 7.11 × 10⁴ J/mol.

Explanation:

Let's consider the Arrhenius equation.

$lnk=lnA-\frac{Ea}{R} .\frac{1}{T}$

where,

k is the rate constant

A is a collision factor

Ea is the activation energy

R is the ideal gas constant

T is the absolute temperature

The plot of ln k vs 1/T is a straight line with lnA as intercept and -Ea/R as slope. Then,

$\frac{-Ea}{R} =-8.55 \times 10^{3} K^{-1} \\Ea= 8.55 \times 10^{3} K^{-1} \times 8.314 \frac{J}{K.mol} =7.11 \times 10^{4} J/mol$

8 0

3 years ago

What does Le Châtelier's principle say about upsetting a system at

never [62]

Answer:

Explanation:

7 0

2 years ago

What happens to the electrons in an ionic bond?

Anton [14]

Answer:

In an ionic bonds, the metal loses electrons to become a positively charged cation, In which the nonmetal accepts those electrons to become a negatively charged anion.

Explanation:

3 0

3 years ago

Read 2 more answers

A gas sample is found to contain 39.10% carbon, 7.67% hydrogen, 26.11% oxygen, 16.82% phosphorus, and 10.30% fluorine. If the mo

Doss [256]

Answer:

C6H14O3F

Explanation:

The first step is to divide each compound by its molecular weight

Carbon

= 39.10/12

= 3.258

Hydrogen

= 7.67/1

= 7.67

Oxygen

= 26.11/16

= 1.63

Phosphorous

= 16.82/31

= 0.542

Flourine

= 10.30/19

= 0.542

The next step is to divide by the lowes value

3.258/0.542

= 6 mol of C

7.67/0.542

= 14 mol of H

1.63/0.542

= 3 mol of O

0.542/0.542

= 1 mol of P

0.542/0.542

= 1 mol of F

Hence the molecular formula is C6H14O3F

5 0

2 years ago

1 C8H10(l) +21/2O2(g) → 8CO2(g) + 5H2O(g), Hcomb= ? Hf for C8H10(l) = +49.0kJ/mol C8H10(l) Use the balanced combustion reaction

nikklg [1K]

Answer:

$H_{comb}=-4406kJ/mol$

Explanation:

Hello,

In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

$H_{comb}=8*\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{10}}$

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

$H_{comb}=8*(-393.5kJ/mol)+5(-241.8kJ/mol)-49.0kJ/mol\\\\H_{comb}=-4406kJ/mol$

Best regards.

4 0

3 years ago