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3 years ago

10

“Gasoline boils at a relatively low temperature (about 150°C). The kerosene is removed at around 200°C, followed by diesel oil a

t 300°C and fuel oil at around 370°C.” What topic is the teacher most likely talking about?

1 answer:

ANEK [815]3 years ago

3 0

Well its Organic Chemistry... and is the Fractional Distillation of Crude Oil

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In the laboratory, a volume of 100 mL of sulfuric acid (H2SO4) is recorded. How many g are there of the liquid if its density is

gladu [14]

Answer:

$\large \boxed{\text{183 g}}$

Explanation:

$\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\1.83 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{100 cm}^{3}}\\\\m & = & \text{183 g}\\\end{array}\\\text{There are $\large \boxed{\textbf{183 g}}$ of sulfuric acid.}$

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2 years ago

Question: Which is the last step in excavation of the skeletal remains?

Kay [80]

Answer: B:

Explanation:

This is the most reasonable answer

5 0

3 years ago

Read 2 more answers

6. If the broken wire of the top branch

vagabundo [1.1K]

Answer: When two light bulbs are connected in parallel, which is true? A. The total resistance is less than the resistance of either bulb alone. B. The Voltage provided

Explanation:

3 0

1 year ago

Sulfuryl dichloride is formed when sulfur dioxide reacts with chlorine.

zubka84 [21]

<u>Answer:</u> The value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

<u>Explanation:</u>

For the given chemical reaction:

$SO_2(g)+Cl_2(g)\rightarrow SO_2Cl_2(g)$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f_{(product)}]-\sum [n\times \Delta H^o_f_{(reactant)}]$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(1\times \Delta H^o_f_{(SO_2Cl_2(g))})]-[(1\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(Cl_2(g))})]$

We are given:

$\Delta H^o_f_{(SO_2Cl_2(g))}=-364kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-296.8kJ/mol\\\Delta H^o_f_{(Cl_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times (-364))]-[(1\times (-296.8))+(1\times 0)]=-67.2kJ/mol=-67200J/mol$

The equation used to calculate entropy change is of a reaction is:

$\Delta S^o_{rxn}=\sum [n\times \Delta S^o_f_{(product)}]-\sum [n\times \Delta S^o_f_{(reactant)}]$

The equation for the entropy change of the above reaction is:

$\Delta S^o_{rxn}=[(1\times \Delta S^o_{(SO_2Cl_2(g))})]-[(1\times \Delta S^o_{(SO_2(g))})+(1\times \Delta S^o_{(Cl_2(g))})]$

We are given:

$\Delta S^o_{(SO_2Cl_2(g))}=311.9J/Kmol\\\Delta S^o_{(SO_2(g))}=248.2J/Kmol\\\Delta S^o_{(Cl_2(g))}=223.0J/Kmol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(1\times 311.9)]-[(1\times 248.2)+(1\times 223.0)]=-159.3J/Kmol$

To calculate the standard Gibbs's free energy of the reaction, we use the equation:

$\Delta G^o_{rxn}=\Delta H^o_{rxn}-T\Delta S^o_{rxn}$

where,

$\Delta H^o_{rxn}$ = standard enthalpy change of the reaction =-67200 J/mol

$\Delta S^o_{rxn}$ = standard entropy change of the reaction =-159.3 J/Kmol

Temperature of the reaction = 600 K

Putting values in above equation, we get:

$\Delta G^o_{rxn}=-67200-(600\times (-159.3))\\\\\Delta G^o_{rxn}=28380J/mol=28.38kJ/mol$

Hence, the value of $\Delta G^o$ of the reaction is 28.38 kJ/mol

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3 years ago

Describe how to determine the rate law for an<br> overall reaction that involves multiple steps.

Zarrin [17]

Answer:

The sum of each elementary step in a reaction mechanism must yield the overall reaction equation. From the rate law of the rate-determining step it must agree with the experimentally determined rate law. The rate-determining step is the slowest step in a reaction mechanism. Because it is the slowest, it determines the rate of the overall reaction.

Explanation:

5 0

2 years ago